text:
\section{Motivation: Kanter's random variable}
The study of stable random variables has known a considerable growth during approximately the last five decades, since they are shown to be an efficient model for various phenomena occuring in biology, quantum physics, market finance (\cite{Zolo1}). Unfortunately, their densities has no closed formulas except in few cases. Consider for instance, the stable random variable $X_{\alpha}$ of index $0 < \alpha < 1$ and of asymmetry parameter equal one, known also as the positive stable variable since it is supported by $(0,\infty)$ (\cite{Ibr}, p.50). Its Laplace transform is given by $e^{-t^{\alpha}}, t > 0$ and its density admits the following expansion ((2.41) p.54 in \cite{Ibr}, (14.31) p.88 in \cite{Sato}, see also \cite{Zolo2}):
\begin{equation*}
-\frac{1}{\pi x} \sum_{k=1}^{\infty} \frac{(-1)^k}{k!} \sin(k\pi \alpha) \Gamma(1+k\alpha) \frac{1}{x^{k\alpha}}, \quad x > 0.
\end{equation*}
The positivity of this series is not trivial and follows for instance from the integral representation displayed p.74 in \cite{Zolo1}. Amazing and not trivial as well is the complete monotonicity of the density of $(X_{\alpha})^{-r}, r \geq \alpha/(1-\alpha)$ (in particular its infinite divisibility, \cite{Sato}) which follows from the following representation (\cite{PSS}): there exists a standard exponential random variable $L$ and an infinitely divisible random variable $V_{\alpha,r}$, both variables being independent and
\begin{equation*}
(X_{\alpha})^{r} \overset{d}{=} \frac{e^{V_{\alpha,r}}}{L}.
\end{equation*}
Recall that the proof of this fact relies on a suitable integral representation of the Gamma function and that the random variable $V_{\alpha,r}$ is characterized by its Laplace transform. For $r = \alpha/(1-\alpha)$, the above representation is more precise and traces back implicitely to Chernin and Ibragimov's paper \cite{Chernin} (see also Ch.II in \cite{Ibr}). In fact (\cite{Kanter} p.703)
\begin{equation*}
(X_{\alpha})^{\alpha/(1-\alpha)} \overset{d}{=} \frac{a_{\alpha}(U)}{L},
\end{equation*}
where $U$ is a uniform random variable on $(0,\pi)$ and
\begin{equation*}
a_{\alpha}(u) = [b_{\alpha}(u)]^{1/(1-\alpha)}, b(u) = [c_{\alpha}(u)]^{\alpha}[c_{1-\alpha}(u)]^{(1-\alpha)}, c_{\alpha}(u):= \frac{\sin(\alpha u)}{\sin(u)}.
\end{equation*}
We shall refer to $a_{\alpha}(U)$ as the Kanter's random variable. From the injectivity of the Laplace transform, one gets
\begin{equation*}
V_{\alpha,1-\alpha} \overset{d}{=} \log[a_{\alpha}(U)].
\end{equation*}
Mysteriously, the function $a_{\alpha}$ is intimately related to the density of the free stable law of index $0 < \alpha < 1$ and asymmetry parameter $\rho=1$ derived p.1053 in \cite{Berco}, that we shall call shortly the positive free stable law. More precisely, if $U$ is defined on some probability space $(\Omega,\mathscr{F},\mathbb{P})$, then
\begin{equation*}
\mathbb{P}(a_{\alpha}(U) \leq u) = \frac{1}{\pi} a_{\alpha}^{-1}(u), \quad u \geq a_{\alpha}(0) =  (1-\alpha)\alpha^{\alpha/(1-\alpha)}
\end{equation*}
since $a_{\alpha}$ is strictly increasing (\cite{Kanter} p.704), while the density of the positive free stable distribution takes the form (\cite{Berco}):
\begin{equation}\label{Biane}
\frac{1}{\pi x} \frac{\sin(a_{1-\alpha}^{-1}(x))\sin(\alpha a_{1-\alpha}^{-1}(x))}{\sin((1-\alpha) a_{1-\alpha}^{-1}(x))},
\end{equation}
for $x \geq a_{1-\alpha}(0)$. That is why we thought it is reasonable to start with seeking an expression for the density of $a_{\alpha}(U)$ as a special function and more generally for that of $e^{V_{\alpha,r}}$. The easiest way to proceed is to invert the Mellin transform of the density of $e^{-V_{\alpha,r}}$ given by (\cite{PSS} p.292):
\begin{equation*}
\mathbb{E}(e^{-sV_{\alpha,r}}) = \frac{\Gamma(rs/\alpha +1)}{\Gamma(s+1)\Gamma(sr +1)}, \quad \Re(s) > -\alpha/r \geq -(1-\alpha).
\end{equation*}
However, the Mellin's inversion formula
\begin{equation}\label{Mellin}
\frac{1}{2i\pi y}\int_{-i\infty}^{+i\infty} \frac{\Gamma(rs/\alpha +1)}{\Gamma(s+1)\Gamma(sr +1)} y^{-s} ds,
\end{equation}
applies for any $r > \alpha/(1-\alpha)$ (we take $\Re(s) = 0$ as a path of integration) and does not when $r=\alpha/(1-\alpha)$. The latter fact is seen from the following estimate of $|\Gamma(a+ib)|, a,b \in \mathbb{R}, |b| \rightarrow \infty$ displayed in formula 1.24. p.4 from \cite{HMS}:
\begin{equation}\label{EST}
|\Gamma(a+ib)| \sim \sqrt{2\pi} |b|^{a-1/2}e^{-a-\pi |b|/2}, \quad |b| \rightarrow \infty
\end{equation}
so that the absolute value of the integrand in \eqref{Mellin} is equivalent to $|s|^{-1/2}$ (up to a constant). Note however that when $r > \alpha/(1-\alpha)$, the integral displayed in \eqref{Mellin} is nothing else but a so-called Fox H-function (\cite{HMS} p.3, see below):
\begin{equation*}
\frac{1}{2i\pi y}\int_{-i\infty}^{+i\infty} \frac{\Gamma(rs/\alpha +1)}{\Gamma(s+1)\Gamma(sr +1)} y^{-s} ds = \frac{1}{y} H_{2,1}^{1,0}\left[y|_{(1,r/\alpha)}^{(1,1),(1,r)}\right]
\end{equation*}
so that the density of $e^{V_{\alpha,r}}$ reads
\begin{equation*}
\frac{1}{y} H_{2,1}^{1,0}\left[\frac{1}{y}|_{(1,r/\alpha)}^{(1,1),(1,r)}\right], \quad y > 0, \, r > \frac{\alpha}{1-\alpha}.
\end{equation*}
The main result of this paper states then that in the pathological case $r=\alpha/(1-\alpha)$, the density of $a_{\alpha}(U)$ is a H-function too represented through a path integral resembling to \eqref{Mellin} but with a different contour (Hankel path). When $\alpha$ approaches zero, this density tends to zero pointwisely, a fact that agrees with Cressie's result implying that the Kanter's random variable becomes degenerate in the limit $\alpha \rightarrow 0^+$. When $1-\alpha = 1/j$ for some integer $j \geq 2$, the derived H-function reduces to the so-called Meijer G-function so that $a_{\alpha}(U)$ is closely related to a product of suitably chosen $j-1$ independent Beta random variables. The latter fact resembles to Williams respresentation of positive stable variables (\cite{Will}). Coming to free probability theory, the distribution function of $a_{\alpha}(U)$ is computed yielding an explicit expression for the density of a positive free stable law. Using the free analog of Zolotarev's duality (\cite{Berco}), a similar result holds for a free stable law with index $1 < \alpha < 2$ and asymmetry parameter $\rho = 0$. The paper is closed with supplying analytic arguments that explain the occurence of $a_{\alpha}$ in both probabilistic settings. It turns out that it is resumed in the inverse formula for the Fourier transform of the density of $X_{\alpha}$ together with deformations of paths of integration. Nevertheless, we think that a group theoretical argument exists but seems to be very hidden. For sake of completeness, some basic needed facts on the H-function are collected in the next section. A good reference to them is the monograph \cite{HMS}.

\section{On the Fox H-function}
The Fox H-function is defined as a Mellin-Barnes integral:
\begin{equation*}
H_{p,q}^{m,n}\left[z|_{(b_i,B_i)_{1 \leq i \leq q}}^{(a_i,A_i)_{1 \leq i \leq p}}\right] = \int_L \Theta(s)z^{-s} ds
\end{equation*}
where $1 \leq m \leq q, 0\leq n \leq p, a_i,b_i \in \mathbb{R}, A_i,B_i > 0$,
\begin{equation*}
\Theta(s) = \frac{\prod_{i=1}^m \Gamma(b_i + B_is)\prod_{i=1}^n\Gamma(1-a_i - A_is)}{\prod_{i=m+1}^q \Gamma(1- b_i - B_is)\prod_{i=n+1}^p\Gamma(a_i + A_is)},
\end{equation*}
the principal determination of the power function is taken (though it is not necessary here), an empty product being equal one and $L$ is a suitable contour separating the poles of both products of the meromorphic (Gamma) functions in the numerator of $\Theta$ (\cite{HMS} p.3,4). The choice of $L$ and the domain of convergence of the Mellin-Barnes integral defining the H-function for each contour depend on the parameters $a_i,b_i,A_i,B_i$. Eight cases are discussed in \cite{HMS} p.4. and we will only make use of five of them in the sequel that we shall refer to whenever needed:
\begin{itemize}
\item[i)] $q \geq 1$,
\begin{equation*}
\mu:= \sum_{i=1}^q B_i - \sum_{i=1}^p A_i > 0,
\end{equation*}
then the H-function exists in the punctured complex plane.
\item[ii)] $p \geq 1, \mu = 0$, the H-function exists in the domain $|z| > \beta$ where
\begin{equation*}
\beta:= \prod_{i=1}^p A_i^{-A_i} \prod_{i=1}B_i^{B_i}
\end{equation*}
and $L = L_{+\infty}$ described in \cite{HMS} p.3.
\item[iii)] Let $\Omega$ be the real number defined by
\begin{equation*}
\Omega:= \sum_{i=1}^n A_i + \sum_{i=1}^m B_i -  \sum_{i=n+1}^p A_i - \sum_{i=m+1}^q B_i.
\end{equation*}
Then the Fox H-function exists for all $z \neq 0$ such that $|\arg(z)| < \pi \Omega/2, \Omega > 0$ and $L$ is the infinite semi-circle $\gamma-i\infty, \gamma + i \infty$ for a suitable real $\gamma$.
\item[iv)] When $q \geq 1, \mu=0$, the H-function exists in the open disc $|z| < \beta$ and $L= L_{-\infty}$ is described in \cite{HMS} p.3.
\item[v)] Let
\begin{equation*}
\delta:= \sum_{i=1}^q b_i - \sum_{i=1}^pa_i + \frac{p-q}{2}.
\end{equation*}
If $\mu = 0,\delta < -1$, then the H-function is well-defined for complex numbers lying on the circle $|z| = \beta$.
\end{itemize}
The most crucial fact when looking at these definitions is that whenever more than one definition make sense, they lead to the same value of the integral. In fact, the integral is evaluated by means of Cauchy's residue Theorem applied to truncated contours together with asymptotic estimates at infinity, and the contour of integration can be deformed into another one provided that the deformation is admissible (\cite{Dieud}, \cite{Whitt}, see also CH.V. in \cite{Erd} for a similar discussion on Meijer's G-function which fits the Fox H-function when $A_i = B_j = 1, 1 \leq i \leq p, 1 \leq j \leq q$).

