text:
\section{Introduction and statement of the main result.}
In \cite{B-S} J.Br\"uning\& R.Seeley have studied asymptotic
expansions of integrals :
\begin{equation}
H(z)=\int\limits_{0}^{\infty} \sigma (xz,x)dx,\text{ }
z\rightarrow \infty,
\end{equation}
where $\sigma(x,\xi)$ is a singular symbol. The result of \cite{B-S} is quite remarkable, in particular because it
can be directly applied to spectral analysis. Many asymptotic questions can be reduced to the study of the
previous problem but it is also interesting to consider a
generalization :
\begin{equation}\label{the problem}
I(z)=\int\limits_{X} g(z f(x),x)dx,\text{ }
z\rightarrow \infty.
\end{equation}
where $g: \mathbb{R}\times X \rightarrow \mathbb{R}$, $f:X\rightarrow \mathbb{R}$
are smooth and $X$ is a smooth differentiable manifold equipped with the $C^{\infty}$ density $dx$.
As a motivation, we remark that the trace formula for
certain linear self-adjoint operators can be stated in the form of Eq. (\ref{the
problem}). In the semi-classical regime when $f$ is a quadratic form this problem was involved in \cite{BPU}.
See also \cite{Cam1} for the case of $f$ with an isolated and degenerate
singularity associated to an homogeneous definite jet.

\textit{General assumptions.} Throughout this work, we will
assume that $|f|$ is strictly positive outside of a compact set
and that the Fourier transform $\hat{g}$ w.r.t. $t$ exists with
$\partial_t^k \hat{g}(t,x)\in L^1(\mathbb{R}\times X)$, $\forall k$.
For this reason $g$ will be called a symbol. $\hfill{\square}$
\begin{remark}\rm{This assumption on $g$ is strong but can be weakened.
More general conditions can be found in \cite{HOR1}. Mainly, this condition will be used
to reach integrals with compact supports.}
\end{remark}
As $z\rightarrow \infty$, the asymptotic behavior of $I(z)$ is related to the critical fiber :
\begin{equation}
\mathfrak{S}=f^{-1}(\{0\})=\{ x\in X \text{ / } f(x)=0\}.
\end{equation}
This can easily be viewed with the Fourier inversion formula :
\begin{equation}\label{problem osci}
I(z)= \int\limits_{X} \int\limits_{\mathbb{R}} e^{i z t f(x)}
\hat{g}(t,x)dtdx,\text{ } z\rightarrow \infty,
\end{equation}
where $\hat{g}(t,x)$ is the normalized Fourier transform of $g$ w.r.t. $t$ :
\begin{equation}
\hat{g}(t,x)=\frac{1}{2\pi}\int\limits_{\mathbb{R}} e^{-i\tau t}
g(\tau,x)d\tau,
\end{equation}
In Eq. (\ref{problem osci}), the stationary points w.r.t. $t$ are
precisely given by $\mathfrak{S}$, i.e. $I(z)$ is asymptotically
supported by $\mathfrak{S}$. Since $f$ is smooth $\mathfrak{S}$ is closed and
according to the general assumptions above we obtain :
\begin{equation}
(H_0) \textit { The fiber }\mathfrak{S} \textit{ is compact}.
\end{equation}
$(H_0)$ is not absolutely necessary, e.g. if $g$
decreases fast enough near the boundary of $X\cap \mathfrak{S}$. But, to simplify, we will
only consider the compact case. To obtain an easier formulation of the problem we recall
an elementary result.
\begin{lemma}\label{non stationary}
If $\partial^k_t\hat{g}\in L^1(\mathbb{R}\times X)$, $\forall k\in\mathbb{N}$, modulo terms $\mathcal{O}(z^{-\infty})$,
asymptotics of Eq. (\ref{the problem})
are not changed by assuming that
$\hat{g}$ is compactly supported near
$\mathfrak{S}$.
\end{lemma}\label{compact-form}
\textit{Proof.} With $\mathfrak{S}$ compact we choose a cut-off
function $\Theta\in C_0^{\infty}(X)$ such that $\Psi=1$ near
$\mathfrak{S}$ and $0\leq \Psi \leq1$. We shall estimate
the error integral :
\begin{equation}
E(z)=\int\limits_{X} \int\limits_{\mathbb{R}} e^{i z t f(x)}
\hat{g}(t,x)(1-\Psi(x))dtdx,\text{ } z\rightarrow \infty.
\end{equation}
With $L=-(i/zf(x))\partial_t$, we have
$L^k e^{i z t f(x)}=e^{i z t f(x)}$, $\forall k \in\mathbb{N}$.
By integration by parts and since $|f(x)|\geq C$ on $\mathrm{supp}(1-\Psi)$, we obtain :
\begin{equation}
|E(z)|\leq (Cz)^{-k} || \partial_t^{k} \hat{g}(t,x)
||_{\mathrm{L}^1(\mathbb{R}\times X)}=\mathcal{O} (z^{-k}), \text{ } \forall k
\in\mathbb{N}.
\end{equation}
This gives the desired result, with our hypothesis on $g$.
$\hfill{\blacksquare}$\medskip\\
Lemma \ref{compact-form} allows to consider only integrals with compact support
w.r.t. $x$ which simplifies all questions of convergence. Notice that we can weaken the condition
on $g$ to $\partial^k_t\hat{g}\in L^1(\mathbb{R}\times X)$, $\forall k\leq k_0$ with an error $\mathcal{O}(z^{-k_0})$.
We are mainly interested in the situation where $\mathfrak{S}$ has an isolated singularity.
If $x_0\in\mathfrak{S}$ is such a critical point, let $\Theta\in C_0^\infty(X)$ be a cut-off microlocally
supported near $x_0$. We split-up our integral as
$I(z)=I_r(z)+I_s(z)$, where :
\begin{gather}
I_r(z)=\int\limits_{X}  g(z f(x),x)(1-\Theta)(x)dx, \label{regularpart}\\
I_s(z)=\int\limits_{X} g(z f(x)t,x) \Theta(x)dx. \label{singpart}
\end{gather}
This procedure can be extended with finitely many critical points on $\mathfrak{S}$.
The regular part $I_r$ can be treated by the generalized stationary phase method,
with non-degenerate normal Hessian, which we recall in section 2.
Since $I_s$ is a local object and the main
contributions below concern invariant objects, there is no
loss of generality to assume that $\mathrm{supp}(\Theta)$ is an open of
$\mathbb{R}^n$, $n=\dim (X)$. For $x_0\in \mathfrak{S}$ a singularity of finite order we can write the germ of $f$ as :
\begin{equation}\label{germ}
f(x)=f_k(x)+\mathcal{O}(||(x-x_0)||^{k+1}),
\end{equation}
where $f_k\neq 0$ is homogeneous of degree $k\geq 2$ w.r.t.
$(x-x_0)$. The first elementary result concerns extremum attached to such
homogeneous germs :
\begin{theorem} \label{main1}
If $f$ has a local extremum $x_0$ on $\mathfrak{S}$ whose jet is given by Eq.
(\ref{germ}) (a fortiori $k$ is even), we obtain a full-asymptotic expansion :
\begin{equation}
I_s(z)\sim\sum\limits_{j\in\mathbb{N}} c_j z^{-\frac{j}{k}}.
\end{equation}
If $\dim(X)=n$, the leading term is given by :
\begin{equation} \label{sing part for extremum}
I_s(z)=z^{-\frac{n}{k}} \left\langle t_e^{\frac{n-k}{k}}, g(t,x_0)
\right\rangle \frac{1}{k}\int\limits_{\mathbb{S}^{n-1}}
|f_k(\theta)|^{-\frac{n}{k}} d\theta
+\mathcal{O}(z^{-\frac{n+1}{k}}),
\end{equation}
with $t_e=\max(t,0)$ if $x_0$ is a minimum and $\max(-t,0)$ for a
maximum.
\end{theorem}
The reader can observe that Theorem \ref{main1} includes the case of a non-integrable
singularity on $\mathfrak{S}$ for $k>n$. Accordingly, always for $k>n$,
the contribution of the critical point is bigger than the regular contribution (see section 2).
\begin{remark} \rm{The result stated in Eq. (\ref{sing part for extremum}) includes the case
$k=2$ and the integral on the sphere can be expressed in terms of
special functions. Under ad-hoc conditions, Theorem \ref{main1} can be extended to a singularity which is a sum
of positively homogeneous singularities, e.g. $f(x_1,x_2)=||x_1||^4+||x_2||^6$.
}
\end{remark}
The case of non-extremum degenerate critical points is more difficult. Since this problem can be
very complicated in general position we impose :\medskip\\
\textit{$(H_1)$ $\mathfrak{S}$ has a unique critical point $x_0$. Moreover, $f_k$ defined in Eq. (\ref{germ}) is
non-degenerate in the sense that :}
\begin{equation}\nabla f_k \neq 0 \text{ }on \text{ }\mathbb{S}^{n-1}\cap
\{f_k=0\}.
\end{equation}
By homogeneity, $(H_1)$ is equivalent to "$f_k$ has an isolated
singularity". Observe that this condition is very close to H\"ormander's real principal type condition
for distributions. We define the integrated density of $f_k$ on the sphere (or co-area) as :
\begin{equation}
\mathrm{LVol}(w)=\int\limits_{\{f_k(\theta)=w\}} |dL|(\theta),\text{ }dL(\theta)\wedge df_k(\theta)=d\theta.
\end{equation}
Where $dL$ is the $n-2$ dimensional Liouville measure induced by $f_k$ on $\mathbb{S}^{n-1}$,
i.e. the Riemannian density induced by $f_k$ on the standard density of $\mathbb{S}^{n-1}$.
Note that $(H_1)$ insures that $\mathrm{LVol}(w)$ is well defined and smooth near the origin.
\begin{theorem}\label{main2}
Under the previous assumptions and if $x_0$ satisfies $(H_1)$, the
singular part of our integral admits a full asymptotic expansion :
\begin{equation}
I_s(z)\sim \sum\limits_{j=0}^{\infty} c_{j}z^{-\frac{j}{k}}
+\sum\limits_{j=0}^{\infty} d_j z^{-j}\log(z).
\end{equation}
a) If $k>n$ (non-integrable singularity), the leading term is :
\begin{equation}
I_s(z)=C_0 z^{-\frac{n}{k}}
+\mathcal{O}(z^{-\frac{n+1}{k}}\log(z)),
\end{equation}
where the distributional coefficient $C_0$ is given by :
\begin{equation*}
\frac{1}{k}(\left\langle t_{+}^{\frac{n}{k}-1}, g(t,x_0)\right\rangle
\int\limits_{\{f_k \geq 0\}}
|f_k(\theta)|^{-\frac{n}{k}}d\theta  +\left\langle
t_{-}^{\frac{n}{k}-1}, g(t,x_0)\right\rangle
\int\limits_{\{f_k \leq 0\}}
|f_k(\theta)|^{-\frac{n}{k}}d\theta).
\end{equation*}
b) If $k$ divides $n$ the leading term is logarithmic :
\begin{equation}
I_s(z)= D_0 z^{-\frac{n}{k}} \log (z)
+\mathcal{O}(z^{-\frac{n}{k}}),
\end{equation}
with $n=kp$, we obtain :
\begin{equation*}
D_0=\frac{1}{k}
\left ( \frac{d^{p-1}}{dw^{p-1}}\mathrm{Lvol}(0) \right ) \int\limits_{\mathbb{R}} |t|^{p-1}g(t,x_0)dt.
\end{equation*}
c) If $n>k$ and $n/k \notin \mathbb{N}$ (integrable singularity) we obtain the same result as in a) but
with the modified distributions :
\begin{equation*}
\left\langle t_{+}^{\frac{n}{k}-1}, g(t,x_0)\right\rangle
\left\langle \frac{\tilde{d}^n}{\tilde{d}w^n} w^{n-\frac{n}{k}}_{+}, \mathrm{Lvol} \right\rangle
+\left\langle t_{-}^{\frac{n}{k}-1}, g(t,x_0)\right\rangle
\left\langle \frac{\tilde{d}^n}{\tilde{d}w^n} w^{n-\frac{n}{k}}_{-}, \mathrm{Lvol} \right\rangle,
\end{equation*}
where the derivatives w.r.t. $w$ are normalized distributional derivatives.
\end{theorem}
The meaning of normalized derivative it that one choose the normalization :
\begin{equation}\label{normalized derivatives}
\left\langle \frac{\tilde{d}^n}{\tilde{d}w^n} w^{n-\frac{n}{k}}_{\pm}, f(w)\right\rangle :=
\left\langle w^{-\frac{n}{k}}_{\pm}, f(w)\right\rangle,
\end{equation}
for all $f\in C_0^\infty$ with $f=0$ in a neighborhood of the origin.
The distributional bracket involving $\mathrm{Lvol}$ is detailed in section 3.
Results c) and b) for $p\geq 2$ are not intuitive and are certainly difficult to be
reached without geometry. In particular, for applications to oscillatory integrals (see below) one has to work
in the dual since both Fourier transforms w.r.t. $t$ in c) and b) are distributional.
In c), the $n$-th derivative is arbitrary and the result is the same for any normalized derivative of order greater than
$\mathrm{E}(n/k)$.
\begin{remark} \rm{Results a) and b) for $p=1$ are certainly interesting for spectral analysis
since these contributions are bigger than $I_r(z)=\mathcal{O}(z^{-1})$. As in Theorem \ref{main1},
non-integrable singularities have a dominant contribution. Hence, the leading term of $I(z)$ is always an invariant.}
\end{remark}
\textbf{Application to oscillatory integrals.}\\
A typical application of Theorems \ref{main1}\&\ref{main2} can be the asymptotic expansion of
distributional traces of quantum propagators. Hence, it is interesting to remark that our results
can be extended to asymptotic integrals :
\begin{equation*}
\tilde{I}(z)=\int\limits_{X} G(z,zf(x),x)dx,\text{ }z\rightarrow
+\infty,
\end{equation*}
if $G$ admits an asymptotic expansion with a priori estimates,
i.e. :
\begin{gather*}
G(z,t,x)=\sum\limits_{j=0}^{l} z^{-\alpha_j}g_j(t,x)+R_l(z,t,x),\\
\forall k\in\mathbb{N}^* : ||R_k(z,t,x)||_{L^1(\mathbb{R}\times
X)}=\mathcal{O}(z^{-(\alpha_k+\varepsilon)}),\text{ } \varepsilon >0,
\end{gather*}
where $(\alpha_j)_j$ is a strictly increasing sequence and $R$
controlled by uniform estimates. Similarly, we can consider
expansions in term of $z^{-\alpha_j}\log(z)^m$. This notion of graduation w.r.t. $z$ allows to
apply our results but, to simplify the exposition, in this work
we will just consider the case of an integral of a symbol
$g(t,x)$. We can treat degenerate oscillatory integrals :
\begin{equation}\label{IO}
O(z)=\int\limits_{\mathbb{R}\times X} e^{iz t f(x)} a(t,x,z)dtdx, \text{ }z\rightarrow +\infty,
\end{equation}
providing that $f$ satisfies the conditions of Theorem \ref{main1} or \ref{main2}.
An important application in quantum mechanics is the case $X=T^*\mathbb{R}^n$ where, after some technical
modifications, the localized (distributional) trace of $h$-pseudors :
\begin{equation*}
\mathrm{Tr } u_h (A_h-E):= \mathrm{Tr} \int\limits_{\mathbb{R}}  \hat{u}(t) e^{\frac{i}{h}t(A_h-E)} dt,
\text{}  \hat{u}\in C_0^\infty(\mathbb{R}), \text{ } E\in\mathbb{R},
\end{equation*}
can be written as a locally finite sum of oscillatory integrals :
\begin{equation}\label{IO quantum}
\int\limits_{\mathbb{R}\times T^*\mathbb{R}^n} e^{\frac{i}{h}( S(t,y,\eta)-\left\langle y,\eta \right\rangle -tE)}
b(h,t,y,\eta) dtdyd\eta,
\end{equation}
where $b(h,\bullet)\sim \sum h^{-k} b_k$ satisfies a priori estimates as above and $S$ is the local generating
function of the group of diffeomorphism of the principal symbol of $A_h$.
Here $z=h^{-1}$ is the parameter and, after a
discussion based on classical mechanics, Eq. (\ref{IO quantum}) can be reformulated as in Eq. (\ref{IO}) where
$\mathfrak{S}$ is the energy surface of level $E$. For more details, we refer to \cite{BPU,Cam0,Cam1}.

We recall now basics on homogeneous transformations, some of them will also be used below.
The Melin transform of a function $h$ is defined as :
\begin{equation}
M[h](\xi)=\int\limits_{0}^{\infty} h(t) t^{\xi-1}dt.
\end{equation}
Generally $M[h](\xi)$ is analytic, for
example, in the strip $\Re(\xi)\in]a,b[$ with :
\begin{equation*}
a= \inf\limits_{x\in\mathbb{R}} \{ \int\limits_{0}^\infty
|h(t)|t^{x-1} dt <\infty \},\text{ }
b=\sup\limits_{x\in\mathbb{R}} \{ \int\limits_{0}^\infty
|h(t)|t^{x-1} dt <\infty \},
\end{equation*}
and when $M[h](c+iy)\in L^1(\mathbb{R},dy)$ we have the inversion
formula :
\begin{equation}
h(t)=\frac{1}{2i\pi} \int\limits_{c-i\infty}^{c+i\infty} t^{-\xi}
M[h](\xi)d\xi.
\end{equation}
By elementary changes of path we obtain that :
\begin{equation*}
M[e^{\pm i t}](\xi)= e^{\pm i \pi \frac{\xi}{2}}\Gamma(\xi),\text{ }
\mathrm{Re}(\xi)\in ]0,1[.
\end{equation*}
We write  $O(z)=O_{+}(z)+O_{-}(z)$, where :
\begin{equation*}
O_{\pm}(z)=\int\limits_{\{\pm tf(x)>0\}} e^{iztf(x)}a(t,x)dtdx.\\
\end{equation*}
Melin's inversion formula leads to the distributional formulation :
\begin{equation*}
O_{\pm}(z)=\frac{1}{2i\pi} \int\limits_{c-i\infty}^{c+i\infty}
e^{\pm i\xi\frac{\pi}{2}} \Gamma(\xi) z^{-\xi}
\int\limits_{\{\pm tf(x)>0\}}
|t f(x)|^{-\xi} a(t,x)dtdx d\xi,
\end{equation*}
where $0<c<\delta$ for $\delta>0$ depending on $f$.
If no elementary method works, one can construct explicit meromorphic extensions
of distributions :
\begin{gather*}
\left\langle T_{f}^{+}(\xi),a\right\rangle=e^{+i\xi\frac{\pi}{2}} \Gamma(\xi) z^{-\xi}\int\limits_{\{tf(x)>0\}}
(tf(x))^{-\xi} a(t,x)dxdt,\\
\left\langle T_{f}^{-}(\xi),a\right\rangle=e^{-i\xi\frac{\pi}{2}} \Gamma(\xi) z^{-\xi}\int\limits_{\{tf(x)<0\}}
|tf(x)|^{-\xi} a(t,x)dxdt,
\end{gather*}
and use Cauchy's residue formula to get asymptotic w.r.t.
$z\rightarrow \infty$. This method is described in \cite{WON}
when $f$ is monomial, which is a generic situation when $f$ is
analytic on $\mathrm{supp}(a)$ by Hironaka's theorem of resolution
of singularities \cite{Hir}. Also in many cases it is possible to reach this
setting by a non-analytic change of coordinates. This approach is very interesting but a great disadvantage
is that there is 4 domains, a sequence of poles (of order 3) which does not contribute at all in the expansion and
the complex factors $\exp(\pm i\pi \xi /2) \Gamma(\xi)$ lead to long calculations in presence of multiple poles.\\
In fact we will reach Theorem \ref{main1}\&\ref{main2} more directly and by a method which avoids
Fourier analysis until the last step of the proof.
\section{Case of $\mathfrak{S}$ regular and non-degenerate critical points.}
If $f$ is regular on $\mathfrak{S}$, then $\mathfrak{S}$ is a compact and smooth submanifold.
\begin{prop}Under the previous assumptions and if $\mathfrak{S}$ is a regular
surface, $I(z)$ admits a full asymptotic expansion in powers of
$z^{-1}$ with :
\begin{equation}
I(z)=\frac{1}{z} \int\limits_{x\in\mathfrak{S}}
\int\limits_{\mathbb{R}}g(t,x)
dtd_{\mathfrak{S}}(x)+\mathcal{O}(z^{-2}),
\end{equation}
where $d_{\mathfrak{S}}$ is the invariant surface measure of
$\mathfrak{S}$. The same result holds for the integral $I_r(z)$
with insertion of the cut-off in the integral.
\end{prop}
By the implicit functions theorem and compactness, we pick some
coordinates $y$, defined near $\mathfrak{S}$, and a finite
partition of unity $\Omega_j$ covering $\mathfrak{S}$, such that
$f$ is diffeomorphic to the coordinate $y_1$ in
$\mathrm{supp}(\Omega_j)$. By Lemma \ref{non stationary} we obtain
:
\begin{gather*}
I(z)=\sum I_j(z)+\mathcal{O}(z^{-\infty}),\\
I_j(z)=\int\limits_{\mathbb{R}^2}\int\limits_{\mathbb{R}^{n-1}}
e^{i z t y_1} y^{*}(\Omega_j(x) \hat{g}(t,x)|Jy|) dtdy_1dy_2
...dy_n,
\end{gather*}
where $y^{*}$ and $|Jy|$ are respectively the pullback by $y$ and
the multiplication by the standard Jacobian. Hence, we have to estimate an oscillatory integral with
quadratic phase. The stationary phase Lemma, see e.g. Lemma
7.7.3 of \cite{HOR1} vol.1, provides a full-asymptotic expansion :
\begin{equation}
\int\limits_{\mathbb{R}^2}e^{i z t y_1} y^{*}(\Omega_j(x)
\hat{g}(t,x)|Jy|) dtdy_1=\sum\limits_{l=0}^{N-1} C_l
z^{-(l+1)} +\mathcal{O}(z^{-(N+1)}).
\end{equation}
Since $\mathrm{sign}(ty_1)=0$, the expansion is real with leading term :
\begin{equation}
I_j(z)=\frac{2\pi}{z} \int\limits_{\mathbb{R}^{n-1}}
(y^{*}(\Omega_j(x) \hat{g}(t,x)|Jy|))(0,0,y_2,...,y_n)
dy_2...dy_n+\mathcal{O}(z^{-2}).
\end{equation}
Geometrically, since $\mathfrak{S}$ is locally given by $y_1=0$,
this corresponds to integration on $ \mathfrak{S}$ w.r.t. the
canonical $n-1$ dimensional measure $d_{\mathfrak{S}}$. By
summation over the indices $j$ we obtain the integral on the whole
surface, i.e. :
\begin{equation}\label{density-regul.}
I(z)\sim \frac{2\pi}{z} \int\limits_{x\in\mathfrak{S}} \hat{g}(0,x)
d_{\mathfrak{S}}(x)=\frac{1}{z} \int\limits_{x\in\mathfrak{S}}
\int\limits_{\mathbb{R}}g(t,x)
dtd_{\mathfrak{S}}(x).
\end{equation}
\begin{remark}\rm{Eq. (\ref{density-regul.}) is a particular (flat) case
of stationary phase formulas with a smooth compact manifold of critical points
and non-degenerate transverse Hessian. $d_{\mathfrak{S}}$ is the Liouville-measure of classical
mechanics or Guelfand-Leray-measure in theory of singularities.
The oscillatory representation of delta-Dirac
distributions, by mean of Schwartz kernels, provides a very natural definition of this object.}
\end{remark}
Now, assume that $f$ admits a single non-degenerate critical point
on $\mathfrak{S}$. As usually, non-degenerate means that
$Q(\xi)=\frac{1}{2}\left\langle \xi,d^2f(x_0)\xi\right\rangle$ is a non-degenerate quadratic form. After perhaps a
reduction of the cut off $\Theta$, the Morse Lemma, with coordinates $w$, provides a simpler problem :
\begin{equation} \label{IO cubic}
F(z)=\int\limits_{\mathbb{R}\times\mathbb{R}^n} e^{izt Q(w)}
w^{*} (\hat{g}(t,x) \Theta(x)|Jw|(x)) dw.
\end{equation}
This kind of asymptotic problem was precisely studied in
\cite{BPU}. The result is :
\begin{prop}
If $Q$ has at least one even index of inertia then :
\begin{equation}
F(z)\sim \sum\limits_{j=0}^{\infty} c_j z^{-1-\frac{j}{2}}.
\end{equation}
But, if $Q$ has both indices of inertia odd then we have :
\begin{equation}
F(z) \sim \sum\limits_{j=0}^{\infty} \sum\limits_{m=0,1} c_{j,m}
z^{-1-j}\mathrm{log}(z)^m.
\end{equation}
\end{prop}
The method used in section 3.2 allows to find this result. Our
proof is not simpler and that's why we refer to Proposition 3.4
and Theorem 3.5 of \cite{BPU} for a detailed proof and discussion
of the coefficients. This discussion is necessary because the exponential distributions
of Eq. (\ref{IO cubic}) possess spherical symmetries.
\section{Proof of the main results.}
\subsection{Critical points attached to a local extremum.}
We start by the case where $f$ admits a local minimum $x_0$ on the
fiber $\mathfrak{S}$. We split-up our integral $I(z)$ via Eqs. (\ref{regularpart},\ref{singpart}).
$I_r$ can be treated as in section 2 and from now we concentrate our attention on $I_s$.
With the extremum condition, $x_0$ is isolated on $\mathfrak{S}$
and $\mathfrak{S}\cap\mathrm{supp}(\Theta)=\{x_0\}$ for $\mathrm{supp}(\Theta)$ small
enough.To simplify notations we identify $x_0$ with the origin and we recall that we consider :
\begin{equation}
f(x)=f_{k}(x)+\mathcal{O}(||x||^{k+1}),
\end{equation}
where $f_k$ is homogeneous of degree $k$ and definite positive.
Using polar coordinates, the Taylor formula shows that :
\begin{equation}
f(r\theta)=r^k (f_k(\theta)+R(r,\theta)),
\end{equation}
with $R(0,\theta)=0$ and where we note again $f_k(\theta)$ the
restriction of $f_k$ to $\mathbb{S}^{n-1}$. If $\mathrm{supp}(\Theta)\subset B(0,r_0)$
is chosen small enough we have :
\begin{equation*}
f_k(\theta)+R(r,\theta)\neq 0,\text{ } \forall
\theta\in\mathbb{S}^{n-1}, \text{ } \forall r \in [0,r_0[.
\end{equation*}
Hence, with the homogenous coordinates $v=(u,\theta)$ where :
\begin{equation}
u(r,\theta)=r (f_{k}(\theta)+R(r,\theta))^{\frac{1}{k}},
\end{equation}
we can express our integral as :
\begin{equation}\label{reduced form mini}
I_s(z)=\int\limits_{\mathbb{R}_{+}} G(z u^k,u)du.
\end{equation}
The new symbol $G$ is obtained by pullback and integration :
\begin{equation}\label{amplitude 1}
G(t,u)=\int\limits_{\mathbb{S}^{n-1}} v^*(g(t,r\theta)\Theta(r\theta)r^{n-1}|J(v)|)d\theta.
\end{equation}
The existence of the asymptotic expansion is a consequence of :
\begin{lemma}\label{lemma for extremum}
For $a$ in
$C_0^{\infty}(\mathbb{R}\times\mathbb{R}_{+})$, the following asymptotic expansion holds :
\begin{equation}
J(z)=\int\limits_{\mathbb{R}_{+}}
a(zu^k,u)du \sim \sum\limits_{j\geq 0}
z^{-\frac{j+1}{k}}d_j(a),\text{ }z\rightarrow \infty,
\end{equation}
where the $d_j$ are universal distributions given by :
\begin{equation}
d_j=\frac{1}{k}
\frac{1}{j!}(\tau_{+}^{\frac{j+1-k}{k}}\otimes
\delta_0^{(j)}(u)).
\end{equation}
\end{lemma}
\textit{Proof.} The crucial point is that $u^k$ is increasing on $\mathbb{R}_+$. We have :
\begin{equation*}
J(z)=z^{-\frac{1}{k}}
\int\limits_{0}^{\infty}
a(u^k,\frac{u}{z^{\frac{1}{k}}})du.
\end{equation*}
A Taylor expansion w.r.t. the second argument at the origins gives
:
\begin{equation}
a(u^k,\frac{u}{z^{\frac{1}{k}}})=\sum\limits_{l=0}^{N}
\frac{1}{l!} z^{-\frac{l}{k}} u^l \frac{\partial^l
a}{\partial u^l}(u^k,0)+z^{-\frac{N+1}{k}}R_{N+1}(u,z),
\end{equation}
where $R_{N+1}(u,z)$ is integrable w.r.t. $u$ with $L^1$ norm
uniformly bounded in $z$. By a new change of variable we obtain :
\begin{equation}
J(z)=\frac{1}{k} \sum\limits_{l=0}^{N} \frac{1}{l!}
z^{-\frac{1+l}{k}} \int\limits_{0}^{\infty} \frac{\partial^l