\begin{nota}
In the sequel, the complex variable $z$ will take only strictly positive values thereby $s \mapsto z^{-s}$ is defined through the real-valued logarithm function.
\end{nota}
\section{The density of $a_{\alpha}(U)$}
According to Kanter's representation, there exists a density $h_{\alpha}$  supported by $a_{\alpha}(0,\pi) = (a_{\alpha}(0),\infty)$  such that
\begin{equation*}
-\frac{1-\alpha}{\alpha\pi x} \sum_{k=1}^{\infty} \frac{(-1)^k}{k!} \sin(k\pi \alpha) \Gamma(1+k\alpha) x^{k(1-\alpha)}  = \int_0^{\infty}e^{-xy} y h_{\alpha}(y)dy.
\end{equation*}
The main result of this paper is stated as:
\begin{pro}
The density of the Kanter's variable is expressed as:
\begin{align*}
h_{\alpha}(y) = \frac{1}{y}H^{1,0}_{2,1}\left[\frac{1}{y}|_{(1,1/(1-\alpha))}^{(1,\alpha/(1-\alpha)),(1,1)}\right] = \frac{2(1-\alpha)}{y} H^{1,0}_{2,1}\left[\frac{1}{y^{2(1-\alpha)}}|_{(1,2)}^{(1,2\alpha),(1,2(1-\alpha))}\right]
\end{align*}
for $y \in [a_{\alpha}(0), \infty[$.
 \end{pro}
{\it Proof}: We shall start from the density of $(X_{\alpha})^{-\alpha}$ which has the expansion
\begin{equation*}
-\frac{1}{\alpha\pi x} \sum_{k=1}^{\infty} \frac{(-1)^k}{k!} \sin(k\pi \alpha) \Gamma(1+k\alpha) x^{k}.
\end{equation*}
Using the mirror formula satisfied by the Euler's Gamma function (\cite{Erd} p.3),
\begin{equation*}
\Gamma(1+ k\alpha)\Gamma(-k\alpha) = - \frac{\pi}{\sin(k\alpha \pi)},
\end{equation*}
the density of $X_{\alpha}^{-\alpha}$ is transformed to
\begin{equation*}
\frac{1}{\alpha x} \sum_{k=1}^{\infty} \frac{(-1)^k}{k!} \frac{x^k}{\Gamma(-k\alpha)}
\end{equation*}
and the infinite summation may start from $k=0$ since $z=0$ is a pole of the Gamma function. According to A.6 p.218 in \cite{HMS}, this is expressed through the H-function as follows
\footnote{It is a special kind of what V. M. Zolotarev called incomplete hypergeometric function in \cite{Zolo2}.}:
\begin{align*}
-\frac{1}{\alpha\pi x} \sum_{k=1}^{\infty} \frac{(-1)^k}{k!} \sin(k\pi \alpha) \Gamma(1+k\alpha) x^{k} & = \frac{2}{\alpha x}H_{1,1}^{1,0}\left[x^2|_{(0,2)}^{(0,2\alpha)}\right]
\end{align*}
for all $x > 0$ (note that the H-function in the RHS exists for all $z \neq 0$ in the complex plane by i) since  $q \geq 1, \mu = 2-2\alpha > 0$). Therefore,  the density of
$(X_{\alpha})^{-\alpha/(1-\alpha)}$ reads
\begin{equation*}
\frac{2(1-\alpha)}{\alpha x}H_{1,1}^{1,0}\left[x^{2(1-\alpha)}|_{(0,2)}^{(0,2\alpha)}\right].
\end{equation*}
According to iii), one has
\begin{equation*}
H_{1,1}^{1,0}\left[x^{2(1-\alpha)}|_{(0,2)}^{(0,2\alpha)}\right] = \frac{1}{2i\pi} \int_{\gamma -i\infty}^{\gamma+i\infty} \frac{\Gamma(2s)}{\Gamma(2\alpha s)}  \frac{ds}{x^{2(1-\alpha)s}}
\end{equation*}
where $\Re(s) = \gamma > 0$. With the help of the Gamma integral
\begin{align*}
\frac{2(1-\alpha)}{\alpha x^2}H_{1,1}^{1,0}\left[x^{2(1-\alpha)}|_{(1,2)}^{(1,2\alpha)}\right] &= \frac{2(1-\alpha)}{2\alpha i\pi} \int_{\gamma-i\infty}^{\gamma+i\infty} \frac{\Gamma(2s)}{\Gamma(2\alpha s)\Gamma(2+2(1-\alpha)s)}  \int_0^{\infty}e^{-xy} y^{2(1-\alpha)s+1} dyds
\\& = \frac{2(1-\alpha)}{2 i\pi} \int_{\gamma-i\infty}^{\gamma+i\infty} \frac{\Gamma(2s+1)}{\Gamma(2\alpha s+1)\Gamma(2+2(1-\alpha)s)}  \int_0^{\infty}e^{-xy} y^{2(1-\alpha)s+1} dyds.
\end{align*}
In order to change the order of integration, one needs to check that
\begin{equation*}
\int_{-\infty}^{+\infty} \frac{|\Gamma(2is +2\gamma+1)|}{|\Gamma(2i\alpha s+ 2\alpha \gamma + 1)\Gamma(2(1-\alpha)s+ 2(1-\alpha)\gamma + 2)|}  \int_0^{\infty}e^{-xy} y^{2(1-\alpha)\gamma} dyds < \infty.
 \end{equation*}
Performing an integration with respect to $y$, it amounts to check
\begin{equation*}
\int_{-\infty}^{+\infty} \frac{|\Gamma(2is +2\gamma+1)|}{|\Gamma(2i\alpha s+ 2\alpha \gamma + 1)\Gamma(2i(1-\alpha)s+ 2(1-\alpha)\gamma + 2)|} ds < \infty
\end{equation*}
which follows again from the estimate of $|\Gamma(a+ib)|, |b| \rightarrow \infty$ (see \eqref{EST}). Thus, Fubini's Theorem yields
\begin{align}
\int_0^{\infty}e^{-xy}yh_{\alpha}(y)dy &= 2(1-\alpha) \int_0^{\infty}xe^{-xy}  \int_{\gamma-i\infty}^{\gamma+i\infty} \frac{\Gamma(2s+1)}{\Gamma(2\alpha s+1)\Gamma(2+2(1-\alpha)s)}
y^{2(1-\alpha)s+1}ds dy \nonumber
\\& = 2(1-\alpha) \int_0^{\infty}xe^{-xy} yH^{1,0}_{2,1}\left[\frac{1}{y^{2(1-\alpha)}}|_{(1,2)}^{(1,2\alpha),(2,2(1-\alpha))}\right]dy. \label{E1}
\end{align}
Note that the H-function displayed in \eqref{E1} may be represented for $y > (1-\alpha)\alpha^{\alpha/(1-\alpha)}$ as a path integral through the loop $L_{-\infty}$ according to iv), it vanishes for $0 < y < (1-\alpha)\alpha^{\alpha/(1-\alpha)}$ since according to ii) the contour is a loop $L_{+\infty}$ that contains no pole of the meromorphic function $s \mapsto \Gamma(1+2s)$, and it takes a finite value at $z=(1-\alpha)\alpha^{\alpha/(1-\alpha)}$ regarding v) since $\mu=0, \delta= -3/2 < -1$ (in particular, it vanishes there by continuity). Hence, an integration by parts shows that the density of the Kanter's random variable is
\begin{align*}
h_{\alpha}(y) &= \frac{2(1-\alpha)}{y} \frac{d}{dy}\left\{yH^{1,0}_{2,1}\left[\frac{1}{y^{2(1-\alpha)}}|_{(1,2)}^{(1,2\alpha),(2,2(1-\alpha))}\right]\right\}
\\& = \frac{2(1-\alpha)}{y} \frac{d}{dy}\int_{L_{-\infty}} \frac{\Gamma(2s+1)}{\Gamma(2\alpha s+1)\Gamma(2+2(1-\alpha)s)} y^{2(1-\alpha)s+1}ds
\\& = \frac{2(1-\alpha)}{y} \int_{L_{-\infty}} \frac{\Gamma(2s+1)}{\Gamma(2\alpha s+1)\Gamma(2(1-\alpha)s+1)} y^{2(1-\alpha)s}ds
\\& = \frac{2(1-\alpha)}{y} H^{1,0}_{2,1}\left[\frac{1}{y^{2(1-\alpha)}}|_{(1,2)}^{(1,2\alpha),(1,2(1-\alpha))}\right]
\\& =  \frac{1}{y} \int_{L'_{-\infty}} \frac{\Gamma(s/(1-\alpha)+1)}{\Gamma(\alpha s/(1-\alpha)+1)\Gamma(s+1)} y^{s}ds =
 \frac{1}{y}H^{1,0}_{2,1}\left[\frac{1}{y}|_{(1,1/(1-\alpha))}^{(1,\alpha/(1-\alpha)),(1,1)}\right],
\end{align*}
where $L'_{-\infty}$ is the image of $L_{-\infty}$ under the map $s \mapsto 2(1-\alpha)s$. The density $h_{\alpha}$ is well defined for $y > (1-\alpha)\alpha^{\alpha/(1-\alpha)}$ by iv) and vanishes for $0 < y  < (1-\alpha)\alpha^{\alpha/(1-\alpha)}$ according to ii) (see also 1.33 p.6 in \cite{HMS}). $\hfill \blacksquare$

\section{Special indices}
\subsection{The $1/2$-stable case}
When $\alpha=1/2$, the density of the positive stable random variable reads (\cite{Zolo1} p.66)
\begin{equation*}
\frac{1}{2\sqrt{\pi}} x^{-3/2} e^{-1/(4x)} {\bf 1}_{\{x > 0\}}.
\end{equation*}
According to Kanter's representation, the image of the above density under the map $x \mapsto 1/x$, that is
\begin{equation*}
\frac{1}{2\sqrt{\pi x}}e^{-x/4} {\bf 1}_{\{x > 0\}} = \frac{1}{2\pi}\sqrt{\frac{\pi}{x}}{\bf 1}_{\{x > 0\}} e^{-x/4},
\end{equation*}
should be the Laplace transform of $y \mapsto yh_{1/2}(y)$ which takes the form (after the use of the mirror formula):
\begin{equation*}
yh_{1/2}(y) = \frac{1}{2\pi^2} \sum_{k=0}^{\infty}\sin^2[(k +1)\pi/2] \Gamma^2[(k+1)/2] \frac{(-1)^k}{k!} \frac{1}{y^{(k+1)/2}}, \, y > 1/4.
\end{equation*}
Note that $x \mapsto x^{-1/2}$ is completely monotone since $x \mapsto x^{1/2}$ is a Bernstein function, and that the representing measure is easily computed using the Gamma integral as
\begin{equation*}
\frac{dy}{\sqrt{\pi y}} {\bf 1}_{\{ y > 0\}}.
\end{equation*}
Now, since $\sin (n\pi) = 0$ for any integer $n$, then
\begin{align*}
yh_{1/2}(y) &= \frac{1}{2\pi^2} \sum_{k=0}^{\infty} \frac{\sin^2[(2k +1)\pi/2] \Gamma^2[(2k+1)/2]}{(2k!)}\frac{1}{y^{(2k+1)/2}}
\\& = \frac{1}{2\pi^2 y^{1/2}} \sum_{k=0}^{\infty} \frac{ \Gamma^2(k+1/2)}{\Gamma(2k+1)} \frac{1}{y^{k}}
\end{align*}
for $y > 1/4$. Using Legendre's duplication formula (\cite{Erd} p.5)
\begin{equation*}
\sqrt{\pi} \Gamma(2k+1) = 2^{2k} \Gamma(k+1/2) \Gamma(k+1),
\end{equation*}
one gets
\begin{align*}
yh_{1/2}(y)  &= \frac{1}{2\pi y^{1/2}}\sum_{k=0}^{\infty} \frac{(1/2)_k}{k!} \frac{1}{(4y)^{k}} = \frac{1}{2 \pi y^{1/2}} \frac{1}{\sqrt{1-1/(4y))}} \\
 & =  \frac{1}{2\pi} \frac{1}{\sqrt{y-1/4}}, \quad y > 1/4,
 \end{align*}
which is nothing but
\begin{equation*}
\frac{1}{2\pi}\frac{1}{\sqrt{y}} {\bf 1}_{\{ y > 0\}} \star \delta_{1/4}(dy)
\end{equation*}
where $\star$ is the classical convolution of positive measures.