a}{\partial u^l}(\tau,0) \tau^{\frac{l+1-k}{k}} d\tau
+\mathcal{O}(z^{-\frac{N+1}{k}}).
\end{equation}
These coefficients are well defined since
$|\tau|^{\frac{l+1-k}{k}}\in L^1_{\mathrm{loc}}(\mathbb{R})$ for all
$l\in\mathbb{N}$. $\hfill{\blacksquare}$
\begin{remark}\label{remark maximum} \rm{This expansion holds also for a pullback by
$-u^k$ if we replace $\tau_{+}^{\alpha}$ by $\tau_{-}^{\alpha}$.
This allows to treat the case of a local maximum in Theorem
\ref{main1}.} Also for an application to oscillatory integrals we obtain a nice formulation
via the Fourier transform of the distributions $\tau_{\pm}^{\alpha}$, which avoids any "regularization".
\end{remark}
\textit{Proof of Theorem \ref{main1}.} We treat the case of a
local minimum. We apply Lemma \ref{lemma for extremum} to the
integral of Eq. (\ref{reduced form mini}) to prove the existence
of the asymptotic expansion and it remains to express
invariantly the leading term. With the polar coordinates $G$ vanishes up to the
order $n-1$. Consequently, we have :
\begin{equation*}
I_s(z)=\frac{z^{-\frac{n}{k}}}{k} \frac{1}{(n-1)!} \left\langle
t_{+}^{\frac{n-k}{k}}\otimes \delta_0 ^{n-1},
G \right\rangle+\mathcal{O}(z^{-\frac{n+1}{k}}).
\end{equation*}
Starting form Eq. (\ref{amplitude 1}) and since :
\begin{equation*}
|Jv|(0,\theta)=|f_k(\theta)|^{-\frac{1}{k}},
\end{equation*}
by elementary manipulations on delta-Dirac distributions, we obtain that :
\begin{equation}\label{formula for C}
I_s(z)= z^{-\frac{n}{k}}\left\langle t_{+}^{\frac{n-k}{k}},g(t,0)\right\rangle
\frac{1}{k}\int\limits_{\mathbb{S}^{n-1}}|f_k(\theta)|^{-\frac{n}{k}}
d\theta+\mathcal{O}(z^{-\frac{n+1}{k}}).
\end{equation}
Finally, for a local maximum we replace the distributions $t_{+}^{\frac{n-k}{k}}$ by $t_{-}^{\frac{n-k}{k}}$.
$\hfill{\blacksquare}$
\begin{remark}
\rm{The leading term is invariantly defined
since Eq. (\ref{formula for C}) involves the evaluation $g(t,z_0)$
and $f_k$ is an invariant since the $(k-1)$-jet in $z_0$ is flat.}
\end{remark}
\textbf{On the integral on the sphere.}\medskip\\
The integrals on $\mathbb{S}^{n-1}$ of Eq. (\ref{formula for C}) can be
reformulated. For example in dimension 2 these are elliptic
integrals. In higher dimensions we define :
\begin{equation}
I(f_k)=\int\limits_{\mathbb{R}^n} e^{-|f_k|(x)}dx,
\end{equation}
since $f_k$ is homogeneous we obtain :
\begin{equation*}
I(f_k)=\int\limits_{\mathbb{R}_{+}\times \mathbb{S}^{n-1}} e^{-r^k
f_k(\theta)} r^{n-1} dr d\theta=\int\limits_{0}^{\infty} e^{-u^k}
u^{n-1} du \int\limits_{\mathbb{S}^{n-1}}
|f_k(\theta)|^{-\frac{n}{k}} d\theta.
\end{equation*}
Hence our integral is given by :
\begin{equation}
\frac{1}{k}\int\limits_{\mathbb{S}^{n-1}}
|f_k(\theta)|^{-\frac{n}{k}} d\theta=\frac{I(f_k)}{\Gamma (n/k)}.
\end{equation}
$I(f_k)$ can be computed, by elementary methods, as a product of
gamma factors or hypergeometric functions. We give an elementary
example in section 4.
\subsection{Case of non-extremum critical points.}
This problem is more complicated and we recall that we just treat
here singularities given by $(H_1)$. As before, we just have to
study $I_s(z)$ and, in local coordinates, we identify
the critical point with the origin. We define $C(f_k)$ as the trace on the unit sphere of the conical set
of the zeros of $f_k$, i.e. :
\begin{equation}
C(f_k)=\{ \theta \in \mathbb{S}^{n-1} \text{ / } f_k(\theta)=0\}
=\mathbb{S}^{n-1}\cap \{x\in\mathbb{R}^n \text{ / } f_k(x)=0\}.
\end{equation}
With $(H_1)$, $C(f_k)$ is a compact smooth submanifold of $\mathbb{S}^{n-1}$ of codimension 1.
To perform a blow-up of the singularity we use polar
coordinates $(r,\theta)$ and the next lemma gives a resolution of the
singularity w.r.t. $C(f_k)$.
\begin{lemma}\label{normal form}
In a micro-local neighborhood of the origin there exists local
coordinates $y$, on the blow-up of the critical point, such that :
\begin{gather*}
f(x) \simeq \left\{
\begin{matrix}
y_{1}^{k},\text{ in all directions where }f_{k}(\theta )>0,\\
-y_{1}^{k},\text{ in all directions where }f_{k}(\theta )<0,\\
y_{1}^{k}y_{2},\text{ locally near }C(f_{k}).
\end{matrix}
\right.
\end{gather*}
\end{lemma}
\noindent\textit{Proof.} By Taylor, there exists $R$ continuous in $r=0$ such that
:
\begin{equation}
f(x)\simeq f(r\theta)=r^k (f_k(\theta)+R(r,\theta)).
\end{equation}
If $\theta_0\notin C(f_k)$ and $\theta$ is close to $\theta_0$  we
simply choose :
\begin{gather*}
(y_{2},...,y_{n})(r,\theta) =(\theta _{1},...,\theta _{n-1}), \\
y_{1}(r,\theta)=r|f_{k}(\theta )+R(r,\theta)|^{\frac{1}{k}}.
\end{gather*}
In these coordinates the phase becomes $y_{1}^{k}$ if
$f_{k}(\theta_0)$ is positive (resp. $-y_{1}^{k}$ for a negative
value) and the Jacobian satisfies $|J y|(0,\theta )=|f_{k}(\theta
)|^{\frac{1}{k}}\neq 0$ locally. Now, let $\theta_0\in C(f_k)$. Up to a permutation, we can suppose that
$\partial_{\theta_{1}}f_{k}(\theta_{0})\neq 0$. We accordingly choose the
new local coordinates :
\begin{gather*}
(y_{1},y_{3},...,y_{n})(r,\theta)=(r,\theta
_{2},...,\theta _{n-1}), \\
y_{2}(r,\theta)=f_{k}(\theta)+R(r,\theta).
\end{gather*}
Since we have
$|Jy|(0,\theta_0)=|\partial_{\theta_{1}}f_{k}(\theta_{0})|\neq 0$,
lemma follows. \hfill{$\blacksquare$}
\begin{remark}\rm{These coordinates are
admissible and this leads to an adapted system of charts via a
partition of unity on $\mathbb{S}^{n-1}$ introduced below. Also
the condition $(H_1)$ insures the existence of a canonical measure on $C(f_k)$.}
\end{remark}
To use Lemma \ref{normal form} we introduce an adapted partition of
unity on $\mathbb{S}^{n-1}$. We pick cut-off functions
$\Psi_{j}\in C_0^{\infty}(\mathbb{S}^{n-1})$, $0\leq
\Psi_j \leq 1$, $\sum \Psi_j =1$ in a tubular neighborhood of $C(f_k)$,
with supports chosen so that normal forms of Lemma \ref{normal
form} exist, for $r$ small enough, in a conic neighborhood of
$\mathrm{supp}(\Psi_j)$. By compactness this set of functions can
be chosen finite and we obtain a partition of unity on
$\mathbb{S}^{n-1}$ by adding $\Psi_{0}=1-\sum \Psi_{j}$ to our
family. The support of $\Psi_{0}$ is not connected and we
define $\Psi_{0}^{+}$, with $f_k(\theta)>0$ on
$\mathrm{supp}(\Psi_{0}^{+})$, and similarly we define
$\Psi_{0}^{-}$ where $f_k<0$, so that
$\Psi_{0}=\Psi_{0}^{+}+\Psi_{0}^{-}$. If we accordingly split up
$I_s(z)$ we obtain :
\begin{gather*}
I_{s}^{\pm}(z)=\int\limits_{\mathbb{R\times R}_{+}\times
\mathbb{S}^{2n-1}} \Psi_{0}^{\pm}(\theta) g( zf(r\theta),r\theta )\Theta(r\theta)r^{n-1}drd\theta\\
=\int\limits_{\mathbb{R}_{+}} G_{0}^{\pm}(\pm z y_1^k, y_{1})dy_{1},
\end{gather*}
respectively for the directions where $f_k(\theta)>0$ and
$f_k(\theta)<0$, also :
\begin{gather*}
I_{s}^{0,j}(z)=\int\limits_{\mathbb{R\times R}_{+}\times
\mathbb{S}^{n-1}} \Psi_{j}(\theta) g(z f(r\theta) ,r\theta )\Theta(r\theta)r^{n-1}drd\theta \\
=\int\limits_{\mathbb{R}_{+}\times \mathbb{R}}
G_{j}(zy_1^ky_2, y_1,y_2)dy_1 dy_2,
\end{gather*}
for the set $C(f_k)$. The new symbols are
respectively given by :
\begin{gather}
G_{0}^{\pm}(t,y_{1})=\int y^{\ast
}(\Psi_{0}^{\pm}(\theta)g(t,r\theta
)\Theta (r\theta)r^{n-1}|Jy |)dy_{2}...dy_{n}, \label{ampli1}\\
G_{j}(t,y_{1},y_{2})=\int y ^{\ast }(\Psi_{j}(\theta)g(t,r\theta
)\Theta (r\theta)r^{n-1}|Jy |)d y_{3}...dy_{n}. \label{ampli2}
\end{gather}
Hence, the singular part of our integral can be written as a
finite sum :
\begin{equation}
I_s(z)=I_s^{-}(z)+I_s^{+}(z)+\sum\limits_{j} I_s^{0,j}(z),
\end{equation}
where each term of the r.h.s. will be treated by elementary
methods. Note that $I_s^{-}(z)$ and $I_s^{+}(z)$ can be treated as in the previous section.
\begin{remark}\label{degres amplitude}
\rm{Since $y_{1}(r,\theta ) =r$, our new symbols satisfy
$G_{j}(t,y_1,y_2)=\mathcal{O}(y_1^{n-1})$, near $y_1=0$. Since
the asymptotic expansion involves delta-Dirac distributions w.r.t $y_1$, cf. Lemma \ref{Theo IO 2eme carte} below,
the dimension will cause a shift in the expansion.}
\end{remark}
For $a\in C_0^{\infty}(\mathbb{R}\times \mathbb{R}_+\times\mathbb{R} )$, we define the family of elementary fiber integrals :
\begin{gather}
I_{n,k}(z)=\int\limits_{0}^{\infty }(\int\limits_{\mathbb{R}}a(zy_1^k y_2,y_1,y_2)dy_2)y_1^{n-1}dy_1.
\end{gather}
\begin{lemma}\label{Theo IO 2eme carte}
There exists a sequence of distributions $(D_{j,p})$ such that :
\begin{equation}
I_{n,k}(z)\sim \sum\limits_{p=0,1} \sum\limits_{j\in\mathbb{N},\text{ } j\geq n} D_{j,p}(a) z^{-\frac{j}{k}} \log(z)^p,
\text{ as } z\rightarrow \infty,
\end{equation}
where the logarithms only occur when $(j/k)$ is an integer. As concerns the leading term,
if $(n/k)\notin \mathbb{N}^{*}$ we obtain :
\begin{equation}\label{equation for I(nk)}
I_{n,k}(z)= z^{-\frac{n}{k}}d(a)+\mathcal{O}(z
^{-\frac{n+1}{k}}\mathrm{log}(z)),
\end{equation}
with :
\begin{gather*}
d(a)=C_{n,k} \int\limits_{0}^{\infty}\int\limits_{0}^{\infty} t^{\frac{n}{k}-1}
y_2^{n-\frac{n}{k}} \left( \partial^{n}_{y_2} a(t,0,y_2)+\partial^{n}_{y_2} a(-t,0,-y_2)\right) dy_2dt.
\end{gather*}
But when $n/k=p\in\mathbb{N}^{*}$, we have :
\begin{equation*}
I_{n,k}(z)= \frac{1}{k} z^{-\frac{n}{k}}\log(z)
\int\limits_{\mathbb{R}} |t|^{p-1} \partial_{y_2}^{p-1} a(t,0,0)dt
+\mathcal{O}(z^{-\frac{n}{k}}).
\end{equation*}
\end{lemma}
\begin{remark}\rm{The remainder of Eq. (\ref{equation for I(nk)}) can be optimized to $\mathcal{O}(z
^{-\frac{n+1}{k}})$ when $(n+1)/k$ is not an integer, as shows the proof below.}
\end{remark}
\noindent\textit{Proof.} By a standard density argument we can assume that the amplitude is of the form
$a(s,y_1,y_2)=f(s)b(y_1,y_2)$. The justification is that our coefficients below are computed by
continuous linear functionals, i.e. distributions. We define the Melin transforms of $f$ as :
\begin{equation}
M_{\pm}(\xi)= \int\limits_{0}^{\infty} s^{\xi-1} f(\pm s)ds.
\end{equation}
We split-up $I_{n,k}$ as $I_{+}$ and $I_{-}$ by separating integrations $y_2>0$ and $y_2<0$. Via Melin's
inversion formula, we accordingly obtain :
\begin{equation}\label{analytic1}
I_{+}(z)=
\frac{1}{2i\pi }\int\limits_{\gamma} M_{+}(\xi) z^{-\xi}
\int\limits_{\mathbb{R}_{+}^2} (y_1 y_2^{k})^{-\xi} b(y_1,y_2) y_1^{n-1} dy_1dy_2 d\xi,
\end{equation}
where $\gamma=c+i\mathbb{R}$ and $0<c<k^{-1}$. Similarly we have :
\begin{equation}\label{analytic2}
I_{-}(z)=
\frac{1}{2i\pi }\int\limits_{\gamma} M_{-}(\xi) z^{-\xi}
\int\limits_{\mathbb{R}_{+}^2} (y_1 y_2^{k})^{-\xi} b(y_1,-y_2) y_1^{n-1} dy_1dy_2 d\xi,
\end{equation}
The existence of a full asymptotic expansion is a direct consequence of :
\begin{lemma}\label{poles extensions}
The family of distributions $\xi\mapsto (y_1 y_2^{k})^{-\xi}$ on $C_0^{\infty}(\mathbb{R}_{+}^2)$ initially defined
in the domain $\Re(\xi)< k^{-1}$ is meromorphic on $\mathbb{C}$ with poles :
$\xi_{j,k} = j/k$, $j\in\mathbb{N}^{*}$. These poles are of order 2 when $\xi_{j,k}\in \mathbb{N}^*$ and of order 1
otherwise.
\end{lemma}
\textit{Proof.} We form the Bernstein-Sato polynomial $b_k$ attached to our problem :
\begin{gather*}
T(y_2y_1^{k})^{1-\xi}:=\frac{\partial}{\partial y_2}
\frac{\partial^{k}}{\partial y_1^{k}}
(y_2y_1^{k})^{1-\xi}=b_k(\xi)(ty_2y_1^{k})^{-\xi}, \\
b_k(\xi)=(1-\xi)\prod\limits_{j=1}^{k}(j-k\xi).
\end{gather*}
If $\Re(\xi)<k^{-1}$, $(k+1)$-integrations by parts yield :
\begin{equation*}
\int\limits_{\mathbb{R}_{+}^2} (y_1 y_2^{k})^{-\xi} f(y_1,y_2)dy_1dy_2=\frac{(-1)^{k+1}}{b_k(\xi)}
\int\limits_{\mathbb{R}_{+}^2} (y_1 y_2^{k})^{1-\xi} (Tf)(y_1,y_2)dy_1dy_2.
\end{equation*}
Now the integral in the r.h.s. is analytic in $\Re (\xi)<1+k^{-1}$. After $m$ iterations
the poles, with their orders, can be read off the rational functions :
\begin{equation}
\mathfrak{R}_m(\xi) =\prod\limits_{p=1}^m \frac{1}{b_k(\xi-p)}.
\end{equation}
This gives the result since $m$ can be chosen arbitrary large. $\hfill{\blacksquare}$
\begin{remark}\rm{Starting from Eqs. (\ref{analytic1},\ref{analytic2}) it is possible
to obtain an asymptotic expansion as $z\rightarrow 0^{+}$.
By shifting the path $\gamma$ to the left, only the poles of the Melin transforms
contribute and we obtain an expansion in power of $z$.
This can be predicted by a Taylor expansion of the associated oscillatory integral but the improvement
is that one can establish sharp integral remainders.}
\end{remark}
Hence, the following functions are meromorphic on $\mathbb{C}$ :
\begin{equation}
\mathfrak{g}^{\pm}(\xi)=\int\limits_{\mathbb{R}_{+}^2} (y_1 y_2^{k})^{-\xi} b(y_1,\pm y_2)dy_1dy_2.
\end{equation}
A classical result, see e.g. \cite{Bl-Hand}, is that $M_{\pm}(c+ix)\in \mathcal{S}(\mathbb{R}_x)$
when $c \notin -\mathbb{N}$. If we shift the path of integration $\gamma$ to the right in our
integral representation, Cauchy's residue method provides the asymptotic expansion.
In fact for any $d>c$, outside of the poles, we have :
\begin{gather*}
I_{+}(z)=\int\limits_{c+i\mathbb{R}} z^{-\xi} M_{+}(\xi) \mathfrak{g}^{+}(\xi) d\xi\\
= \sum\limits_{c<\xi_{j,k}<d} \mathrm{res}(z^{-\xi}M_{+}\mathfrak{g})(\xi_{j,k})+
\int\limits_{d+i\mathbb{R}} z^{-\xi} M_{+}(\xi) \mathfrak{g}^{+}(\xi) d\xi.
\end{gather*}
Since $d$ is not a pole the last integral can be estimated via :
\begin{equation}
|\int\limits_{d+i\mathbb{R}} z^{-\xi} M_{+}(\xi) \mathfrak{g}^{+}(\xi) d\xi|\leq C(f,b) z^{-d}=\mathcal{O}(z^{-d}),
\end{equation}
where, for each $d$, the constant $C$ involves the $L^1$-norm of a finite number derivatives of $b$.
This will indeed lead to an asymptotic expansion with precise remainders.
Applying this method to $I_{+}(z)$ and $I_{-}(z)$ we obtain the existence of a full asymptotic expansion of the form :
\begin{equation}
I(z) \sim \sum\limits_{p=0,1} \sum\limits_{j\in\mathbb{N}^*} C_{j,p} z^{-\frac{j}{k}} \log(z)^p.
\end{equation}
Moreover, by Lemma \ref{poles extensions}, these logarithms only occur when $j/k$ is integer.\medskip\\
Now, we compute the leading term of this expansion for our particular problem. To avoid unnecessary discussions and
calculations below, we remark that we can commute the polynomial weight via :
\begin{gather}
T ((y_1 y_2)^{1-\xi} y_1^{n-1})=\mathfrak{b}(\xi)(y_2y_1^{k})^{-\xi}y_1^{n-1}, \\
\mathfrak{b}(\xi)=(1-\xi)\prod\limits_{j=1}^{k}(j-k\xi+n-1).
\end{gather}
By iteration, we obtain that the poles are the rational numbers :
\begin{equation*}
\xi_{p,j,k,n}=p+\frac{j+n-1}{k},\text{ } j\in[1,...,k], \text{ } p\in\mathbb{N}.
\end{equation*}
For all $\alpha-\beta>1$, $\beta\in\mathbb{N}^{*}$, we have :
\begin{equation*}
\int\limits_{0}^{\infty} \partial_r^{\beta}  (r^\alpha f(r)dr) =0, \text{ }\forall f\in C_{0}^{\infty}.
\end{equation*}
If we apply this to the integral w.r.t. $y_1$, there is no contribution before :
\begin{equation}
\xi_{0}= \frac{n}{k}.
\end{equation}
To compute the first effective residue we must distinguish out
the case where $\xi_0$ is an integer or not. The optimal number of iterations to reach $\xi_0$
is $\mathrm{E}(n/k)+1$ but, by analytic continuation, any number bigger than this one is acceptable.
A fortiori we can use $n$ iterations and our starting point will be :
\begin{equation}
z^{-\xi}M_{+}(\xi) (-1)^{n(k+1)} \mathfrak{B}_n (\xi)
\int\limits_{\mathbb{R}_{+}^{2}} (y_1^k y_2)^{n-\xi} y_1^{n-1} T^n b(y_1,y_2) dy_1dy_2,
\end{equation}
with :
\begin{equation}
\mathfrak{B}_n(\xi)=\prod\limits_{l=0}^{n-1} \frac{1}{\mathfrak{b}(\xi-l)}.
\end{equation}
\textbf{Case of $\xi_0$ simple pole.}\\
In this case our residue is simply given by :
\begin{equation}
C z^{-\frac{n}{k}} M_{+}(\frac{n}{k})
\int\limits_{\mathbb{R}_{+}^{2}} (y_1^k y_2)^{n-\frac{n}{k}} y_1^{n-1} T^n b(y_1,y_2) dy_1dy_2,
\end{equation}
with :
\begin{equation}
C =\lim\limits_{\xi\rightarrow \frac{n}{k}} (-1)^{n(k+1)}(\xi-\frac{n}{k})\mathfrak{B}_n(\xi).
\end{equation}
In particular we can compute the integral w.r.t. $y_1$ via :
\begin{equation*}
\int\limits_{0}^{\infty} y_1^{kn-1} \partial^{kn}_{y_1}  (\partial^{n}_{y_2} b(y_1,y_2)) dy_1
=(-1)^{kn}(kn-1)!\partial^{n}_{y_2} b(0,y_2).
\end{equation*}
A similar result holds for $I_{-}$ and we obtain :
\begin{gather}
I_{+}(z)=z^{-\frac{n}{k}} C_{n,k}  M_{+}(\frac{n}{k})
\int\limits_{0}^{\infty} y_2^{n-\frac{n}{k}} (\partial^{n}_{y_2} b)(0,y_2)dy_2+R_1(z),\\
I_{-}(z)=z^{-\frac{n}{k}} C_{n,k}  M_{-}(\frac{n}{k})
\int\limits_{0}^{\infty} y_2^{n-\frac{n}{k}} (\partial^{n}_{y_2} b)(0,-y_2)dy_2+R_2(z).
\end{gather}
Here $C_{n,k}$ is the canonical constant :
\begin{equation}\label{canonical regularization}
C_{n,k}=\frac{1}{k} \prod\limits_{j=1}^n \frac{-1}{j-\frac{n}{k}}.
\end{equation}
Also, according to the analysis above, each remainder is of order $\mathcal{O}(z^{-\frac{n+1}{k}})$
if $(n+1)/k \notin \mathbb{N}$ and $\mathcal{O}(z^{-\frac{n+1}{k}}\log(z))$
otherwise.\medskip\\
\textbf{Case of $\xi_0$ double pole.}\\
If $h$ is meromorphic with a pole of order 2 in $\xi_0$ we have :
\begin{equation*}
\mathrm{res} (h)(\xi_0)=\frac{1}{2} \lim\limits_{\xi\rightarrow \xi_0} \frac{\partial}{\partial \xi}
(\xi-\xi_0)^2 h(\xi).
\end{equation*}
Applying this principle to our residue we obtain, via Leibnitz's rule, that :
\begin{equation}
I_{+}(z)=B \log(z)z^{-\frac{n}{k}} +\mathcal{O}(z^{-\frac{n}{k}}).
\end{equation}
We can compute the distribution $B$ as before and we find :
\begin{gather*}
B=-\frac{1}{2} D_{n,k} M_{+}(\frac{n}{k})
\int\limits_{0}^{\infty} y_2^{n-\frac{n}{k}} (\partial^{n}_{y_2} b)(0,y_2)dy_2,\\
D_{n,k}=(-1)^{n(k+1)} \lim\limits_{\xi \rightarrow \frac{n}{k}} (\xi-\frac{n}{k})^2 \mathfrak{B}_n(\xi).
\end{gather*}
Since $p=n/k$ is an integer, by integration by parts we obtain :
\begin{equation}
\int\limits_{0}^{\infty} y_2^{n-p} (\partial^{n}_{y_2} b)(0,y_2)dy_2
=(-1)^{n-p+1}(n-p)!  \partial_{y_2}^{p-1} b(0,0).
\end{equation}
Since a similar result holds for $I_{-}$, we obtain the desired result by gathering all the constants and summation.
Finally, we can extend our formulas since all coefficients in the expansion are of the form :
\begin{equation*}
\left\langle T^j ,f\otimes b\right\rangle=\left\langle T^j_1,f\right\rangle \left\langle T^j_2,b\right\rangle,
\text{ }T^j_{1,2}\in \mathcal{D}'.
\end{equation*}
By linearity and continuity, the result holds for a symbol $a(t,y_1,y_2)$.
\hfill{$\blacksquare$}\medskip\\
Taking Remark \ref{degres amplitude} into account, to avoid
unnecessary calculations we define :
\begin{gather}
G_{0}^{\pm}(t,y_1)=y_1^{n-1}\tilde{G}_{0}^{\pm}(t,y_1),\\
G_j(t,y_1,y_2)=y_1^{n-1}\tilde{G}_{j}(t,y_1,y_2).
\end{gather}
\textbf{Directions where $f_k(\theta)\neq 0$.}\\
By Lemma \ref{lemma for extremum}, the first non-zero coefficient, obtained for $l=n-1$, is :
\begin{equation*}
\frac{z^{-\frac{n}{k}}}{k}\frac{1}{(n-1)!}\left\langle
t_{+}^{\frac{n-k}{k}}\otimes \delta
_{0}^{(n-1)},G_{0}^{+}(t,y_{1})\right\rangle =\frac{z^{-\frac{n}{k}}}{k}\int\limits_{\mathbb{R}}
t_{+}^{\frac{n-k}{k}}
\tilde{G}_{0}^{+}(t,0)dt.
\end{equation*}
By construction, we have :
\begin{equation*}
\tilde{G}_{0}^{+}(t,0)=\int\limits_{\mathbb{S}^{n-1}}g(t,0)\Psi_{0}^{+}(\theta)|f_{k}(\theta
)|^{-\frac{n}{k}}d\theta.
\end{equation*}
A similar computation gives the contribution of
$\mathrm{supp}(\Psi_{0}^{-})$, and we obtain :
\begin{gather} \label{positive}
I_s^{+}(z)= z^{-\frac{n}{k}} \left\langle
t_{+}^{\frac{n-k}{k}}, g(t,0) \right\rangle\frac{1}{k}
\int\limits_{\mathbb{S}^{n-1}}\Psi_{0}^{+}(\theta)|f_{k}(\theta
)|^{-\frac{n}{k}}d\theta+\mathcal{O} (z^{-\frac{n+1}{k}}),\\
\label{negative}%