\subsection{Relation to product of independent Beta variables}
Let
\begin{equation*}
1-\alpha = \frac{1}{j}, \quad \alpha = \frac{j}{j-1}
\end{equation*}
for some integer $j \geq 2$, then
\begin{equation*}
\mathbb{E}\left[\frac{1}{[a_{1-1/j}(U)]^s}\right] = \frac{\Gamma(js +1)}{\Gamma(s+1)\Gamma((j-1)s +1)} \quad \Re(s) >  -1/j.
\end{equation*}
The multiplication Theorem (generalization of Legendre's duplication formula, \cite{Erd} p.4)
\begin{equation*}
\Gamma(js+1) = \frac{1}{(\sqrt{2\pi})^{j-1}} j^{js+1/2}\prod_{k=1}^{j} \Gamma\left(s + \frac{k}{j}\right)
\end{equation*}
yields
\begin{equation*}
\mathbb{E}\left[\frac{1}{[a_{1-1/j}(U)]^s}\right] = \frac{1}{\sqrt{2\pi}} \sqrt{\frac{j}{j-1}} \frac{j^{js}}{(j-1)^{(j-1)s}}  \prod_{k=1}^{j-1} \frac{\Gamma\left(s + k/j\right)}{\Gamma(s+k/(j-1))}.
\end{equation*}

Now, let $\beta_1, \beta_2, \cdots, \beta_{j-1}$ be independent Beta random variables of the first kind such that the density of $\beta_k$ reads:
\begin{equation*}
\frac{1}{B(k/j+1, k/(j(j-1)))} t^{k/j}(1-t)^{k/(j-1) - k/j - 1}{\bf 1}_{(0,1)}(t)
\end{equation*}
where $B(\cdot,\cdot)$ stands for the Beta function. According to formula (4.11) p.122 in \cite{HMS},
\begin{align*}
\mathbb{E}[(\beta_1 \cdots \beta_{j-1})^{s-1}] &= \prod_{k=1}^{j-1}\frac{\Gamma(k/(j-1)+1)}{\Gamma(k/j+1)} \prod_{k=1}^{j-1}\frac{\Gamma(s+ k/j)}{\Gamma(s+ k/(j-1))}
\\& = \frac{1}{\sqrt{2\pi}} \sqrt{\frac{j-1}{j}} \frac{j^{j}}{(j-1)^{(j-1)}} \frac{\Gamma(j-1)}{\Gamma(j)} \prod_{k=1}^{j-1}\frac{\Gamma(s+ k/j)}{\Gamma(s+ k/(j-1))}
\\& = \frac{1}{\sqrt{2\pi}} \sqrt{\frac{1}{j(j-1)}} \frac{j^{j}}{(j-1)^{(j-1)}} \prod_{k=1}^{j-1}\frac{\Gamma(s+ k/j)}{\Gamma(s+ k/(j-1))}
\end{align*}
again by the multiplication Theorem. It follows that
\begin{equation*}
\mathbb{E}\left[\frac{1}{[a_{1-1/j}(U)]^s}\right] = j  \left[\frac{j^{j}}{(j-1)^{(j-1)}}\right]^{s-1}\mathbb{E}[(\beta_1 \cdots \beta_{j-1})^{s-1}].
\end{equation*}
When $s \neq 0$, the last equality may be written as
\begin{equation*}
\int_0^{\infty} u^{s-1} \mathbb{P}(a_{1-1/j}(U) < 1/u) du =  j  \mathbb{E}\left[\left(\frac{j^{j}}{(j-1)^{(j-1)}}V\beta_1 \cdots \beta_{j-1}\right)^{s-1}\right]
\end{equation*}
where $V$ is uniformly distributed in $(0,1)$ and is independent from the $\beta_k$s.

\begin{nota}
The relation to product of independent Beta variables is not surprising since the H-function involved in the density of $h_{1-1/j}$ reduces, when $1-\alpha = 1/j$ to a Meijer's G-function (\cite{HMS} p.123). It somewhat matches with Williams result (\cite{Will}) if one keeps in mind the Gamma-Beta algebra. In fact, $1/X_{\alpha}$ is distributed, when $\alpha = 1/j, j \geq 2$ as a product of $j-1$ Gamma random variables $\gamma_k, 1 \leq k \leq j-1$ with densities
\begin{equation*}
\frac{1}{\Gamma(k/j)}t^{k/j - 1} e^{-t} {\bf 1}_{\{t > 0\}}, \quad 1 \leq k \leq j-1.
\end{equation*}
Moreover, since
\begin{equation*}
[a_{\alpha}(\theta)]^{(1-\alpha)/\alpha} = a_{1-\alpha}(\theta),
\end{equation*}
then Williams result implies
\begin{equation*}
j^{j} \gamma_1\cdots \gamma_k \overset{d}{=}  \frac{1}{X_{1/j}} \overset{d}{=} \frac{L^{j-1}}{a_{1-1/j}(U)}.
\end{equation*}
Note also that, for $\alpha = 1/(jd), j \geq 2d, d \geq 1$, a more general representation of positive stable laws is due to T. Simon (\cite{Simon}) and involves independent Gamma and Beta variables.
\end{nota}

\section{Free positive stable distribution}
Free stable distributions may be defined in a similar way as in the classical setting, when substituting the classical convolution of probability measures by Voiculescu's free convolution (see \cite{Berco}). As mentioned in the introduction, the inverse function (in the composition's sense) of $a_{1-\alpha}$ provides an explicit expression of the density of a stable law of index $0< \alpha < 1$ and asymmetry parameter $\rho=1$ in the free probability setting. More precisely, if $V$ is a uniform random variable on $(0,1)$ defined on some probability space $(\Omega,\mathscr{F},\mathbb{P})$, then
\begin{equation*}
 \frac{1}{\pi}a_{1-\alpha}^{-1}(x) = \mathbb{P}(a_{1-\alpha}(\pi V) \leq x)

\end{equation*}
for $x > \alpha(1-\alpha)^{(1-\alpha)/\alpha}$. When $\alpha = 1/2$, it simplifies to
\begin{align*}
a_{1/2}^{-1}(x) &= \frac{1}{2}\int_{1/4}^x \frac{dy}{y\sqrt{y-1/4}} = \frac{1}{2}\int_0^{x-1/4} \frac{dy}{(y+1/4)\sqrt{y}}
\\& = 2\int_{0}^{x-1/4} \frac{2dy}{1+4y^2} = 2\arctan\left[2\sqrt{x-\frac{1}{4}}\right].
\end{align*}
Using the trigonometric identity
\begin{align*}
\sin(\theta) = 2\frac{\tan(\theta/2)}{1+\tan^2(\theta/2)}
\end{align*}
one recovers the density given by Biane (\cite{Berco})
\begin{align*}
\frac{2}{\pi x} \sin[a_{1/2}^{-1}(x)] &= \frac{2}{\pi x} \frac{\tan(a_{1/2}^{-1}(x)/2)}{1+\tan^2(a_{1/2}^{-1}(x)/2)} = \frac{2}{\pi x} \frac{\sqrt{4x-1}}{1 + 4(x-1/4)} =  \frac{\sqrt{4x-1}}{2\pi x^2}.
\end{align*}
For general $\alpha \in (0,1)$, we use the integral representation
\begin{equation*}
H^{1,0}_{2,1}\left[\frac{1}{y^{2\alpha}}|_{(1,2)}^{(1,2\alpha),(1,2(1-\alpha))}\right] = \frac{1}{2i\pi}\int_{L_{-\infty}}\frac{\Gamma(2s+1)}{\Gamma(2\alpha s+1)\Gamma(1+2(1-\alpha)s)} y^{2\alpha s} ds.
\end{equation*}
Now, assume for a while that Fubini's Theorem applies so that
\begin{align*}
 \frac{1}{\pi}a_{1-\alpha}^{-1}(x) &= \frac{2\alpha}{2i\pi}\int_{L_{-\infty}}\frac{\Gamma(2s+1)}{\Gamma(2\alpha s+1)\Gamma(1+2(1-\alpha)s)} \int_{\alpha(1-\alpha)^{(1-\alpha)/\alpha}}^{x}
 y^{2\alpha s-1} dy ds \\&
= \frac{2}{2i\pi}\int_{L_{-\infty}}\frac{\Gamma(2s)}{\Gamma(2\alpha s+1)\Gamma(1+2(1-\alpha)s)} \left[x^{2\alpha s} - (\alpha^{2\alpha}(1-\alpha)^{2(1-\alpha)})^s\right]ds.
 \end{align*}
Since the poles of $s \mapsto \Gamma(2s)$ are $s= -k/2, k \in \mathbb{N}$ and simple, and since the pole $s=0$ lies outside $L_{-\infty}$, then Cauchy's Residue Theorem yields
\begin{equation*}
\frac{1}{2i\pi}\int_{L_{-\infty}}\frac{\Gamma(2s)}{\Gamma(2\alpha s+1)\Gamma(1+2(1-\alpha)s)} x^{2\alpha s} ds = H_{2,1}^{1,0}\left[\frac{1}{x^{2\alpha}}|_{(0,2)}^{(1,2\alpha),(1,2(1-\alpha))}\right]  - \textrm{Res}(x,0)
\end{equation*}
where $\textrm{Res}(x,0)$ is the residue of the meromorphic function
\begin{equation*}
s \mapsto \frac{\Gamma(2s)}{\Gamma(2\alpha s+1)\Gamma(1+2(1-\alpha)s)} x^{2\alpha s}
\end{equation*}
at $s=0$ for fixed parameter $x >  \alpha(1-\alpha)^{(1-\alpha)/\alpha}$. This residue is easily computed as
\begin{equation*}
\lim_{s \rightarrow 0} \frac{s\Gamma(2s)}{\Gamma(2\alpha s+1)\Gamma(1+2(1-\alpha)s)} x^{2\alpha s} = \frac{1}{2}, \quad x > \alpha^{2\alpha}(1-\alpha)^{2(1-\alpha)}.
\end{equation*}
Since the last H-function is defined on the whole real line, in particular at $\alpha(1-\alpha)^{(1-\alpha)/\alpha}$ by the virtue of v) since $\mu=0$ and $\delta < -3/2 < -1$, then
\begin{align*}
\frac{1}{\pi}a_{1-\alpha}^{-1}(x) = 2H_{2,1}^{1,0}\left[\frac{1}{x^{2\alpha}}|_{(0,2)}^{(1,2\alpha),(1,2(1-\alpha))}\right] - 2H_{2,1}^{1,0}\left[\frac{1}{\alpha^{2\alpha}(1-\alpha)^{2(1-\alpha)}}|_{(0,2)}^{(1,2\alpha),(1,2(1-\alpha))}\right]
 \end{align*}
Regarding the continuity of the H-function at $\alpha(1-\alpha)^{(1-\alpha)/\alpha}$, it vanishes there (since it vanishes for $x < \alpha(1-\alpha)^{(1-\alpha)/\alpha}$) and one gets
\begin{align*}
\frac{1}{\pi}a_{1-\alpha}^{-1}(x) = 2H_{2,1}^{1,0}\left[\frac{1}{x^{2\alpha}}|_{(0,2)}^{(1,2\alpha),(1,2(1-\alpha))}\right], x > \alpha(1-\alpha)^{(1-\alpha)/\alpha}
\end{align*}
and zero otherwise. Finally, in order to fit in the density of the free positive stable law displayed in \eqref{Biane}, we can rewrite the last H-function as
\begin{equation*}
\frac{1}{\alpha} H_{2,1}^{1,0}\left[\frac{1}{x}|_{(0,1/\alpha)}^{(1,1),(1,(1-\alpha)/\alpha)}\right] = \frac{1}{1-\alpha} H_{2,1}^{1,0}\left[\frac{1}{x^{\alpha/(1-\alpha)}}|_{(0,1/(1-\alpha))}^{(1,\alpha/(1-\alpha),(1,1)}\right].
\end{equation*}
Note that the duality relation given in \cite{Berco} allows to derive the density of a free stable distribution of index $1 < \alpha < 2$ and asymmetry coefficient $\rho=0$ (the case of an asymmetry parameter $\rho = 1$ is obtained by the simple variable change $x \mapsto -x$).
Coming back to the validity of Fubini's Theorem, one has to prove, after integrating with respect to the variable $y$, the convergence of
\begin{equation*}
\int_{L_{-\infty}}\frac{|\Gamma(2s+1)|}{|\Gamma(2\alpha s+1)\Gamma(1+2(1-\alpha)s)|}  \left[x^{2\alpha \Re(s)} - (\alpha^{2\alpha}(1-\alpha)^{2(1-\alpha)})^{\Re(s)}\right] \frac{|ds|}{|\Re(s)|}.
\end{equation*}
This is easily cheked from the estimate of $\Gamma(a+ib)$ for large $a$ (formula 1.23 p.4. in \cite{HMS}):
\begin{align*}
|\Gamma(a+ib)|& \sim \sqrt{2\pi} |a|^{a-1/2}e^{-a- a(1-\textrm{sgn}(b))/2}, \quad |a| \rightarrow \infty
\end{align*}
which shows that the integrand is equivalent to
\begin{equation*}
 \frac{C}{|\Re(s)|^{3/2}}e^{-(\alpha\log(\alpha) + (1-\alpha)\log(1-\alpha))\Re(s)}, \, |\Re(s)| \rightarrow \infty
 \end{equation*}
for some constant $C = C(\Im(s))$ depending only on $\Im(s)$.