I_s^{-}(z) = z^{-\frac{n}{k}} \left\langle
t_{-}^{\frac{n-k}{k}}, g(t,0) \right\rangle
\frac{1}{k}\int\limits_{\mathbb{S}^{n-1}}\Psi_{0}^{-}(\theta)|f_{k}(\theta
)|^{-\frac{n}{k}}d\theta +\mathcal{O} (z^{-\frac{n+1}{k}}).
\end{gather}
\textbf{Microlocal contribution of $C(f_k)$.}\newline%
Here we examine the contribution of terms $I_{s}^{0,j}(z)$. According to the analysis above,
we must distinguish out the case $k$ divides $n$.\medskip\\
\textit{a) Case of $k>n$, non-integrable singularity on $\mathfrak{S}$.}\\
Here $n/k \in ]0,1[$, so that the singularity on the blow-up is integrable. Via Lemma \ref{Theo IO 2eme carte}, the contribution of $I_{s}^{0,j}(z)$
is given by :
\begin{equation*}
\frac{1}{k}z^{-\frac{n}{k}}\int\limits_{\mathbb{R}_{+}^2} |t|^{\frac{n}{k}-1}
|y_2|^{-\frac{n}{k}} \left(\tilde{G}_{j}(t,0,y_2)+\tilde{G}_{j}(-t,0,-y_2)\right)dt dy_2
+\mathcal{O}(z^{-\frac{n+1}{k}}\mathrm{log}(z)).
\end{equation*}
Reminding that $y_2(t,0,\theta)=f_k(\theta)$, we obtain :
\begin{equation*}
\int\limits_{\mathbb{R}_{+}}
|y_2|^{-\frac{n}{k}} \tilde{G}_{j}(t,0,y_2)dy_2=g(t,0)\int\limits_{\{f_k(\theta)\geq0\}}|f_k
(\theta) |^{-\frac{n}{k}}\Psi_j(\theta) d\theta.
\end{equation*}
Since $\Psi_0^{\pm}$, $\Psi_j$ is a partition of unity on $\mathbb{S}^{n-1}$, by summation of all local
contributions $I_s(z)$ is asymptotically equivalent to :
\begin{equation*}
\frac{z^{-\frac{n}{k}}}{k}  \left ( \left\langle
t_{+}^{\frac{n}{k}-1},g(t,0) \right\rangle
\int\limits_{\{ f_k \geq 0\}} |f_k
(\theta) |^{-\frac{n}{k}}d\theta
+ \left\langle t_{-}^{\frac{n}{k}-1},g(t,0)\right\rangle
\int\limits_{\{ f_k \leq 0\}} |f_k (\theta)
|^{-\frac{n}{k}}d\theta \right ).
\end{equation*}
Note that none of these coefficients are
equal unless $g$ or $f_k$ are symmetric. $\hfill{\blacksquare}$\medskip\\
\textit{b) Case of $p=n/k$ integer.}\\
Here the contribution of $I_{s}^{0,j}(z)$ is dominant and we obtain :
\begin{equation*}
I_{s}^{0,j}(z)\sim \frac{1}{k}\log (z)z^{-p}
\int\limits_{\mathbb{R}} |t|^{p-1} \partial^{p-1}_{y_2}
\tilde{G}_j(t,0,0)dt+\mathcal{O}(z^{-p}).
\end{equation*}
Unless $p=1$, there is no way to take the limit directly, and the geometric properties are still hidden in the Jacobian.
But we will reach the result by the Schwartz kernel technic. Clearly, it is enough to evaluate our derivative
and to integrate w.r.t. $t$. With $s=(s_1,s_2)\in\mathbb{R}^2$, we write the evaluation as :
\begin{equation*}
\partial^{p-1}_{y_2} \tilde{G}_j(t,0,0)=\frac{1}{(2\pi)^2} \int e^{i\left\langle
s,(y_1,y_2)\right\rangle} (is_2)^{p-1}
\tilde{G}_j(t,y_1,y_2) dy_1dy_2 ds.
\end{equation*}
Here we have used an oscillatory Schwartz kernel for $\delta_{y_1} \otimes \delta^{p-1}_{y_2}$.
This integral representation allows to inverse our diffeomorphism to obtain :
\begin{equation*}
\partial^{p-1}_{y_2}
\tilde{G}_j(t,0,0)=\frac{1}{(2\pi)^2} \int e^{i\left\langle
s,(r,y_2(r,\theta))\right\rangle} (i s_2)^{p-1} t^{p-1}
g(t,r\theta) \Psi_j(\theta) dtdr d\theta ds.
\end{equation*}
Extending the integrand by 0 for $r<0$, the normalized integral w.r.t. $(r,s_1)$ provides $\delta_r$.
By construction $y_2(0,\theta)=f_k(\theta)$, hence :
\begin{equation}
\partial^{p-1}_{y_2}
\tilde{G}_j(t,0,0)=g(t,0)\frac{1}{(2\pi)} \int\limits_{\mathbb{R}\times\mathbb{S}^{n-1}}
e^{i u f_k(\theta)} (i u)^{p-1} \Psi_j(\theta) d\theta du .
\end{equation}
This Fourier integral makes sense with $\mathbb{S}^{n-1}$ compact. We recall the density :
\begin{equation}
J_j(w)=\int \limits_{\{f_k(\theta)=w\}} \Psi_j(\theta) dL_w(\theta),
\end{equation}
where $dL_w$ is the density induced by the Leray-form $dL_{f_k}$ : $d f_k \wedge dL_{f_k}(\theta)=d\theta$.
Note that all these objects can be constructed by mean of local coordinates under the only condition
that $\mathrm{supp}(\Psi_j)$ is small enough near $C(f_k)$. Since $f_k$ is continuous on $\mathbb{S}^{n-1}$,
each $J_j(w)$ is compactly supported and smooth near the origin. The sum
over all the $\Psi_j$ gives the geometric contribution :
\begin{equation}
\frac{1}{(2\pi)} \int\limits_{\mathbb{R}^2} e^{i uw} (i u)^{p-1} \sum\limits_j J_j(w) dwdu=\frac{d^{p-1}\mathrm{Lvol}}{d w^{p-1}}(0).
\end{equation}
By integration w.r.t. $t$ we obtain the result for a double pole in general position.
\begin{remark}
\rm{The case $n=k$ is directly accessible. In this case we obtain :
\begin{equation*}
I_s(z)=\frac{1}{k} \frac{\log(z)}{z} \mathrm{LVol}(0)\int\limits_{\mathbb{R}} g(t,0)dt+\mathcal{O}(z^{-1}).
\end{equation*}
Here $\mathrm{LVol}(0)$ is the Liouville volume of $C(f_k)$. Note that $I_s$ dominates $I_r$.}
\end{remark}
\textit{c) $k<n$ and simple pole, integrable singularity on $\mathfrak{S}$.}\\
Finally, we treat the case of a simple pole with a non-integrable singularity on $\mathbb{S}^{n-1}$.
For the positive part, the distributional coefficients are :
\begin{equation}
\left \langle \mu^j_{+}, g\right\rangle = C_{n,k} \int\limits_{0}^{\infty} \int\limits_{0}^{\infty}
t^{\frac{n}{k}-1} y_2^{n-\frac{n}{k}} (\partial^{n}_{y_2} \tilde{G}_j)(t,0,y_2)dy_2dt.
\end{equation}
Clearly, we can use the same globalization technic as above.
The sum of all $\mu^j_{+}$, completed with the main term of $J_s^{+}$, provides :
\begin{equation}
\left\langle T_{+}, g\right\rangle =C_{n,k}\int\limits_{0}^{\infty}|t|^{\frac{n}{k}-1}g(t,0)dt
\int\limits_{0}^{\infty} w^{n-\frac{n}{k}} \frac{ \partial^{n} L(w)}{\partial w^n}  dw,
\end{equation}
where the integral w.r.t. $w$ is absolutely convergent since the measure $L(w)$ is compactly supported.
A similar result holds for the directions where $f_k(\theta)<0$
and we obtain the result stated in Theorem \ref{main2}.$\hfill{\blacksquare}$\medskip\\
Now we detail the construction of the distributional bracket of the part c) of Theorem \ref{main2}.
Let be $\chi\in C_0^{\infty}$, $0\leq \chi \leq 1$ on $\mathbb{R}$, chosen such that
$\chi=1$ near the origin and $\chi(u)=0$ for $|u|\geq \varepsilon$. If $\varepsilon>0$ is small enough,
$\mathrm{Lvol}$  is smooth on $]-\varepsilon, \varepsilon[$.
We write the geometric contribution as :
\begin{equation*}
\left\langle T,\mathrm{Lvol} \right\rangle= \left\langle T, \chi \mathrm{Lvol} \right\rangle
+\left\langle T, (1-\chi)\mathrm{Lvol}  \right\rangle
\end{equation*}
Away from the origin, e.g. for $u>0$, we obtain directly :
\begin{equation}
C_{n,k}\left\langle \frac{d^n}{du^n} u_{+}^{n-\frac{n}{k}}, (1-\chi)(u)\mathrm{Lvol}(u) \right\rangle
=\frac{1}{k}\int\limits_{\{f_k(\theta) >0\}} (1-\chi)(f_k(\theta))|f_k(\theta)|^{-\frac{n}{k}}d\theta.
\end{equation}
Note that Eq. (\ref{canonical regularization}) for $C_{n,k}$ justifies the
normalization of Eq. (\ref{normalized derivatives}).
On $\mathrm{supp}(\chi)$, we use the local regularity of $\mathrm{Lvol}(u)$ near $u=0$ and
integrations by parts to conclude. Finally, to obtain a totally rigorous treatment, we remark that
since $C(f_k)$ is compact and $f_k$ is continuous we can choose our partition of unity such that $\sum\Psi_j=1$ for
$|f_k(\theta)|\leq 2\varepsilon$. $\hfill{\blacksquare}$
\begin{remark}\rm{The key point here is that we can put in duality the
distributions $\partial^n_u |u|_{\pm}^{n-\alpha}$ and $\mathrm{Lvol}(u)$ since their singular supports are disjoints.}
\end{remark}
\section{Elementary examples.}
\textbf{A family of extremum.} If $p$ is an even integer and $a=(a_1,...a_n)\in (\mathbb{R}_{+}^{*})^n$,
we consider a $n$ dimensional integral obtained by integration in the fiber of :
\begin{equation}
f_{a,p}(x)=\sum \limits_{j=1}^n a_j x_j^p
\end{equation}
For example :
\begin{equation*}
\int\limits_{\mathbb{R}^n} e^{-z f_{a,p}(x)}dx=
\prod\limits_{j=1}^{n} (\int\limits_{\mathbb{R}} e^{-z a_j x^p} dx)=(2 \Gamma
(1+\frac{1}{p}))^n z^{-\frac{n}{p}}\prod\limits_{j=1}^{n}a_j^{-\frac{1}{p}},
\end{equation*}
with $g(t,x)=e^{-t}$ symbol of order 0 on $\mathbb{R}^n$. Since we have here a local minimum, the
distributional factor is simply :
\begin{equation}
\int\limits_{0}^{\infty} t^{\frac{n-p}{p}} e^{-t}dt= \Gamma(n/p).
\end{equation}
The Jacobian of the standard polar coordinates gives the value of the generalized elliptic
integrals :
\begin{equation} \label{elliptic}
\int\limits_{\mathbb{S}^{n-1}}
(f_{a,p}(\theta))^{-\frac{n}{p}}d\theta=\frac{1}{p}\frac{(2 \Gamma
(1+\frac{1}{p}))^n}{\Gamma(n/p) } \prod\limits_{j=1}^{n}a_j^{-\frac{1}{p}}.
\end{equation}
Note that Eq. (\ref{elliptic}) can be analytically continued with a phase $\sum a_j|x_j|^\alpha$, $\alpha>0$.\medskip\\
\textbf{Conical singularities.} In 2 dimensions
let be :
\begin{equation*}g(t,u)=\frac{e^{-||u||^2}}{(1+t^2)}.
\end{equation*}
With $f(x,y)=x^2-y^2$ we have $\mathfrak{S}=\{(x,x)\} \cup \{(-x,x)\}$. $\mathfrak{S}$ is not compact but
the exponential decrease of $g$ will compensate. Accordingly let be :
\begin{equation}
I(z)=\int\limits_{\mathbb{R}^2} \frac{e^{-(x^2+y^2)}}{1+z^2
(x^2-y^2)^2}dxdy.
\end{equation}
Since $n/k=1$, we expect a logarithm for the leading term. Since :
\begin{equation*}
\int\limits_0^{2\pi} \frac{d\theta}{1+\alpha \cos^2
(2\theta)}=\frac{2\pi}{\sqrt{1+\alpha}},\text{ } \forall \alpha
>-1,
\end{equation*}
by passage in polar coordinates we obtain :
\begin{equation*}
I(z)=2\pi \int\limits_{0}^{\infty} e^{-r^2} \frac{rdr}{\sqrt{1+z^2
r^4}}=\pi \int\limits_{0}^{\infty} e^{-u} \frac{du}{\sqrt{1+z^2
u^2}} \sim \pi \frac{\log(z)}{z}.
\end{equation*}
Always in dimension 2, a singularity of degree 4 will generally not give a logarithmic leading term.
For example, we have :
\begin{equation*}
\tilde{I}(z)=\int\limits_{\mathbb{R}^2}\frac{e^{-(x^2+y^2)}}{1+z^2(x^4-y^4)^2}dxdy
=\pi\int\limits_{0}^{\infty} e^{-u} \frac{du}{\sqrt{1+z^2
u^4}} \sim \frac{\pi}{\sqrt z} \int\limits_{0}^{\infty} \frac{du}{\sqrt{1+u^4}}.
\end{equation*}

redpajama_set_name: RedPajamaArXiv


text:
Q: php code coverage on request execution I am working on a PHP migration project which is very complex and huge .
I want to know is there any tool which help me to see line of codes got executed when one page is rendered on the browser. This will help me to find codes in use and check business logics & migrate the one required.
Like code coverage report help us in unit test. Is there something available to know code executed when a particular request executed.

A: If you enable XDebug profiler you can use that to get code coverage data and use that later to identify unused code. Please find more details here: http://xdebug.org/docs/code_coverage
If you need to get that formatted in a nicer way you can use PHP-Code-Coverage lib that PHPUnit is using to generate the reports. You can find an example how to us it here: https://github.com/sebastianbergmann/php-code-coverage

redpajama_set_name: RedPajamaStackExchange


text:
Debbie: Breast Cancer Survivor
Dedicated to tomorrow's new beginning
Debbie Ellis, a breast cancer survivor with a stressful job, has been a smoker for over 20 years. Since 1994, Debbie has used cigarettes as a way to relieve her stress. And although she did quit once, for almost four years, she relapsed.
At one of her follow-up visits, her oncologist asked her an important question. "Would you like help to stop smoking?"
When Debbie answered "yes", she was referred to the Tobacco Treatment Program, a program offered to all patients treated at Smilow Cancer Hospital and its Care Centers throughout Connecticut.
The Tobacco Treatment Program helps patients quit through evidence-based treatment options, including medications and behavioral strategies. It combines counseling and drug treatment tactics and is structured to meet the individual needs of each patient.
"I was assigned to Margaret, a counselor who works for Smilow," noted Debbie. "She has been just fabulous. She has helped me so much."
The program works by identifying what triggers a person to want a cigarette and includes a nurse practitioner who works with patients individually. Debbie identified her biggest trigger as stress from her job.
"I work for the State of Connecticut as a court monitor. All day, I monitor and record courtroom proceedings and prepare transcripts to maintain a record of the proceedings. I sometimes need to understand the complex terminology and be able to produce overnight and expedited transcripts when requested. Most days there are over 200 cases on a docket per day. My job is often stressful and demanding."
"Margaret made me look at other times I wanted a cigarette … times that I was most in need, such as after dinner and first thing in the morning. Also places that were triggers, such as when I'm driving in my car."
"Then, Margaret had me write down some ideas of what I could do, both at work and at home, instead of having the cigarette. For instance, sucking on a sugar-free candy, licking a lollypop, or using a stress ball or a toothpick. Margaret gave me a small prayer shawl I could keep in my pocket, knowing it is there for me to hold onto when I am feeling anxious has helped me a lot."
"We did all of this at Smilow … it's such a useful program for the Hospital to offer for free as part of my follow-up care."
The Program at Smilow operates on the principles of the U.S. Public Health Service Clinical Practice Guidelines. All treatment components are evidence-based. All patients are treated with FDA approved first line medications for smoking cessation and smoking cessation counseling that stresses appropriate coping skills. The program utilizes the appropriate treatment based on each patient's individual situation and history with smoking.
"The most important thing I needed to do," noted Debbie, "was to set a date for quitting. I procrastinated quite a bit. We talked about my lungs, how they worked and what they did. That was very informative for me – to understand what I was doing to my body. With Margaret's help I finally set my Quit date – March 1, 2019."
"I am so proud that I made that date. I fulfilled my own pledge by quitting on March 1. I continue to take Chantix (a prescription smoking cessation aid), twice a day, morning and night. It has some side effects like weird dreams, but I can deal with those. What is most important is that it continues to help me deal with my stress."
The program also offers follow-up messages to help patients after they have quit.
"I get text messages to check in and ask me how I'm doing," said Debbie. "If I say I have had any problem, such as that I am feeling stressed, I get an immediate text back with ideas of what I can do to deal with it. Those texts have been so useful to me, keeping me on the right path."
"I can't say enough about how this program has helped me. The people involved are so generous with their time and so supportive to all of us. And for Smilow to offer this important program as part of my follow-up care has been a real blessing for me. It has changed my life and I am forever grateful."
{"namedChunks":["modules-survivor-story-details-components-SurvivorStoryDetails-survivorStoryDetails"]}

redpajama_set_name: RedPajamaCommonCrawl


Contemporary villa with splashing marom architects that highlight the modern luxury villas they design home and interior ideas, this modern villa the device is designed with a wealth of ideas and keen sense design interior colors decor contemporary. Find and save ideas about modern interior design on see more home bedroom.
Modern villa design ideas home decorating structure style interior contemporary, modern living room design ideas brooklyn based interior blog as she included a vintage rug in her chic and, find and save ideas about villa design on see more modern houses villas. X modern villa design a beautiful house that optimizes on the location style and creative contemporary interior ideas, join our million followers in being up to date on architecture and interior design this contemporary has a ideas. Interior design apartments if you plan to build or rearrange your current villa can find many creative ideas how contemporary. Find the best modern home design ideas but just what constitutes interior less is more when it comes to contemporary so don.


Walking is a complimentary privilege to Full members only.
Sports members receive 5 rounds per calendar year at cart fee of $20/round.
Clubhouse members receive 3 rounds per calendar year at cart fee of $20/round.
*Cart packages available for full members.
Call the golf shop for weekly specials, events, and outing packages!


text:
Home / News / Citizens uninspired by new money
Citizens uninspired by new money
MOST Zimbabweans feel the new currency which will start circulating today will not change anything as long as government has not implemented economic and political reforms to generate market confidence.
BY EVERSON MUSHAVA
The Reserve Bank of Zimbabwe (RBZ) is today introducing $5 and $2 notes, as well as the $2 bond coin as part of measures to ease cash shortages and bring back the local currency, which was scrapped in 2009 and replaced by a multi-currency regime.
The new notes look exactly the same as the old bond notes with added security features, while the $2 bond coin looks like the $1 bond coin already in circulation.
But Garikai Mbudaya, a Warren Park resident, said although the new notes would increase liquidity, it would likely increase inflation like it used to happen during the hyperinflationary era under former RBZ governor Gideon Gono.
"This will not change anything much because as long as there is no confidence in the government, there will be hardly any change in the market," Mbudaya said.
"The biggest losers will, however, be the EcoCash agents who had been selling money at a premium. The new currency will push the premium down a bit, but not for long."
Commenting on the similarity of the new notes to the bond notes already in circulation, Mbudaya said that was the manifestation of the confusion in President Emmerson Mnangagwa's government.
He said whether by design or accident, the similarities on the new and old currency would not help inspire confidence, describing the new money as a packet of hot air to replace a bottle of the same.
Joseph Mlambo, a Bikita resident, said the currency being introduced today would hardly change anything as far as the country's economic situation is concerned.
"The new currency, which will not have market confidence, is not what the economy needs, therefore, it cannot be a solution," the 23-year-old Mlambo said.
"Since the currency will be almost identical to the bonds that are already in circulation, it won't induce a mental difference to the currency users. So this implies that the currency will as well suffer from the ongoing wave of inflation."
A resident of Harare's Epworth satellite town, Vivian Mwayera, said she does not care whether the country introduces new money today or tomorrow, she only wants Mnangagwa to fix the economy.
"We are suffering. We will not eat the new money. We want the President to address the economy first and then introduce the new money. We were good under the US dollar regime than now," Mwayera said.
Political analyst Ernest Mudzengi said Zimbabwe does not need a new currency, but economic and political reforms to boost confidence.
"We need the economy to produce, agriculture and manufacturing sectors must produce so that we can export. Printing of money will not help, it is not a panacea, and we need economic and political reforms and address the issue of confidence. Things like the land issue of land should be addressed once and for all," he said.
Mudzengi said the similarities and differences of the new money would be inconsequential.
An economist, who refused to be named because his company does not allow him to speak to the media, said: "The new currency will not survive simply because it is not backed by anything, apart from policies from an untrusted government. It will not survive. The denomination is too small for the prices in the country. The biggest note is $5 and yet bread costs $15."
Last week, speakers at the Institute of Chartered Accountants summer school held in Harare, accused government of fuelling economic instability through inconsistent currency reforms and too many regulations that do not inspire confidence.
Tafadzwa Bandama, the Confederation of Zimbabwe Industries chief economist, said the plethora of legislation rolled out by the government was causing confusion in the economy.
MMC Capital executive director Itai Chirume weighed in, saying money creation by the government poisoned the investment environment in the country.
In 2016, the RBZ introduced bond notes whose value was at par with the US dollar until October last year when the currency reforms started.
Since then, the government has issued a number of statutory instruments as it navigates the transition from a multi-currency system to the mono-currency regime.
These new policies have, however, failed to take the country out of the economic woods.
Previous Machete wars claim 105 lives: Report
Next 'Technology fuelling child sexual abuse'
2020 year of fighting slavery: ZCTU
Court overturns Mliswa defamation victory
AFM congregants clash in Marondera
BY JAIROS SAUNYAMA Scores of Apostolic Faith Mission (AFM) church worshippers in Marondera were yesterday …
wind: 3km/h NNE
Global study spotlights peace as driving force to achieve SDGs
SHDF rolls out 'green' plan for small enterprises
South Africa aim for whitewash
Senegal and Japan keep World Cup knockout hopes alive with 2-2 draw