\subsection{Behaviour for small indices: Cressie's result and its free analog}
In this section, we investigate the limiting behaviour of $h_{\alpha}$ as $\alpha \rightarrow 0$. In the classical setting, the density tends to zero pointwisely therefore agrees somehow with Cressie's result we recall below. The latter implies that $a_{\alpha}(U)$ becomes degenerate when $\alpha$ approaches zero. In the free setting, the limiting density is an infinite positive measure whose image under the map $x \mapsto 1/x$ is the Haar measure on the compact interval $[0,1]$. Start with the series expansion of $h_{\alpha}$ is given by (A.6 p. 218 in \cite{HMS})
\begin{equation*}
h_{\alpha}(y) = (1-\alpha)\sum_{k=0}^{\infty} \frac{1}{\Gamma(1- \alpha - k\alpha )\Gamma(\alpha-k(1-\alpha))} \frac{(-1)^{k}}{k!} \frac{1}{y^{(k+1)(1-\alpha) +1}}
\end{equation*}
for $y > (1-\alpha)\alpha^{\alpha/(1-\alpha)}$. The mirror formula
\begin{equation*}
\Gamma(\alpha-k(1-\alpha))\Gamma((1-\alpha)(k+1)) = \frac{\pi}{\sin((1-\alpha)(k+1)\pi)}
\end{equation*}
transforms $h_{\alpha}$ to
\begin{equation*}
h_{\alpha}(y) = \frac{1-\alpha}{\pi} \sum_{k=0}^{\infty} \frac{\sin[(1-\alpha)(k+1)\pi] \Gamma[(1-\alpha)(k+1)]}{\Gamma(1- \alpha - k\alpha )} \frac{(-1)^k}{k!} \frac{1}{y^{(k+1)(1-\alpha) +1}}.
\end{equation*}
The defining term of the last series converges as $\alpha \rightarrow 0$ to
\begin{equation*}
\frac{(-1)^k \sin((k+1)\pi) }{y^{k+2}} = 0
\end{equation*}
for any integer $k \geq 0$ while the support $((1-\alpha)\alpha^{\alpha/(1-\alpha)}, \infty)$ of $h_{\alpha}$ decreases to $[1,\infty)$ as $\alpha \rightarrow 0$.  Since
\begin{equation*}
 \frac{\Gamma[(1-\alpha)(k+1)]}{\Gamma(1- \alpha - k\alpha )k!} \quad \rightarrow \quad 1
 \end{equation*}
as $\alpha \rightarrow 0$ and it defines a continuous function on the variable $\alpha$, then Lebesgue's convergence Theorem ensures the convergence of $h_{\alpha}(y)$, for fixed $y > 1$, to zero as the positive index $\alpha$ does. According to \cite{Cressie}, (see also \cite{Brockwell}), a stable variable with shape parameter $c=1$ and zero shift $\gamma = 0$ satisfies
\begin{equation*}
(X_{\alpha})^{\alpha} \overset{d}{\rightarrow} \frac{1}{L}, \quad \alpha \rightarrow 0
\end{equation*}
so that $a_{\alpha}(U)$ converges in distribution to the Dirac measure $\delta_1$. For a positive free random variable, we start from the image of its density function under the map $x \mapsto x^{\alpha}$:
\begin{equation*}
\frac{a_{1-\alpha}^{-1}(x^{1/\alpha})}{\pi x}\frac{\sin(\alpha a_{1-\alpha}^{-1}(x^{1/\alpha}))}{\alpha a_{1-\alpha}^{-1}(x^{1/\alpha})} \frac{\sin(a_{1-\alpha}^{-1}(x^{1/\alpha}))}{\sin((1-\alpha) a_{1-\alpha}^{-1}(x^{1/\alpha}))},
\end{equation*}
for $x > [a_{1-\alpha}(0)]^{\alpha} = \alpha^{\alpha}(1-\alpha)^{1-\alpha} \geq 1/2$. Then, we expand
\begin{align*}
\frac{a_{1-\alpha}^{-1}(x^{1/\alpha})}{\pi} &=  2 H_{2,1}^{1,0}\left[\frac{1}{x^{2\alpha}}|_{(0,2)}^{(1,2\alpha),(1,2(1-\alpha))}\right]
\\& = \sum_{k=0}^{\infty} \frac{1}{\Gamma(1- k\alpha )\Gamma(1- k(1-\alpha))} \frac{(-1)^{k}}{k!} \frac{1}{x^k}.
\end{align*}
With the help of the mirror formula,
\begin{equation*}
\Gamma(1-k(1-\alpha))\Gamma((1-\alpha)k) = \frac{\pi}{\sin(k(1-\alpha)\pi)}, \quad k \geq 1,
\end{equation*}
one easily derives
\begin{align*}
\lim_{\alpha \rightarrow 0} \frac{1}{\pi}a_{1-\alpha}^{-1}(x^{1/\alpha}) = 1 + \frac{1}{\pi}
\lim_{\alpha \rightarrow 0}\sum_{k=1}^{\infty} \frac{\sin(k(1-\alpha)\pi)\Gamma(k(1-\alpha))}{\Gamma(1- k\alpha)} \frac{(-1)^{k}}{k!} \frac{1}{x^k} = 1
 \end{align*}
for fixed $x > 1$. Since $\alpha^{\alpha}(1-\alpha)^{1-\alpha} \rightarrow 1$ as $\alpha \rightarrow 0$, then
\begin{equation*}
\lim_{\alpha \rightarrow 0} \frac{a_{1-\alpha}^{-1}(x^{1/\alpha})}{\pi x}\frac{\sin(\alpha a_{1-\alpha}^{-1}(x^{1/\alpha}))}{\alpha a_{1-\alpha}^{-1}(x^{1/\alpha})} \frac{\sin(a_{1-\alpha}^{-1}(x^{1/\alpha}))}{\sin((1-\alpha) a_{1-\alpha}^{-1}(x^{1/\alpha}))} = \frac{1}{x}
\end{equation*}
for any $x > 1$. Note that Scheff\'e's Lemma does not apply as in the classical setting, nevertheless the image of
\begin{equation*}
\frac{dx}{x}{\bf 1}_{\{x > 1\}}
\end{equation*}
under the map $x \mapsto 1/x$ is the Haar measure on $[0,1]$, namely:
\begin{equation*}
\frac{dx}{x}{\bf 1}_{\{0 < x < 1\}}.
\end{equation*}

\section{On the function $a_{\alpha}$}
As stated in the introductory part, the occurence of $a_{\alpha}$ in both probabilistic settings is resumed in the integral representation of the density of $X_{\alpha}$ together with deformations of integration paths. Start with the inversion formula for the Fourier transform to see that the density of $X_{\alpha}$ reads (\cite{Ibr} p.49):
\begin{equation*}
\frac{1}{\pi} \Re \int_0^{\infty}e^{-itx} e^{-t^{\alpha}e^{-i\alpha\pi/2}} dt = \frac{1}{\pi} \Re \int_0^{\infty}e^{itx} e^{-(it)^{\alpha}} dt.
\end{equation*}
The path of integration may be deformed into the positive half imaginary-line yielding (\cite{Zolo1})
\begin{align*}
\frac{1}{\pi} \Re \int_0^{\infty}e^{itx} e^{-(it)^{\alpha}} dt &= - \frac{1}{\pi}\Im  \int_0^{\infty}e^{-tx} e^{-e^{i\alpha\pi}t^{\alpha}} dt
\\& = \frac{1}{\pi}\Im  \int_0^{\infty}e^{-tx} e^{-e^{-i\alpha\pi}t^{\alpha}} dt.
\end{align*}
Performing the variable change $t=x^{1/(1-\alpha)}u$, the last integral transforms to
\begin{equation*}
\frac{x^{1/(1-\alpha)}}{\pi} \Im \int_0^{\infty}e^{-x^{\alpha/(1-\alpha)} (u + e^{-i\alpha \pi}u^{\alpha})} du = \frac{x^{1/(1-\alpha)}}{\pi} \Im \int_0^{\infty}e^{-x^{\alpha/(1-\alpha)} (u + \phi_{1-\alpha}(u))} du
\end{equation*}
where $\phi_{1-\alpha}(u): = e^{-i\pi \alpha} u^{\alpha}$. Following \cite{Chernin} (see also Ch.II \cite{Ibr}), the path of integration may be taken as the curve
\begin{align*}
\mathcal{C}_{\alpha} &:= \{u = re^{i\theta}, \theta \in [-\pi,0], \Im (u +\phi_{1-\alpha}(u)) = 0\}
\\& = \left\{u = re^{i\theta}, \theta \in [-\pi,0], r = \left[\frac{\sin(\alpha (\pi-\theta))}{\sin \theta}\right]^{1/(1-\alpha)}\right\}
\\& =  \left\{u = -re^{i\theta}, \theta \in [0,\pi], r = \left[\frac{\sin(\alpha \theta)}{\sin \theta}\right]^{1/(1-\alpha)}\right\}.
\end{align*}
Then for $u = -re^{i\theta} \in \mathcal{C}_{\alpha}$,
\begin{align*}
\Re(u+\phi_{1-\alpha}(u)) &= -r\cos \theta + r^{\alpha} \cos(\alpha \theta)
\\& = r^{\alpha}[-r^{1-\alpha}\cos \theta + \cos(\alpha\theta)]
\\& = \left[\frac{\sin(\alpha \theta)}{\sin \theta}\right]^{\alpha/(1-\alpha)} \frac{\sin((1-\alpha)\theta)}{\sin(\theta)}
\end{align*}
and $a_{\alpha}(\theta)$ shows up. Now, consider $F_{\alpha}:= 1/G_{\alpha}$ where $G_{\alpha}$ is the Stieltjes  transform of the positive free stable distribution (\cite{Berco}). Then, $F_{\alpha}$ is a one-to-one correspondence from the upper half-plane $\mathbb{C}$ onto its image. Let $F_{\alpha}^{-1}$ be the reciprocal function of $F_{\alpha}$, then (\cite{Berco},\cite{Berco1})
\begin{equation*}
F_{\alpha}^{-1}(z) = \phi_{\alpha}(z) + z
\end{equation*}
where $z$ belongs to the domain $\Im(z+ \phi_{\alpha}(z)) > 0$ and $\Im(z)$ is positive and sufficiently large. The function $\phi_{\alpha}$ is known as the Voiculescu transform of the positive free stable distribution. But since Stieltjes inversion formula involves the imaginary part of $G_{\alpha}$ when $z \in \mathbb{C}^+$ approaches the real line, one rather focuses on the domain
\begin{equation*}
\Omega_{\alpha} := \{z, \, \Im(1/z+ \phi_{\alpha}(1/z) > 0\}
\end{equation*}
for $z \in \mathbb{C}^-$ near the origin of the complex plane. This is a Jordan domain, that is delimited by a Jordan curve, whose boundary $\partial \Omega_{\alpha}$ is obviously
\begin{align*}
\partial \Omega_{\alpha} &= \left\{re^{i\theta}, \theta \in [-\pi,0],\, r = \left[\frac{\sin(\pi+ \theta)}{\sin ((1-\alpha)(\pi+\theta))}\right]^{1/\alpha}\right\}
\\& =  \left\{-re^{i\theta}, \theta \in [0,\pi],\, r = \left[\frac{\sin\theta}{\sin ((1-\alpha) \theta))}\right]^{1/\alpha}\right\}.
\end{align*}
This is the image of $\mathcal{C}_{1-\alpha}$ under the inversion map $z \mapsto 1/\overline{z}$. By Caratheodory's extension Theorem, $G_{\alpha}$ extends to a homeomorphism from the real line onto  $\partial \Omega_{\alpha}$ so that every $x$ lying in the support of the free positive stable variable is given by
\begin{align*}
x &= G^{-1}_{\alpha}(z) ,\quad z \in  \partial \Omega_{\alpha}
\\& = \Re(1/z + \phi_{\alpha}(1/z)).
\end{align*}
The same computations performed in the classical setting lead to the function $a_{1-\alpha}$. \\
{\bf Acknowledgments}: the author would thanks W. Jedidi for making him aware of the paper \cite{PSS} and T. Simon for stimulating discussions held at Lille 1 university about stable random variables. He also gives a special thank to Professor J.J. Loeb for his remarks on the Mellin-Barnes integral defining the Fox H-function and to members of LAREMA at Angers university for their hospitality during his staying on June 2010.