This is an assortment of recording done, some electronic, some well recorded and some ideas.
Caroline Towers was a project of mine around 2014 I had a concept for an action movie rock show. I was intending to tour this as a rock musical. Other priorities meant this never got past concept stage. These recording represent the concept.
Caroline Towers , a violent tale set in the future where tower blocks are miles , no-one goes outside. The tower keeps building and people keep dying only the architect knows how to prevent the carnage…..


text:
Wilson Combat T.R.I.M. – Tactical Rail Interface, Modular
The T.R.I.M. (Tactical Rail Interface, Modular) .308 rail has been…
TR-TRIM-AR10-12-002
The T.R.I.M. (Tactical Rail Interface, Modular) .308 rail has been designed by the Wilson Combat engineering team to be among the lightest, strongest and most versatile compact free-floating rail systems on the market.
•	Designed for .308 AR-pattern rifles and clones with 1-7/16-18" threaded receiver (DPMS/SR-25)
•	Light, Strong and Compact Free-Floating Rail Design with small tube diameter for enhanced comfort and control
•	Machined from 6005A-T5 Aluminum Extrusion
•	Mil-Spec Hard Anodized Black Finish
•	Simple, 5-Minute Installation with Common Hand Tools
•	Easily Removed or Replaced in the Field
•	Quick Custom Positioning of Supplied Rail Segment
•	Large Internal Diameter Covers Low Profile Gas Blocks
•	Dual, Integral Push-Button Sling Attachment Points
•	Available in 12" ($349.95)and 14" ($359.95)
Machined of high-strength 6005A-T5 aircraft aluminum with a Mil-Spec hard anodized finish, the T.R.I.M. rail handguard interface and its robust mounting system does not sacrifice strength, stability or accuracy over bulkier designs. The oversized and tough nitrocarburized 4130 steel barrel attachment nut reduces heat transfer and the ergonomic handguard design offers a sleek, non-abrasive platform for comfortable dynamic shooting.
Light, strong and easy to install-the Wilson Combat T.R.I.M. rail interface solves multiple installation issues our gunsmiths have encountered with other rail systems in over a decade of building thousands of our custom AR rifles. The entire T.R.I.M. interface can be easily end-user installed in five minutes with common hand tools and can be quickly removed or replaced with a different length rail interface in the field with nothing more than a hex key.
The T.R.I.M. attachment points enable the end-user to quickly position the supplied modular Picatinny rail panel on the attached interface in pre-specified locations. A generous inside diameter of 1.69″ accepts low profile gas blocks. Integral push-button Q.D. sling swivel attachment points are located on both sides of the T.R.I.M. for the ultimate in sling mounting versatility.
For more information, visit WilsonCombat.com
Drone Strike Reportedly Kills Pakistani Taliban Leader
The T.R.I.M. (Tactical Rail Interface, Modular) .308 rail has been designed by the Wilson…
by Tactical-Life	/ May 29, 2013
The 5 Best MMA Fighters of All Time to Never...
VIDEO: Learn How to Deal with a Charging Opponent in...


MIRANSHAH: Eleven terrorists were killed and three of their hideouts destroyed on Wednesday as forces successfully continued Operation Zarb-e-Azb being conducted in North Wazirstan Agency.
Sources said that Pakistan Air force fighter jets in their precision attack last night at around 3:00 AM targeted three hideouts killing 11 terrorists at Zoi Dara Sayedgi, Tehsil Shawal in their latest offensive.
Pakistan on June 15 launched a military offensive, code named 'Operation Zarb-e-Azb' against militants in the North Waziristan region of Federally Administered Tribal Areas (FATA). The region said to be the hotbed of militancy in the country.


Well, it seems that our bobcat is showing just how happy he is to be in a cage. It's wonderful to see him getting feisty and tearing up his hospital cage because that means he is getting better every day.
If you have been keeping up with this story, then you know that Thor, the bobcat, was hit by a car straight on to his face. His injuries were almost as incredible as his rescue. Upper and lower jaws, shoulder blade and eye socket were all broken and it is still uncertain if they can save his eye. The wonderful thing is that he is getting better, which makes his dangerous capture (to both him and his rescuers) worth it.
While he is still healing, and his prognosis does look good, this is not always the case. In fact, another bobcat was taken in shortly after Thor and nicknamed, Poseidon. As Big Cat Rescue was cautiously optimistic about Thor's slow progress, Poseidon was a different matter. X-rays were negative, but he was covered in mange which is a skin disease caused by parasitic mites, and can be contagious to humans. He had also been badly attacked by another animal, and he had a BB lodged under his skin, which meant he had been shot. He was severely dehydrated among other issues.
Initially, Poseidon was not responding to treatments. But one night, he made a turn for the better. The staff was thrilled when they saw that he managed to eat 3 ounces of food. However, that little victory was not to be celebrated for all that long. Poseidon passed away a few hours later.
I'm sure that you can imagine the heartache. When you put your heart and soul into another living being and it just doesn't work, it is devastating. At Big Cat Rescue, this is a way of life. There are some that thrive and some that, despite most heroic efforts, will succumb to their injuries or illness.
Linda G's wildcat cam. I'm not crazy, this serval was sedated and about to have surgery.
I am in awe of their work, and yes, just a little bit envious. The closest I got to working with a wild cat was when I worked as a Veterinary Technician at a cat-only animal hospital. One day, we received a frantic call from someone who kept a serval as a pet. This big cat liked to eat tennis balls, which is not a good thing. It became lodged in her digestive tract, and she needed surgery to remove it.
This is a video posted 2 days ago. As you can see, Thor is not only grooming himself, a great sign, but he is tearing up the place. More on Thor in the days to come. In fact, Thor may become even more famous. He has an appointment with a dental specialist on Monday, March 14, hoping to save his broken canine, and may be the first cat in history to have a titanium tooth. Stay tuned!


text:
CEO Live
Breast Cancer Means a Higher Risk of Cardiovascular Disease
Laurie Allen
May 18, 2018 at 3:42 PM May 21, 2018 at 9:51 AM
Here's what those who have had cancer—and their employers—can do to avoid heart complications.
It's a quintessential good news/bad news scenario. Thanks to earlier detection and better treatment options, women diagnosed with breast cancer are living longer.
The flip side is that they are experiencing more cardiovascular complications as a result. Breast cancer survivors, especially women over 65, are more likely to die from cardiovascular disease than breast cancer, according to a scientific statement released by the American Heart Association earlier this year. That underscores the need for managing heart disease risk during and after cancer treatment, the AHA says.
To some degree, the increase in cardiovascular problems isn't surprising, local experts say. "Approximately 90 percent of breast cancer survivors are alive five years after diagnosis," says Dr. Anne Albers, a cardiologist with OhioHealth Heart & Vascular Physicians. "Age is a risk factor for cardiovascular disease, so a longer [cancer] survival rate will increase your risk for heart and vascular disease."
That said, both chemotherapy and radiation exert stress on a woman's heart and vascular system in a variety of ways:
Some drugs, such as anthracycline, can result in abnormal heart rhythms that may be life-threatening in some cases. HER2-targeted therapies, such as Herceptin, can cause weakening of the heart muscle, or heart failure, which occurs when the heart loses its ability to contract normally. Herceptin and other targeted immunotherapies treat so-called HER2-positive breast cancers, which contain abnormally high levels of a protein that stimulates cancer cell growth. Anti-metabolites can cause spasms in the heart's arteries that cause chest pain but could lead to heart attacks as well. Radiation creates collateral damage to surrounding tissue, which can lead to narrowing of the arteries and other problems.
Dr. Laxmi Mehta, director of the Women's Cardiovascular Health Program at the Ohio State University Wexner Medical Center, was one of the authors of the AHA paper. "Patients are overwhelmed when they're diagnosed with two serious conditions. … There has to be a focus on the heart both during and after cancer treatment," she says. "It's imperative to carefully monitor heart functions."
However, she cautions, "We don't want to raise hysteria. We definitely don't want patients stopping their treatment."
Physicians say the medical community is increasingly aware and vigilant when it comes to monitoring a breast cancer patient's cardiovascular status and adjusting treatment accordingly. "We aggressively monitor patients while they are receiving [potentially toxic] medications to minimize these toxicities when possible," says Dr. Raquel Reinbolt, breast medical oncologist with the Ohio State University Comprehensive Cancer Center – Arthur G. James Cancer Hospital and Richard J. Solove Research Institute.
Patients should be reassured that, "When we recommend a treatment for a patient, we really believe the benefits outweigh the risks," says Dr. Jeanna Knoble, a medical oncologist with Zangmeister Cancer Center and co-medical director of the breast cancer program at Mount Carmel Health System.
Reinbolt says "multiple efforts" are under way to prevent, reduce or eliminate heart-related complications. In some cases, physicians can try different medications or administer them differently to try to mitigate serious side effects.
For example, studies have shown that when the drug doxorubicin is infused slowly, rather than all at once, patients may have a lower risk of heart failure. Prone radiation, in which a woman lies stomach-down on the treatment table, delivers a far more targeted dose of radiation, sparing nearby tissue.
Women also may receive so-called "cardio-protective" agents, such as beta blockers or cholesterol-lowering statins, if they are at risk for developing heart and vascular complications.
Because no two women share the same risk factors, or cancer, it's essential to think ahead when planning treatment.
Local physicians stress the importance of communication and collaboration between oncologists and heart specialists in higher-risk patients. "There is much more awareness of the link between cancer treatment and heart disease, and we're seeing a more multidisciplinary approach," says Dr. Laura Gravelin, a cardiac electrophysiologist at Mount Carmel. The growing field of cardio-oncology is evidence of that, she says.
"Cardio-oncology is a conversation" between specialties, says Albers, who notes that in her experience, oncologists have been diligent about ordering echocardiograms and other tests to monitor effects on the heart. "If we see something abnormal, maybe we take a break or consider changing medication or adding medication," she says.
Mehta says specialists are working to develop guidelines to determine which patients would most benefit from testing and ongoing surveillance. "Not all treatments result in cardiovascular complications," she notes.
Reinbolt says physicians and researchers also are trying to better understand why some cancer drugs cause heart issues in certain breast cancer survivors. For example, researchers at OSU are trying to identify a genetic "signature" that may signal that a woman is at higher risk. "We've also evaluated if we can use cardiac imaging, like cardiac MRI, to detect heart toxicities earlier on."
Another key message contained in the AHA scientific statement was that of shared risk factors for both breast cancer and cardiovascular disease, namely: age, obesity, physical inactivity, smoking, poor diet and family history. Another overlapping risk factor is being post-menopausal. Some cancer therapies can trigger premature or early menopause, Reinbolt notes.
She and others say lifestyle changes can have an impact on the risk for breast cancer recurrence as well as cardiovascular disease.
The American Heart Association says adherence to its Life's Simple 7 program is associated with a trend toward lower incidence of breast cancer. It includes:
Being physically active (at least 150 minutes of moderate to vigorous physical activity a week) Achieving and maintaining an ideal body weight Eating a healthy diet, with an emphasis on plant-based foods Avoiding tobacco use Maintaining healthy levels of blood pressure, cholesterol and blood sugar
Albers says it's important for women to know that when it comes to the physical activity, exercise can be spaced out in three, 10-minute intervals a day.
"If you can adhere to these seven things, you really can put yourself in a better position" both in terms of cancer and cardiovascular disease, she adds.
The business sector has a role to play as well, physicians say. Albers says businesses both large and small can make a difference by offering incentives for tracking steps and supporting lifestyle choices such as mindfulness and stress reduction, for example.
Gravelin says companies also can establish lunchtime walking programs, offer healthier choices in cafeterias and vending machines, and partner with gyms to offer discounted memberships. She says she plans to ask her patients about what their employers area doing to support health and wellness.
Women need to be as concerned about their cardiovascular well-being as they are about their breast health, Gravelin believes. "Women are much more afraid of breast cancer, yet heart disease is the number one killer of women, causing one in three deaths."
Laurie Allen is a freelance writer.
Columbus CEO ~ 62 E. Broad St., P.O. Box 1289, Columbus, OH 43216 ~ Do Not Sell My Personal Information ~ Cookie Policy ~ Do Not Sell My Personal Information ~ Privacy Policy ~ Terms Of Service ~ Your California Privacy Rights / Privacy Policy
Breakdowns