redpajama_set_name: RedPajamaArXiv


Gruesome, to say the least — even if the numbers of those burned alive in public auto-da-fes in the name of Christ are exaggerated. But one might add to this list the Protestant persecutions and execution of witches and heretics as well as the many wars fought by various factions of a Church which was founded on love.
And while we must realize that the history of the Christian Church is not the whole of the history of Western Civilization which began with the pre-Socratics, we must wonder why such a civilization that could have produced such horrors — not ignoring the Holocaust that took the lives of millions of Jews, the murder of millions of Russians during the Second World War by the Nazis, and the dropping of bombs on civilians in Germany and Japan by the Allies — should be preserved and allowed to survive. And we mustn't forget the slave trade and the genocide that accompanied the take-over of the North American continent.
Arnold Toynbee tells us that civilizations come and go; they have a lifetime and the expire just as do those who comprise the members of those civilizations. Ours is clearly enduring what Toynbee called a "time of troubles" and seems to be gasping for breath — especially as growing numbers of those historically devoted to preserving Western Civilization in the colleges and universities have turned against it. But why should we be concerned, given the many atrocities that have been committed in the name of civilization when we know that many people who we have always regarded as "uncivilized" seem now to be in many ways superior to those who regard themselves as "civilized" people?
Thus we do wander around bleating like sheep about nothing more important than ourselves. Our art shrinks to the level of the merely personal. We strive to eliminate all suffering because we fail to see that through suffering we might grow as human beings. And this is a sure sign of the "time of troubles" that Toynbee notes is associated with the demise of civilizations. We live shallow lives, and in the process seem to have lost our creative abilities as we worry not about others and about the future but about ourselves now. This suggests that the present civilization may well be replaced by another that will almost certainly have as many atrocities — given the nature of human beings — and fewer of the great works noted above. The best we can hope for, I strongly suspect, is something very closely resembling Huxley's Brave New World peopled by citizens who find their narrow pleasure-seeking lives worth the price of greatness that we must, like it or not, associate with Western Civilization at its height.
In the modern $1.4-trillion luxury economy, bling is out, and high-priced adventures are in. So says a new report from the consultants at BCG, which explores the lifestyle habits of the world's wealthy based on interviews with roughly 1,000 affluent individuals. Cartier jewelry, Hermes handbags, and Tourneau watches are being eclipsed by exclusive getaways to the Maldives and helicopter skiing vacations in Alaska (which, in case you're interested, you can book for about $5,750-a-person, all inclusive). Spending on those sorts of mind-bendingly expensive experiences now accounts for more than half of the luxury market and is growing more than 50 percent faster than sales of luxury goods.
The implications of this report are worth considering. For one thing, if true it gives the lie to ancient wisdom that insists that happiness cannot be bought at any price. Going back as far as the pre-Socratics in the West, the thought was that happiness was a function of inner-peace brought about by contemplation of eternal verities: possessions just divert attention away from what really matters. In the New Testament Christ tells his disciples to rid themselves of earthly possessions and follow him. Eastern wisdom also stresses inner peace as the means to true happiness.
In any event, since the sort of happiness discussed in the above-mentioned study is bought at a very high price, it is restricted to the very few. And it is based on simply asking a sample of the rich if they are happy having fun, which is suspect. The implication here is that the rest of us are doomed to be unhappy. I doubt that. I don't trust what the rich tell us. I doubt their honesty and their perceptiveness. In fact, I side with the ancient wisdom that finds possessions a diversion from what really matters — such things as family and friends and the peace of mind that comes from a life well lived. I take it that was the point of the story, recounted by Herodotus, of Solon's exchange with Croesus of Lydia. I cannot persuade myself that those who buy trinkets and take trips are doing anything but diverting their attention away from what really matters. Of course if you are rich enough you can keep doing that until the day you die and you will never know the sacrifice you have made — until on your death-bed you realize that you have been running away from yourself your whole life and worshipping a false god.
The study disagrees, as the article goes on to say that Psychology tells us that purchasing experiences can actually make us more content. In general, it's better to buy up lots of little experiences — going to a series of concerts, or taking regular classes — than pouring our money into one gigantic splurge. But by shifting their spending away from watches and perfume, the rich may have finally figured out a way to turn their money into peace of mind.
While I suspect any claim that starts with words like "Psychology tells us that…" I dare say that buying "lots of little experiences" might indeed lead to contentment, if not happiness. But, as Aristotle reminds us, it's a question of moderation. For those who can afford the little pleasures, no doubt contentment will follow. But I doubt that those who must go without even the small pleasures of concerts and classes cannot be happy. I have spoken with people who have visited the rag-tag poor in villages in remote parts of the world who insist that the people there are quite happy, indeed some of the happiest people they had ever met. I'll never know for certain, but I also doubt that those with immense wealth who can afford to spend their lives taking trips and buying new homes and fast automobiles are truly happy. In the end, it is the kind of person we are that determines whether or not we are happy — regardless of how many "things" we own or concerts we are able to attend.
Back before I retired from college teaching, I read some Russian literature with one of my favorite honor students as an independent study. At one point I suggested that she read the fifth chapter of the fifth Book of Dostoevsky's Brothers Karamazov, entitled "The Grand Inquisitor." She came back a week later having read the entire book — which is no mean feat as the novel is nearly 800 pages long! It is the last novel Dostoevsky wrote and is generally regarded as the author's major opus. Freud considered it to be the greatest novel ever written. But that chapter in the Fifth Book does stand alone and was even published at one time as a separate book entirely. It deals with human freedom and what Dostoevsky perceived as Christianity's rejection of Christ.
"The Grand Inquisitor" focuses on a dialogue between two of the brothers, Alyosha and Ivan, one of whom is saintly, the other of whom is a radical nihilist. The latter is attempting to shatter his brother's faith in Christ by parading before him a series of episodes that show humans at their very worst. But this particular chapter deals with a parable, "set in Spain, in Seville, in the most horrible time of the Inquisition, when fires blazed every day to the glory of God…" Christ returns, is discovered by the Grand Inquisitor and put into prison where he is questioned relentlessly about why he would dare to come back again after it has taken the Church 1500 years to undo the good that He did on his first coming. Needless to say, Ivan Karamazov, like Dostoevsky himself, was no friend of the Western Church in any of its forms. But unlike Ivan, Dostoevsky was a devout Christian (he saw no contradiction here), and his answer to Ivan comes in the form of the saintly Father Zosima in the novel — and in the form of Christ who sits silent while the Inquisitor presses Him for answers.
In the course of the monologue, the Inquisitor chastises Christ for bringing humans their freedom — which, he says, they don't want. What they want, really, is bread and miracles. Further, they delude themselves into thinking they are free when, in fact, they are not. He goes on to say, "Know, then, that now, precisely now, these people are more certain than ever before that they are completely free, and at the same time they themselves have brought us their freedom and obediently laid it at our feet." What the Church has done in the intervening 1500 years, he maintains, is to take the burden of human freedom upon themselves, with its awful sense of moral responsibility, and given in return what people really want — miracles, mystery, and authority.
In addition to the issue of human freedom, the Inquisitor discusses at length the three temptations of Christ, mentioned by St. Matthew, when Christ rejected all worldly goods and power for spiritual peace. The Inquisitor asks, rhetorically, whether it were possible to imagine a more disturbing confrontation than that between Christ and the devil, "at a time when the future was unknown, but now fifteen centuries later we can see that in these three questions everything was so precisely divined and foretold, and has proved so completely true, that to add to them or subtract anything from them is impossible." Christ said, "no"! whereas His Church has answered "yes"! As Dante noted long before Dostoevsky, a Christianity that seeks to accommodate the goals of the world is bankrupt and has, in effect, succumbed to the Antichrist.
It is a profound chapter in a brilliant novel. But it has been much misunderstood over the years by readers who refuse to take Dostoevsky at his word. His Inquisitor is the embodiment of the Western social reformer whose goal is to make things as easy for humans as possible and alleviate suffering which, Dostoevsky insists, is what makes us truly human. We are only too eager to make the trade the Inquisitor makes for us. We want bread and miracles, and we do not want real freedom with its burden of suffering and moral culpability.
It is interesting to ask in this time of holier-than-thou chest pounding — of the mixing of religion with politics, when hordes of people professing to be Christian condemn others because they are different, and would persecute those who embrace a different ideology — whether any of them could begin to understand the message that Dostoevsky is convinced is at the heart of the New Testament. Or would they deny that it is there at all?


We're delighted to be working with Memphis' James and the Ultrasounds for the first time!
On a given night in midtown Memphis one can find any number of bands. Amid the dim lights, heat and discordant blend of conversation emerging from venues like the Hi Tone, the Buccaneer Lounge and Bar DKDC there has been a particular group distinguishing itself over the course of the last year or so as one of the best up and coming rock acts in the city. Brash yet thoughtful, with sweet guitars and pummelling, yet danceable, Memphis-style rhythm. There isn't quite anything like them in Memphis or elsewhere. If you happen to be one of those who haven't yet heard James and the Ultrasounds, or seen them live… strap in.
The four-piece rock and roll band is the brainchild of James Godwin, a familiar face in the Memphis music scene, having been raised in the Bluff City, playing bass in a number of bands since his teens. His time in experimental rock outfit The New Mary Jane offered him an opportunity to develop what now seems like a telepathic understanding with current Ultrasounds drummer John Argroves, as well as to collaborate with influential Grifters duo David Shouse and Scott Taylor. In the ensuing years Godwin again teamed with Argroves to form a powerful rhythm section for Goner Records legend Jack Oblivian as well as Memphis-by-way-of-Knoxville roots rock singer/guitarist John Paul Keith, whose 2013 Roland Janes-produced release Memphis, Circa 3 A.M. has been lauded as an instant classic.
After touring with Jack O and JPK across the US and Europe, Godwin made time to do some songwriting of his own, and in 2012 recorded his solo debut EP, Lovers and Ghosts, playing all the instruments himself to a four-track recorder. Soon thereafter he put together a band to perform the EP material, bringing in Argroves on drums, Luke White on guitar and David Johnson on bass. The group began to play regionally and have strung together tours up the east coast to New York, returning to Memphis energised and ready to record new material that Godwin had been writing.
This past spring the band planned some recording sessions with another great Memphis multi-instrumentalist, Mark Edgar Stuart, producing. The results of those sessions comprise the band's MadJack Records debut full length album, Bad To Be Here.
More recently, the band released a brand new 7-inch on a new Memphis label, Misspent Records. The 7-inch includes a brand new James and the Ultrasounds track, the hook heavy space-punk jam Robot Love on the A side, with the B side featuring a crunching version of The Cramps' TV Set, from their Phillips Recording cut and Alex Chilton-produced Songs the Lord Taught Us. Look for new releases on both Madjack and Misspent in the coming months, including a special, short-run European press of two early James and The Ultrasounds cuts, Apocalyptic Love and I Get The Shakes.
With these offerings in store for the coming months, James and the Ultrasounds are well poised to be one of the big breakout bands of 2016 and beyond. One way or another, be sure to keep an eye on this band.
Support comes from Bobbie Peru. Bobbie Peru are an eclectic rock band formed by Robert Genovese in Manchester in January 2006. Bert is a singer/songwriter/guitarist who is originally from Waterbury, Connecticut, USA who was signed by Manchester label Storm. The band's music is heavily influenced by indie, garage and post-punk movements. Their sound is an unpredictable mix of stabbing/distorted guitars, social commentary and an abrasive rhythm section.


This study compared lower limb dynamic joint loading after hamstring (HS) and patellar tendon (PT) anterior cruciate ligament (ACL) reconstruction. A three-dimensional motion analysis and force plate system were used to determine sagittal plane kinematics and kinetics in 20 subjects who had undergone ACL reconstruction (10 HS, 10 PT) whilst they performed both horizontal and vertical hopping tasks. Results for both activities showed significantly smaller knee flexion angles and external flexion moments in the operated than the non-operated limbs in PT subjects but no difference between limbs in HS subjects. There was no difference in the summated moment (hip plus knee plus ankle) between limbs for either graft type. The peak vertical ground reaction force was greater in the operated than the non-operated limb in PT subjects during the horizontal hop. These data show functional knee kinematic and kinetic differences between the two graft types that point to an earlier or better functional recovery in patients undergoing ACL reconstruction with HS graft than PT graft.