text:
Fun Facts About The Magic Kingdom!
May 8, 2018 mickeypopsmonorailsocks Did You Know, Disney World, Featured, Magic Kingdom Leave a comment
As the very first park to open in Disney World, Magic Kingdom is arguably the first park people think of when they say "I'm going to Disney World!" With so much to see and do in this magical wonderland, here are some fun facts you can share to impress your friends and family!
When Magic Kingdom Park opened on October 1, 1971 a ticket cost $3.50!
Mr. William Windsor Jr. was the first guest to enter.
Total park attendance for that day was 10,400.
Opening Day Rides and Attractions
Walt Disney World Railroad Main Street U.S.A Station*
Main Street Cinema
Cinderella's Golden Carousel (Prince Charming Regal Carousel)*
Tropical Serenade (Walt Disney's Enchanted Tiki Room)*
Dumbo the Flying Elephant*
Jungle Cruise*
Hall of Presidents*
Mad Tea Party*
Swiss Family Treehouse*
Mr. Toad's Wild Ride
Frontierland Shootin' Arcade*
Country Bear Jamboree*
Grand Prix Raceway (Tomorrowland Speedway)*
Peter Pan's Flight*
It's a Small World*
The Haunted Mansion*
The Dapper Dans*
Davy Crockett Explorer Canoes
Mike Fink Keel Boats
Snow White's Adventures
*All of these rides and attractions are still operating today!
Adventureland is home to many classic rides and attractions, most notably Pirates of the Caribbean and the Jungle Cruise. Walt Disney himself had his hand in designing many of the beloved attractions, despite his death some 5 years before, and the area has a great feel of nostalgia for many guests.
The Pirates of the Caribbean ride was supposed to be a wax museum gallery that guests would walk through instead of a boat ride.
Recently the "We wants the Redhead!" scene, where a woman was being auctioned off by pirates, was changed. She is now a pirate along with the male pirates selling stolen goods.
This is the last attraction Walt Disney had a hand in personally designing before his death.
The famous song, "In the tiki, tiki, tiki, tiki, tiki room" was written by the Sherman Brothers, who also wrote songs for Mary Poppins. The Enchanted Tiki Room has about 150 different animatronics – almost 120 different kids of birds can be heard singing during the show.
Since 1976 the show has been sponsored by Dole, who also makes the most amazing snack in all of Disney World, Dole Whips!
The Jungle Cruise boats are on a track system in water that is only about 4 feet deep!
At the end of the ride you meet Trader Sam, a local head salesman who will trade you two of his heads for one of yours! This ride inspired Disney to open Trader Sam's Grog Grotto which has a location in Disney's Polynesian Village Resort, complete with special effects when you order certain unique alcoholic beverages!
The theme of this land is one that takes you straight into the wild, wild west, complete with two of the most popular "mountain" attractions in the kingdom, Splash Mountain and Big Thunder Mountain Railroad. Here you can ride the "wildest ride in the wilderness" and practice your shootin' skills at the arcade.
Splash Mountain has 950,000 gallons of water constantly running through the attraction, over 100 singing animatronic animals, 3 dips, and a 5 story drop!
There used to be a Walt Disney World Railroad stop here but it was demolished and moved to make room for Splash Mountain!
There are 6 train cars running on Big Thunder Mountain Railroad: U.R. Daring, U.R. Courageous, U.B. Bold, I.M. Fearless, I.M. Brave, and I.B. Hearty.
The Country Bear Jamboree is located in Grizzly Hall and is a 16 minute knee-slappin', rootin' tootin' good time stage show for guests of any age. It was one of the last attractions Walt Disney actually had a hand in bringing to life.
Did You KnowDisney WorldFeaturedMagic Kingdom
Previous Post: Guardians of the Galaxy Roller Coaster Coming to Epcot!
Next Post: It's An Incredible Summer at Disney World!


Brian has recently ventured into making homemade bacon, a skill which I believe will be much appreciated in PNG, as bacon is rare over there. Of course, he did manage to produce much more during his practice than I think we can consume in one month, so we've had to share the bounty!
Caleb and one of his beloved frogs. He checks with us regularly to make sure that his favorite stuffed animals will be able to travel back to PNG with us.
We had some snow and cold weather here this week. Greg has discovered the wonderful pastime of crunching frozen puddles while waiting for the school bus in the mornings. Reminds me of my own childhood!
I also got to check one item off my photography bucket list, which was taking photos of freezing soap bubbles. Conditions weren't ideal, but at least I got a chance to try it!
And I've been doing a lot of packing. Packing up boxes of household things, and also starting to pack the suitcases we'll bring back to PNG as we finish up shopping for things we want to take back with us.
I have to keep telling her - "that is for when you're bigger... it's a kindergarten dress." Many tears seeing all those pretty things packed away.
Wish us luck as we continue to pack. Next on the agenda is to purchase or build a storage shed to store our things that we will leave in the States.


Do you think you could be better in certain situations?
Do you sometimes find it difficult to behave the way you would want?
It's easy to think of things you want to alter in your life.
Perhaps you want to overcome a negative habit, or change the way you react to certain situations, or simply improve your attitude towards something.
But the hard part often comes in actually making these changes.
Because trying to alter deeply embedded patterns of behavior through conscious willpower can be a slow process. And unfortunately it's one that doesn't always work.
And that's often because the pattern we're trying to alter is taking place at an unconscious level in our mind, on a different level from conscious thought.
During unconscious learning, which is the level at which our basic skills, positive learning experiences, and even phobias are formed, we develop powerful new response patterns without being aware of the cause.
And because most of our brain activity actually takes place without our conscious thought being directed at it, it makes sense that working on this level is likely to be the most powerful tool for initiating change.
And the good news is that hypnosis gives us a powerful way to tap into our unconscious mind and alter our unconscious responses so we can notice change on the conscious level.
Be Your Best Self is an audio hypnosis session that will help you make strong, positive change in whichever area of your life you choose.
Download Be Your Best Self and begin positive change in your life today.
Be Your Best Self has been purchased by 139 customers.


Everything You Need To Know about Pakistan vs. South Africa Series 2018-2019!!!
The match season has been started from last few months and many teams participated in this series. All the famous cricket teams of almost all countries are playing in this series and now it's the turn of Pakistan and South Africa team. The Pakistan vs. South Africa would be start in the month of December 2018 and continued to till February 2019. Our Pakistani cricket team tour of South Africa for 3 Test matches, 5 ODI matches and 3 T20 Matches. This series would be continued from 19 December 2018 to 6 February 2019. Here are the complete details of Pakistan vs. South Africa Series 2018 to 2019 are discussed below. Don't forget to watch this thrilling series between the two energetic teams.
Here are the complete matches of Pakistan vs. South Africa Series 2018 to 2019 with venue. The complete series would be played according to the following schedule.
A great competition would be held between Pakistan and South Africa in December 2018 to February 2019. Don't miss to watch this thrilling series.


This is a comprehensive course on Exploit Development in Windows platform. This is a comprehensive course on Exploit Development in Windows platform. The course is designed in such a way to help the beginners. It will help you understand the different domains of software exploitation. The participants will learn about different types and techniques of exploitation, using debuggers to create their own exploits, understand protection mechanism of the Operating Systems and how to bypass them.
This is a comprehensive course on Exploit Development in Windows platform. The course is designed in such a way to help the beginners. It will help you understand the different domains of software exploitation.
The participants will learn about different types and techniques of exploitation, using debuggers to create their own exploits, understand protection mechanism of the Operating Systems and how to bypass them. You will also learn how to write and execute an exploit in the latest Windows operating systems like Windows 7 and 8. This course will be supported by additional reference materials that will help you to understand the concepts and for further reading. The the course content will be updated regularly.
Learn about concepts like Vanilla EIP Overwrite, Egg hunting, SEH overwrite, ROP chains etc.
Learn to use various exploit development tools like debuggers, mona extension etc.
Anyone interested in exploit development.
Do you want to learn how to read and understand existing exploits?
Do need to successfully modify an existing exploit?.
Do you want to write reliable exploits and port them into Metasploit modules?
Do you want to know how shellcode works ?
Do you need to understand the basic concepts of exploit writing?
Are you willing to dig deeper into shellcodes and debugger?
Be the first to review "Windows Exploit Development Megaprimer – WXP" Click here to cancel reply.


text:
module ActionController #:nodoc:
  module Caching
    # Sweepers are the terminators of the caching world and responsible for expiring caches when model objects change.
    # They do this by being half-observers, half-filters and implementing callbacks for both roles. A Sweeper example:
    #
    #   class ListSweeper < ActionController::Caching::Sweeper
    #     observe List, Item
    #
    #     def after_save(record)
    #       list = record.is_a?(List) ? record : record.list
    #       expire_page(:controller => "lists", :action => %w( show public feed ), :id => list.id)
    #       expire_action(:controller => "lists", :action => "all")
    #       list.shares.each { |share| expire_page(:controller => "lists", :action => "show", :id => share.url_key) }
    #     end
    #   end
    #
    # The sweeper is assigned in the controllers that wish to have its job performed using the <tt>cache_sweeper</tt> class method:
    #
    #   class ListsController < ApplicationController
    #     caches_action :index, :show, :public, :feed
    #     cache_sweeper :list_sweeper, :only => [ :edit, :destroy, :share ]
    #   end
    #
    # In the example above, four actions are cached and three actions are responsible for expiring those caches.
    #
    # You can also name an explicit class in the declaration of a sweeper, which is needed if the sweeper is in a module:
    #
    #   class ListsController < ApplicationController
    #     caches_action :index, :show, :public, :feed
    #     cache_sweeper OpenBar::Sweeper, :only => [ :edit, :destroy, :share ]
    #   end
    module Sweeping
      extend ActiveSupport::Concern

      module ClassMethods #:nodoc:
        def cache_sweeper(*sweepers)
          configuration = sweepers.extract_options!

          sweepers.each do |sweeper|
            ActiveRecord::Base.observers << sweeper if defined?(ActiveRecord) and defined?(ActiveRecord::Base)
            sweeper_instance = (sweeper.is_a?(Symbol) ? Object.const_get(sweeper.to_s.classify) : sweeper).instance

            if sweeper_instance.is_a?(Sweeper)
              around_filter(sweeper_instance, :only => configuration[:only])
            else
              after_filter(sweeper_instance, :only => configuration[:only])
            end
          end
        end
      end
    end

    if defined?(ActiveRecord) and defined?(ActiveRecord::Observer)
      class Sweeper < ActiveRecord::Observer #:nodoc:
        attr_accessor :controller

        def before(controller)
          self.controller = controller
          callback(:before) if controller.perform_caching
          true # before method from sweeper should always return true
        end

        def after(controller)
          callback(:after) if controller.perform_caching
          # Clean up, so that the controller can be collected after this request
          self.controller = nil
        end

        protected
          # gets the action cache path for the given options.
          def action_path_for(options)
            Actions::ActionCachePath.new(controller, options).path
          end

          # Retrieve instance variables set in the controller.
          def assigns(key)
            controller.instance_variable_get("@#{key}")
          end

        private
          def callback(timing)
            controller_callback_method_name = "#{timing}_#{controller.controller_name.underscore}"
            action_callback_method_name     = "#{controller_callback_method_name}_#{controller.action_name}"

            __send__(controller_callback_method_name) if respond_to?(controller_callback_method_name, true)
            __send__(action_callback_method_name)     if respond_to?(action_callback_method_name, true)
          end

          def method_missing(method, *arguments, &block)
            return if @controller.nil?
            @controller.__send__(method, *arguments, &block)
          end
      end
    end
  end
end

redpajama_set_name: RedPajamaGithub


Honghe Tea of Jiandou Anxi is a science and technology intensive enterprise that combines R & D, production, processing and marketing in one. We have tea gardens, growing bases, and tea processing plants, also tea and tea set wholesale branches. Zhonghe Tea of Jiandou Anxi's growing base locates in the major Tie'guanyin production areas of Anxi, which enjoys worldwide fame. Altitude of 500-1100 meters, sweet running fountain, shrouded in the mist and cloud, tender climate, clearly seasons, plentiful rainfall, together with pollution free surroundings, we are the optimum place for natural tea --Tie'guanyin.


text:
package org.apache.camel.example.netty;

import io.netty.buffer.ByteBuf;
import io.netty.channel.ChannelHandlerContext;
import io.netty.handler.codec.MessageToByteEncoder;

/**
 * Netty encoder that writes the Camel message to bytes with start and end byte markers.
 */
public class MyCodecEncoder extends MessageToByteEncoder<Object> {

    private static char startByte = 0x0b; // 11 decimal
    private static char endByte1 = 0x1c; // 28 decimal
    private static char endByte2 = 0x0d; // 13 decimal

    @Override
    protected void encode(ChannelHandlerContext channelHandlerContext, Object message, ByteBuf byteBuf) throws Exception {

        byte[] body;
        if (message instanceof String) {
            body = ((String) message).getBytes();
        } else if (message instanceof byte[]) {
            body = (byte[]) message;
        } else {
            throw new IllegalArgumentException("The message to encode is not a supported type: "
                + message.getClass().getCanonicalName());
        }

        byteBuf.writeByte(startByte);
        byteBuf.writeBytes(body);
        byteBuf.writeByte(endByte1);
        byteBuf.writeByte(endByte2);
    }
}