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Found 1 collection related to Lawrence
Lawrence, Jonathan, 1737-1812
Jonathan Lawrence papers, 1765-1810
Jonathan Lawrence (1737-1812) of Newtown, Queens County, Long Island was a New York City merchant, politician and militia officer. The Jonathan Lawrence papers, 1765-1810, contain letters written by Jonathan Lawrence, documents, and two account... more
Jonathan Lawrence (1737-1812) of Newtown, Queens County, Long Island was a New York City merchant, politician and militia officer. The Jonathan Lawrence papers, 1765-1810, contain letters written by Jonathan Lawrence, documents, and two account notebooks pertaining to the mercantile and shipping affairs of Lawrence and members of his family, 1779-1810, with two peripheral business letters, 1765-1766. Letters written by Lawrence comprise three letters to his partner at Lawrence & Tom in Dover, New York regarding shipments sent via the Hudson River, 1783-1784; a letter to Henry Livingston, 1788; and a letter to Captain Peter Hodgkinson, 1793. Documents consist of loose invoices and accounts for Jonathan Lawrence, Lawrence & Tom, and John Ireland, including customs documents for imports by Lawrence from St. Croix and St. Eustatius, 1783-1796 and undated; an order to deliver the brig Seaflower, 1787; and land conveyances by family members for property in Manhattan, 1806 and 1810. A notebook, dated 1779 at Fishkill, contains accounts with the unidentified writer's father and mother, and Thomas Fanning. A notebook for the brig Olive Branch, 1789, contains accounts for Jonathan Lawrence Jr. and Samuel Tibbals, both identified as captains, and others. less

redpajama_set_name: RedPajamaCommonCrawl


The defendant father appealed orders of the trial court modifying custody of the parties' children to "return" sole physical custody to plaintiff mother after the children had spent more than a year in defendant's sole care. The Court later dismissed defendant father's motion to modify custody when he did not appear for a hearing, and later denied the father's renewed motion to modify custody and parenting time. He appealed to the Michigan Court of Appeals.
The COA vacated and remanded, requiring the trial court to hear the father's motion to change custody. This Court of Appeals Opinion concisely and clearly sets out the approach a trial court must take in a contest custody and/or parenting time case. It is an excellent summary of how these cases should proceed.
Witness Tampering: In 2011, Professor Amy Bonomi, Professor and Chair of Human Development and Family Studies at Michigan State University and David Martin, King County (Washington) published the field's first analysis of jail calls that occur between domestic abusers and their victims; outlining a five-stage model describing how abusers manipulate their victims into changing their stories to lessen impending criminal charges. Recently, a new commentary has been published by Bonomi and Martin expanding upon their earlier research. Professor Bonomi has graciously approved the sharing of this information on my blog pre-publication, and a link to the article will be live for a week or two. After that, it's a "pay-per-download."
"As professionals working in the field, a common question we are asked when leading an investigation and providing expert testimony is how abusers involve their children (directly or indirectly) during jail calls.
This is the most moving, most haunting, saddest and most powerful article I have ever read about one of the most horrific things that could happen to a family--forgetting that the baby is in the car, parking the car at work for 8 or 9 hours. And the baby dies.
I read it with tears streaming down my face. This is a Pulitzer Prize-winning quality piece from today's Washington Post Magazine. There are some very useful resources at the end. Please consider sharing this blog post with your clients and your colleagues. As Weingarten makes clear, forgetting that the a baby is still in the car, parking the car, and having the baby die so tragically has happened to so many parents--parents who will have to live with their grief for the rest of their lives.
Today, as I read a blog post by Leo Babuta ["The Unexpected Pleasures of a More Than a Decade of Reading Harry Potter"], I was reminded of the joys of reading creative works written for children of all ages. As a Fifth Grade teacher, the first half hour of every afternoon was devoted to reading to my students. I felt gratified when those children, after listening to and loving a Newberry Award Winner book, would go the the library, check it out and read it again.
When I had children of my own, we started with the Caldecott Award Winners for best picture books of the year, later progressing to the Newberry Award winners. I was more than gratified as my own children learned to love books, love reading, and to be life-long readers of quality literature. Memories flood in here--Richard Scarry books, Robert Kraus books, (especially Pipsqueak Mouse in Shining Armour--now out-of-print). We had treasured moments with books shared, and John and David became fantastic, skilled and discriminating readers. Below, I'll share with you Leo Babuta's wonderful blog post: The Unexpected Pleasures of More Than a Decade of Reading Harry Potter. For parents . . . this is a must read!
In 1999, Colorado mother Jessica Gonzales experienced every parent's worst nightmare when her three young daughters were killed after being abducted by their father in violation of a domestic violence restraining order. Devastated, Jessica sued her local police department for failing to adequately enforce her restraining order despite her repeated calls for help that night. Determined to make sure her daughters did not die in vain, Jessica pursued her case to the US Supreme Court and an international human rights tribunal, seeking to strengthen legal rights for domestic violence victims. Meanwhile, her relationship with her one surviving child, her son Jessie, suffered, as he struggled with the tragedy in his own way.
Home Truth, the documentary film about Jessica Lenahan (formerly Gonzales) will premier on Sunday, June 11 and Monday, June 12 at the Human Rights Watch Film Festival in NYC! For tickets, please visit https://ff.hrw.org/film/home-truth?city=New+York.
Here is the official film description: Shot over the course of nine years, Home Truth chronicles one family's incredible pursuit of justice, shedding light on how our society responds to domestic violence and how the trauma from domestic violence can linger through generations.


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DJ Smith on Stuetzle, Chabot's Ice Time, Training Camp & More
Head Coach DJ Smith joined the TSN 1200 crew this morning to talk about the upcoming, unique NHL season.
By spencerblake Dec 22, 2020, 11:04am EST
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Photo by Joe Sargent/NHLI via Getty Images
This morning, Ottawa Senators Head Coach DJ Smith joined the TSN 1200 crew for his first radio appearance of the season. He talked about training camp, Tim Stützle, the Canadian Division and chewing gum sponsorships.
Let's dive into the good stuff.
On Getting Started
Heading into training camp, Smith says they have a lot of questions. Some of the answers will come from the players, some from the league.
Players have been arriving in Ottawa for the past month, some committing to their 14-day quarantines since arriving from overseas. Players are currently working out in different groups, on and off the ice, and adhering to health and safety guidelines as they go.
"It has that early September feel around the arena, minus the more personal interactions," Smith said, adding that the coaching staff has been hard at work at the Canadian Tire Centre since September.
While the team has an idea of what numbers, routines and schedules will look like, Smith and the rest of the league are awaiting the finer details from the league on what is officially expected as it relates to precautions and safety guidelines.
On the Excitement of a New Season
You can hear the excitement in Smith's voice as he's answering questions about the season ahead.
Smith specifically talked about two groups of people he's excited to see in camp. First, the new players coming in who are looking to get their careers back on track. I would think he's referring to both Matt Murray and Alex Galchenyuk here, as neither individual performed to the level they can last season. Second, he talked about the players coming up and knocking on the door.
"We've got some young exciting prospects of ours that have certainly proven they can do it in the American Hockey League and now they have to show they can transition it to the NHL. That's what it's all about, that's why I was hired to come in and develop these kids and give us a franchise that can be competitive year in, year out."
This is what you want to hear from the Head Coach of your favourite team as they continue through a rebuild. Smith was tagged as the player's coach, the guy who can connect with younger players and get the most out of their performance.
On Tim Stützle
The topic of Tim Stützle was unsurprisingly brought up multiple times in this appearance. First, they talked about where Smith sees Stützle slotting in off the hop.
"[playing Stützle on the wing to start] is for sure [the plan]. Part of development is seeing what the league is like and what you can handle, giving a young guy more to handle every time," Smith said, adding they'll likely see him at centre as the season progresses as well.
Later in the interview, Stützle was brought up again, chiming in on expectations and what Smith wants from his first year in the NHL.
"We want him to be important by the end of the year. You're not going to give anyone anything but there's a reason you're drafted third overall. You have a skillset that makes you a franchise changer and my job is to get the best out of him. By the end of the year, we want this kid to be in pressure situations so that, next year, this is someone who can really get up and get going."
Needless to say, Smith is going to do what he can to ensure Stützle is able to be an impact player for the Sens by the end of this season, hopefully sooner.
On the New Guys
General Manager Pierre Dorion was busy this offseason bringing in fresh faces to round out his 30th place Sens, looking for improvements primarily in experience and physicality. And, in Erik Gudbranson, Austin Watson and Josh Brown, he certainly found what he was looking for.
The big acquisition this offseason, however, came in the form of Evgeni Dadonov.
"[Dadonov] might be the best bumper middle man in the National Hockey League," Smith mentioned. "He's a guy that can make your powerplay go. That was a sore spot for us a year ago. Bringing in Galchenyuk and Dadonov, plus adding [young offensive players], our powerplay is much improved."
One player Smith seems particularly intrigued by is former third overall pick Alex Galchenyuk.
"He will probably start on the wing but with a shortened season, there will be injuries and he'll get some games at centre," Smith mentioned when being asked where he sees Galchenyuk in this lineup. "He had a bad knee injury which hampered him the past year, believes he's back at full throttle. People he was skating with were really impressed with his work ethic."
On the Centre Position
When pressed about the centre position, Smith's answer was music to my ears.
"We don't have veterans down the middle like other teams in our division. We'll have to learn on the fly. Norris has a year under his belt, Brown has two in the American league. Tierney is the veteran. A guy I look to bounce back is Colin White after a bit of an off year."
For me, this is the answer I was looking for. This sounds like a coach who is willing to give his young potential stars like Norris and Brown a real shot at lining up at their natural position this year. While it may be reading too deeply into things, Smith also didn't mention Artem Anisimov once when speaking about the centre position. Hopefully both Norris and Brown will come into camp and earn their spots down the middle.
Specifically, Smith had great things to say about Logan Brown.
"Logan has an outstanding ability to make plays [...] he's 6'6" with hands who can make plays. A guy like that, at some point, it will click in and he'll hit the ground running," Smith said while adding "Brown has as much or more skill than anyone we have. We're going to need help on the powerplay, he's a guy that can provide that for us."
While every spot needs to be earned, it sounds like Smith is hopeful that Brown will secure a spot with the Senators this year. He mentioned multiple times in this appearance how he is looking for an improved powerplay attack and if Brown can make this team, he's a sure fire option to dish the puck on either powerplay unit.
On Goaltending
For Smith, and probably everyone else, newly acquired Matt Murray is the clear number one goalie. With Murray coming in, being a veteran goaltender in his prime, Smith feels really confident about the way the Senators are looking between the pipes this year.
"The surprise of the year last year was [Marcus Hogberg] who came in and gave us quality starts."
While adding the goaltending position can change year to year, Smith feels he has two quality goaltenders who will push each other to perform well while, just as importantly, having a backup goaltender in Hogberg that you can trust to start as much as you need him to.
On the North Division
Smith laments, a bit, about being in what he deems to be the toughest division in the league with the best scorers in the league. When you look at having to face players like Connor McDavid, Leon Draisatl, Auston Matthews and Elias Pettersson more regularly, can you really blame him?
When asked about the baseball format that will see teams play each other two or three times in a row, Smith talks about how this season is going to feel more like a 56 game playoff than anything else.
"What we have to do is play playoff hockey every night."
Smith adds that Dorion's roster acquisitions like Gudbranson, Watson and J. Brown will take the load off players like Stützle, Norris and Brännstrom while firing up the team's "ultimate competitor" in Brady Tkachuk.
When asked about this team's ability to compete for a playoff spot, Smith says what we're all thinking.
"If our young guys can perform at a level we think they can, I think we'll be as competitive as anyone. Can we do it for 56 games in a row? I'm not going to put pressure on those guys and say that."
On Thomas Chabot
Last year's league leader in icetime is heading into another season in which we can expect him to contend for that particular record. On his potential partners, Smith admits that because Chabot plays so much, he's likely to see many partners throughout the season.
"But we'll start him with Gudbranson," he says.
When pressed about the reason for Chabot playing so much last season, Smith recalls the number of injuries they faced, particularly missing veterans like Ron Hainsey and Mark Borowiecki for extended periods of time.
"If you ask [Chabot], he loved it," Smith joked.
You can tell Chabot thrives when he plays a lot but it will be up to the coaching staff to get the most out of the rest of the blueline to ensure they're not putting the top defender at risk by overdoing it.
On Expanded Rosters
With the ability to carry a taxi squad, Smith is asked about what that does for his strategy as a coach and how it builds internal competition.
"You don't want somebody to take a night off and think he gets to keep his job. Whoever plays the hardest and performs the best gets to play every night."
Smith appeared particularly excited about the ability to have more players ready in the wings to ensure his team never takes a night off. If players like Norris, Alex Formenton and Rudolfs Balcers don't make the team out of camp, there's a good chance they'll be added to the taxi squad, especially with the AHL's season in question. Usually, your extra skaters at the NHL level are waiting to sub in for depth roles. With this year's format, Smith has the ability to flip just about any player out for a member of the taxi squad, because it's bound to have a good mix of top and bottom end players, ready to take over for anyone who's underperforming.
If there's no AHL, however, Smith isn't entirely sure what will happen.
"I don't really have an answer for that," Smith admits. "But there will probably be some kind of skill development opportunities. At some point, with all the injuries, prospects are going to get an opportunity to play up here, be on the taxi squad or at least practice with us," Smith added. "Sean Donovan and Jesse Winchester will be working with them, they've been doing it all summer and doing an excellent job."
It sounds like there's some kind of loose plan and there will be an opportunity for those players who don't make the taxi squad to continue playing in some capacity. You do wonder, though, if we'll start seeing players get re-loaned overseas, should the AHL not be able to have a real season.
On Brady Tkachuk
One of the saddest parts for Smith this season appears to be not having fans in the building to watch Tkachuk do what he does best.
"It's a shame we can't have fans because he's a guy who gets you up out of your seat. He's exciting, he gives you energy and he's someone you want to root for."
Smith adds that Tkachuk is getting better every year. He's learning the game more away from the puck, gaining confidence around the net and really understanding what his role on the team is - both on the ice but also as a leader.
On Chewing Gum Sponsorships
We all know Smith is a big gum chewer behind the bench. When asked if he was able to secure a sponsor for the season, he joked that he isn't much of a gum guy in the summer.
The TSN 1200 crew will be working on getting him a sponsor, Wrigley's or something.
"Alright, I appreciate it. I'll let my agent know," Smith joked, as his radio appearance came to a close.


Defranceso+silva's Breathing house is characterized by a floating glass façade of sliding terraces.
Conceived by the Italian architecture lab defrancesco+dilva as a dynamic living system, The breathing house is a three-storey family home whose glass façade is segmented into a series of sliding square terraces. The concept is described by the studio as being similar to a 'treasure flower', a south African plant that blooms in the morning and closes its petals at night.
The system works by means of a large metal grid that serves as an external skeleton for the structure. When closed, the residence resembles a typical 'house' shape. sliding out a terrace, however, extends the internal room into a dynamic balcony area. the façade is slowly broken apart into a pixelated arrangement of moving terraces that provide stunning 360 degree views of the surrounding landscape. The project was recently awarded the director's choice award in the 'Last house on Mulholland' competition, which asks entrants to imagine the innovative use of technology and environmental strategies on properties located beneath the iconic Hollywood sign in Los Angeles.
defranceso+silva has designed the home to float inside its reticular earthquake resistant structure, a feature which allows the frame to conduct cabling and collect rainwater. This water is then guided to a tank at the bottom of the house and filtered to a pure, drinkable standard. Intended as an eco sustainable residence, the windows are made of solar glass cells and provide all the energy needs of the house.
The 'breathing house' is composed of three main floor areas: the studio/working area is located on the ground floor, the living, dining and cinema area on the first floor and the bedrooms and bathrooms are located on the third floor.
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Avaya - Get News & Ratings Daily
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Avaya Holdings Corp Expected to Earn Q1 2020 Earnings of $0.72 Per Share (NYSE:AVYA)
Posted by Wayne Rhoads on Nov 28th, 2019
Avaya Holdings Corp (NYSE:AVYA) – Equities researchers at BWS Financial issued their Q1 2020 EPS estimates for Avaya in a research note issued to investors on Monday, November 25th. BWS Financial analyst H. Khorsand expects that the company will earn $0.72 per share for the quarter. BWS Financial has a "Buy" rating on the stock. BWS Financial also issued estimates for Avaya's Q2 2020 earnings at $0.82 EPS, Q3 2020 earnings at $0.91 EPS, Q4 2020 earnings at $0.95 EPS and FY2020 earnings at $3.40 EPS.
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AVYA has been the subject of a number of other reports. Northland Securities reaffirmed a "buy" rating and issued a $17.00 target price on shares of Avaya in a report on Thursday, November 21st. Citigroup upped their target price on shares of Avaya from $12.00 to $15.00 and gave the company a "neutral" rating in a report on Wednesday, October 9th. Morgan Stanley upped their target price on shares of Avaya from $14.00 to $16.00 and gave the company an "equal weight" rating in a report on Monday, October 7th. Zacks Investment Research raised shares of Avaya from a "hold" rating to a "strong-buy" rating and set a $14.00 target price for the company in a report on Monday. Finally, Guggenheim reaffirmed a "buy" rating and issued a $20.00 target price on shares of Avaya in a report on Tuesday, November 19th. Three research analysts have rated the stock with a hold rating, four have assigned a buy rating and one has issued a strong buy rating to the company. The stock has a consensus rating of "Buy" and an average price target of $18.13.
Shares of AVYA opened at $12.76 on Wednesday. The firm has a fifty day moving average price of $12.35 and a 200 day moving average price of $12.22. Avaya has a 1-year low of $9.72 and a 1-year high of $22.35. The company has a debt-to-equity ratio of 2.10, a current ratio of 1.41 and a quick ratio of 1.35. The stock has a market cap of $1.33 billion, a PE ratio of 3.26 and a beta of 1.54.
Avaya (NYSE:AVYA) last announced its quarterly earnings results on Wednesday, November 20th. The company reported $0.54 earnings per share for the quarter, missing analysts' consensus estimates of $0.78 by ($0.24). Avaya had a negative net margin of 23.24% and a positive return on equity of 13.33%. The company had revenue of $726.00 million during the quarter, compared to analysts' expectations of $745.50 million.
Several large investors have recently made changes to their positions in AVYA. FMR LLC purchased a new stake in shares of Avaya in the first quarter valued at about $20,120,000. First Trust Advisors LP raised its holdings in Avaya by 20.4% during the first quarter. First Trust Advisors LP now owns 146,131 shares of the company's stock worth $2,459,000 after purchasing an additional 24,764 shares in the last quarter. Xact Kapitalforvaltning AB purchased a new position in Avaya during the second quarter worth about $138,000. FNY Investment Advisers LLC purchased a new position in Avaya during the second quarter worth about $222,000. Finally, Strs Ohio raised its holdings in Avaya by 16.3% during the second quarter. Strs Ohio now owns 75,000 shares of the company's stock worth $893,000 after purchasing an additional 10,500 shares in the last quarter.
Avaya Company Profile
Avaya Holdings Corp., through its subsidiaries, provides digital communications products, solutions, and services for businesses worldwide. The company operates through two segments, Products & Solutions, and Services. The Products & Solutions segment offers unified communications and contact center platforms, applications, and devices.
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Jacobs Engineering Group Inc (NYSE:JEC) Forecasted to Post Q1 2020 Earnings of $1.29 Per Share
Grafenia PLC (LON:GRA) Insider Peter R. Gunning Purchases 100,000 Shares


How do counselors/caseworkers sign up for Benefit Kitchen?
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What are Account Numbers and PINs?
How does an agency administrator 'assume' counselors/caseworkers into an agency?
How does a counselor add a client?
How long does it take to do a screening?
Where can I find my Account Number and PIN?
Does Benefit Kitchen work on Android and Apple/iOS devices?
How does a counselor use Benefit Kitchen with clients?
Does my information stay private?


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Konami Sound Cartridge — специальный картридж расширения для бытовых компьютеров стандарта MSX, поставлявшийся в комплекте с играми Snatcher и Super Deform Snatcher (сами игры поставлялись на трёх дискетах формата 3,5 дюйма) фирмы Konami.

Описание
Картридж содержит немного улучшенный вариант микросхемы звукогенератора Konami SCC, имеющий официальное обозначение SCC-I, но обычно неофициально называемый SCC+, и 64 КБ оперативной памяти. Вероятно, дополнительная память была включена в состав картриджа по причине наличия у большинства MSX-совместимых компьютеров в Японии (на момент выхода упомянутых игр) небольшого объёма основной оперативной памяти (обычно 64 КБ).

Микросхема SCC+, использующаяся в картридже, отличается от обычной микросхемы SCC наличием памяти для хранения формы сигнала для всех пяти каналов (по 32 байта для каждого). Обычная микросхема SCC имела память только для четырёх каналов, каналы 4 и 5 использовали одну и ту же память.

Картридж, поставлявшийся в комплекте с игрой Snatcher отличается от поставлявшегося с игрой Super Deform Snatcher. Отличия заключаются только в различном распределении памяти. Возможной причиной этих различий является желание разработчика сделать эти картриджи несовместимыми между собой. Некоторые эмуляторы MSX эмулируют этот вариант картриджа (например, эмулятор MAME).

Большинство игр, входящих в сборники Konami Games Collections, могут использовать картридж, поставлявшийся с игрой Snatcher, для воспроизведения более качественного звука. За исключением этих игр, картридж не поддерживается другими программами.

Микросхема SCC+ выполнена в корпусе DIP48. На корпусе каждой микросхемы имеется маркировка KONAMI 052539 SCC-I 2312P001 JAPAN (отличается от маркировки обычных микросхем SCC), после которой следует код даты и места производства, например 8951EAI. Первые две цифры означают год, следующие две — номер недели в году. Буквы являются кодовым обозначением завода-производителя и партии.

Технические характеристики
 Звукогенератор:
 5 каналов, генерирующие сигнал определяемой пользователем формы:
 Каждый канал имеет 32 байта ОЗУ, в котором хранится описание формы генерируемого сигнала
 16 уровней громкости
 Диапазон генерируемых частот от 27 Гц до 111.25 КГц
 ОЗУ: 64 КБ
 Управление памятью:
 4 окна по 16 КБ в адресном пространстве процессора

Звук генерируется полностью цифровой схемой, и выдаётся в виде параллельного 6-разрядного кода, преобразуемого внешним 11-разрядным ЦАП (простейшая цепная R-2R схема) в аналоговый сигнал.

Технические подробности

Увеличение объёма памяти
Существует возможность увеличить объём оперативной памяти картриджа до 128 КБ. На печатной плате предусмотрено два неиспользованных места для установки микросхем памяти типа 4464 (каждая имеет объём 4x64K). Когда дополнительные микросхемы памяти установлены, оба типа картриджей становятся полностью идентичными, и могут работать как с Snatcher, так и с Super Deform Snatcher.

Несколько неожиданным побочным эффектом расширения памяти становится то, что картридж начинает работать в точности как обычный картридж на основе ПЗУ объёмом 128 КБ и микросхемы SCC. При этом становится возможной загрузка в ОЗУ картриджа любых игр, распространявшихся на картриджах подобного типа (например, Gradius 2, F1 Spirit,
Quarth, и другие) в ОЗУ картриджа, и они будут работать как оригинальные картриджи. Разумеется, при выключении питания информация, загруженная в ОЗУ, теряется, и требуется повторная загрузка. Возможно, что микросхема SCC+ изначально была разработана для внутренних нужд при разработке и тестировании игр в Konami.

Также возможно сделать картридж совместимым с обеими играми без добавления микросхем ОЗУ. Для этого требуется соединить вывод 16 одной из установленных микросхем ОЗУ с контактной площадкой для того же вывода одной из отсутствующих микросхем. При этом установленное ОЗУ станет отображаться в оба банка по 64 КБ.

Использование памяти
Программирование SCC+ осуществляется через порты ввода-вывода, отображённые в адресное пространство памяти, и в целом очень близко к программированию обычных картриджей с ПЗУ и SCC.

Как и на обычных картриджах с использованием микросхемы SCC, память делится на четыре банка по 8 КБ:

 Банк 1  4000h — 5FFFh
 Банк 2  6000h — 7FFFh
 Банк 3  8000h — 9FFFh
 Банк 4  A000h — BFFFh

Для изменения раскладки памяти, требуется произвести запись по одному из следующих адресов:

 Банк 1  5000h — 57FFh
 Банк 2  7000h — 77FFh
 Банк 3  9000h — 97FFh
 Банк 4  B000h — B7FFh

Биты с третьего по седьмой игнорируются. Значения от 0 до 7 выбирают 8-килобайтные страницы в младших 64 КБ ОЗУ картриджа (вариант с игрой Snatcher), значения от 8 до 15 — в старших 64 КБ (как с игрой Super Deform Snatcher).

По умолчанию (после сброса) устанавливается следующее соответствие банков и областей памяти: 1=0, 2=1, 3=2, 4=3. Такое распределение памяти аналогично обычным ПЗУ-картриджам с микросхемой SCC. Если младшие 64 КБ ОЗУ отсутствуют (в случае с Super Deform Snatcher), чтение из соответствующей им области памяти будет возвращать значение FFh, а запись будет игнорироваться.

По умолчанию, содержимое ОЗУ не может быть изменено (запись в него запрещена, режим ПЗУ). Для выбора режима доступа к памяти, а также для выбора режима работы звукогенератора (режим SCC+, в котором каналы 4 и 5 полностью независимы, или режим совместимости с SCC, в котором эти каналы используют общую форму сигнала) существует специальный регистр, называемый регистр режима. Он всегда находится в адресах BFFEh и BFFFh. возможно перевести память в режим ОЗУ, однако в этом режиме нет возможности изменить соответствие банков областям памяти.

 Бит 7  не используется
 Бит 6  не используется
 Бит 5  если установлен, звукогенератор работает в режиме SCC+, иначе в режиме совместимости с SCC
 Бит 4  если установлен, банки с 1 по 3 находятся в режиме ОЗУ, иначе в режиме выбора области памяти
 Бит 3  не используется
 Бит 2  если установлен, и бит 5 сброшен, банк 3 находится в режиме ОЗУ, независимо от значения бита 4
 Бит 1  если установлен, банк 2 находится в режиме ОЗУ, независимо от значения бита 4
 Бит 0  если установлен, банк 1 находится в режиме ОЗУ, независимо от значения бита 4

Ячейки памяти BFFEh и BFFFh не могут быть изменены даже в режиме ОЗУ, так как в них всегда отображается регистр режима.

Игры от Konami всегда используют значения 20h и 3Fh для регистра режима, так как они всегда используют режим SCC+. 30h также будет работать для режима ОЗУ.

Использование звукогенератора
Если банк 3 находится в режиме выбора области памяти, и бит 5 регистра режима сброшен, режим SCC может быть активирован записью значения 3Fh в любой адрес в диапазоне 9000h…97FFh. Набор управляющих регистров аналогичен обычным картриджам с SCC, и отличается только адресами регистров и возможность чтения формы сигнала канала 5. Регистры SCC отображаются памяти в диапазоне адресов от 9800h до 9FDFh.

Если банк 4 в режиме выбора области, и бит 5 регистра режима установлен, режим SCC+ может быть активирован записью любого значения с установленным старшим битом в любой адрес в диапазоне B000h…B7FFh. После этого регистры SCC+ становятся доступными в диапазоне адресов B800h…BFDFh.

См. также
 Konami SCC
 Konami VRC6

Ссылки
 Konami's Sound Cartridge (SCC+) for the MSX — подробное описание картриджа, техническая информация, фотографии

MSX
Микросхемы звукогенераторов

redpajama_set_name: RedPajamaWikipedia


text:
DynaMaxX - Your Gym, Your Home > Blog > Information > TERMS AND CONDITIONS
31st December 2019 DynaMaxX Information
1.1 The purpose of this subscription and the general conditions of sale is to govern the contractual relations between the operating company named "DynaMaxX S.à r.l." and the participant.
1.2 After having read the proposed services, general conditions and rules of procedure, the participant agrees to subscribe with DynaMaxX S.àrl, a registered and non-transferable subscription contract.
1.3 Consequently, the member is the only one authorized to use the devices and all other arrangements made available by DynaMaxX S.à r.l. with unlimited access during opening hours.
1.4 Access to the facilities of DynaMaxX S.à r.l. is based on the subscription plan to which the member subscribes. The subscription formulas provide access to cardio and weight training, group lessons as well as sauna during club hours. DynaMaxX S.à r.l. reserves the right to modify the opening hours and the services offered in order to adapt to the expectations of its customers. If the member does not make use of the possibilities offered by the chosen subscription, no refund will be made.
2. DURATION AND SERVICES
2.1 The right to use the fitness starts on the day of the subscription of the membership and is valid for the duration of the signed membership.
2.2 All subscriptions fully paid in advance, ie semi-annual, annual and biannual subscriptions automatically stop at the end of the specified period/date.
2.3 The semi-annual and annual and bi-annual subscriptions are payable by SEPA Direct Debit and will be automatically renewed for the same term at their expiry, except for the participant terminating their contract by registered letter with one month's notice before the expiry of the term contract.
2.4 The subscription is strictly personal and can not be transferred without the authorization of a member of the Management.
2.5 The member has the possibility to oppose this implied renewal in writing and before the expiry of his subscription.
2.6 DynaMaxX S.à r.l. reserves the right to modify the rates and / or the general conditions of the subscribed contract. DynaMaxX S.à r.l. will inform the member in writing of the new price list and general conditions at least one month before the effective date.
2.7 The new conditions will become effective as of right in the implied renewal of the current subscription.
3. SUBSCRIPTION FEES
3.1 The first month of subscription and registration fees are payable by credit card on the day of subscription. The member also undertakes to sign the SEPA European direct debit authorization on the day of subscription for the balance due on semi-annual and annual subscriptions. The price of subscriptions is determined and indicated on the registration form.
3.2 In case of refusal by the bank of a payment by bank domiciliation, DynaMaxX S.à r.l. reserves the right to demand the reimbursement of the administrative costs – 10€ each time DynaMaxX will have to present the direct debit without result – incurred for the recovery as well as the arrears due.
3.3 In case of non-payment beyond 2 months, the sums due until the expiry of the contract will become immediately due and DynaMaxX S.à r.l. will be able to initiate a recovery procedure. In such a case, the member will be liable to pay an additional indemnity of €50 and interest on arrears of 12% per year. In case of non-payment, DynaMaxX S.à r.l. reserves the right to deny the member access to the facilities until the discharge of his debt.
3.4 The discounts and promotional prices offered by DynaMaxX S.à r.l. are not cumulative, refundable or extendable. Only one discount per subscription will be granted.
3.5 All prices include VAT. DynaMaxX S.à r.l. reserves the right to adjust its prices at any time in the event of a change in the VAT rate.
3.6 Any change of bank domiciliation must be reported by the member, so that there can be no interruption or incident of payment.
4. PROVISIONAL SUSPENSION AND EXTENSION OF THE SUBSCRIPTION
4.1 In case of pregnancy resulting in a specifically mentioned disability not being able to practice sport for more than one month, a documentary proof (medical certificate) must be presented at the reception where a form of suspension, provided for in this effect will be given to the member. The contract will be extended by the time of the suspension.
5. CONDITIONS OF CANCELLATION OF SUBSCRIPTIONS
All subscriptions can only be cancelled in the following situations:
5.1 Premature termination for medical reasons can only be accepted with a medical certificate from a specialist.
5.2 The biannual, annual and semi-annual subscriptions may be terminated by registered letter with one month's notice from the implied renewal of the contract.
5.3 Termination of the contract based on one month's membership is possible at any time with one month prior notice. In this case, the member must pay the difference between the monthly price of the one month subscription and the monthly price of his/her subscription for the past months from the date of signature.
5.4 Terminations or requests relating to contracts must always be sent in writing to DynaMaxX S.à r.l., 111, route de Luxembourg L – 7241 Bereldange. Otherwise, the subscription will be maintained at the agreed conditions. To avoid misunderstandings, a cancellation form is provided for this purpose and available at the reception.
5.5 Any grounds or methods of termination invoked by the participant and other than those mentioned in articles 5.1, 5.2, 5.3, 5.4 and 5.5 (such as a unilateral cessation of monthly payments via direct debit) will be considered as an abusive breach of contract . In this case, all the monthly payments of the commitment period for the subscription fees will be immediately due. If the participant does not voluntarily pay the amounts due, DynaMaxX S.à r.l. may proceed with the recovery of his claim.
5.6 In case of non-compliance by the member of the contract and / or internal regulations, DynaMaxX S.à r.l. reserves the right to terminate the member's subscription contract, withdraw his member card or suspend his access to the club.
5.7 DynaMaxX S.à r.l. also reserves the right to terminate legally and to the exclusive fault of the participant, the contract for the following reasons: attitude or misconduct causing a risk or inconvenience for himself or for others adherents, any incident of payment respectively non-payment of the subscription, noncompliance with the rules of procedure, without this list being considered exhaustive.
5.8 This termination will be done by registered letter to the member and will result in the immediate payment of the sums still due by the member until the end of the commitment period.
6. PROCESSING OF PERSONAL DATA
6.1 Personal data needed for and in relation to the execution of the subscription contract entered into with DynaMaxX such as, but not limited to, name, date of birth, phone number, email address, home address, country, company/school and profession (the "Data") are collected and processed by DynaMaxX as data controller in compliance with the General Data Protection Regulation (EU) 2016/679 of 27 April 2016 and any related Luxembourg regulation.
6.2 The Data are processed by DynaMaxX acting for the purposes (i) of performance of the contractual relationship with the member, (ii) of invoicing the services provided by DynaMaxX (iii) of providing you with information in relation to the gym (such as newsletters, special offers, events, services etc.) and (iv) to comply with any other accountable, statutory and legal obligations of DynaMaxX.
6.3 DynaMaxX may transfer some of your Data to an accounting firm taking care of invoice management.
6.4 DynaMaxX shall only transfer your Data to service providers acting according to DynaMaxX demands in relation with the services provided by DynaMaxX. DynaMaxX undertakes to refrain from transferring Data of the member to third parties, aside from cases where the performance of the contract or the member obliges DynaMaxX to transfer the Data.
6.5 Should the member have any queries concerning the processing of his/her personal data, he/she should address them directly to the person in charge of data protection at DynaMaxX at the following address info@dynamaxx.lu.
6.6 The member has the right to request from DynaMaxX access to, rectification or erasure of his/her Data or restriction of processing. Furthermore, the member has the right to data portability, to the extent possible, according to the provisions of the Law. The requests must be addressed to the following address info@dynamaxx.lu.
6.7 DynaMaxX keeps the Data until the term of the contract and during the applicable archiving legal periods of limitation or any other period provided by the law.
6.8 The member authorizes DynaMaxX S.à r.l. to photograph or film him either individually or in groups during activities, workshops and shows. The member authorizes DynaMaxX S.à r.l. to use and publish his image as part of his visual and audiovisual communication (newspaper, advertising brochures, websites, etc.), except to oppose expressly by registered letter. In case the member considers that the processing of his/her Data does not comply with the applicable laws and regulation, he/she has the right to lodge a complaint with the supervisory authority of its the states of his/her residence, or of the place of the alleged infringement. In Luxembourg, the supervisory authority is the National Commission for Data Protection.
7. ELECTRONIC BRACELET
7.1 After signing the contract, the member is awarded an electronic bracelet. The latter is strictly personal and non-transferable.
7.2 Access to the club is only allowed on presentation of a valid wristband.
7.3 DynaMaxX S.à r.l. reserves the right to refuse access to the club to a member, or even to cancel its access without refund in case of abuse committed with the said electronic bracelet (for example, loan to a third party or substitution of no one).
7.4 The loss or theft of the electronic bracelet must be reported to DynaMaxX S.à r.l. immediately. An amount of € 20.00 will be charged to the member for the delivery of a new bracelet.
7.5 The member may request a new bracelet free of charge once a year due to wear and tear (subject to assessment by a member of the Management).
7.6 If the participant does not respect the contract and / or the internal rules posted in the club, DynaMaxX S.à r.l. reserves the right to break his subscription contract and withdraw or suspend his access to the club.
8. GROUP COURSES
8.1. All courses must be booked by email or at the gym.
8.2. DynaMaxX reserves the right to modify its schedules at its best convenience. These modifications will be announced on the website and on Instagram/Facebook in order to allow the member to find out the times of the group classes. All group lessons are posted on the websites and displayed in the club. Group lesson schedules are not considered by the member as an essential feature of the service to be provided by DynaMaxX.
8.3. The unsubscription to a group lesson must be done via email or by phone and a minimum of two hours before the beginning of the lesson. If on two occasions there has been no unsubscription or if there has been a late unsubscription, the member can no longer book group lessons for one week. In case of a late deregistration without membership with group lessons, a fee of 4€ will be charged for the course.
9. MEDICAL ASSISTANCE
9.1 The member hereby declares that he/she is in good health, that he/she has no medical or other reason that would prevent him or her from participating in a physical effort and that the performance of a physical effort cannot cause any harm to his or her health, safety and/or physical condition.
9.2 No use of the club may be made if he or she suffers from an infection, contagious disease or has injuries such as open wounds, open sores, etc. for which there is a risk that the health, safety and/or physical condition of other members may be damaged.
10. RISK AND LIABILITY
10.1. Neither DynaMaxX, nor its employees or servants can be held responsible for the death, personal injury or illness of members, their children or visitors on the club's territory, except as a result of an intentional act or gross negligence on the part of DynaMaxX or its employees. The use of the equipment provided by DynaMaxX, participation in a program or parts of courses and/or activities of any nature whatsoever is entirely at the member's own risk.
10.2. Neither DynaMaxX, nor its employees or servants can be held responsible for the loss, damage or theft of property within the club's territory, except as a result of an intentional act or gross negligence on the part of DynaMaxX or its employees. Members must always lock their valuables in the lockers provided for this purpose.
10.3. The member is liable for damage caused to the property of DynaMaxX and its employees.
11. LITIGATION
11.1 The law applicable to all disputes arising in connection with the content, interpretation and / or execution of this contract / subscription linking DynaMaxX S.à r.l. to the member is the Luxembourg law.
11.2 The courts of Luxembourg are exclusively competent to settle all disputes arising from the interpretation and / or execution of the contract binding DynaMaxX S.à r.l. to the member.
6th October 2019 Information


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