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Q: regularization of a divergent integral is there any way to regularize the following divergent  integral :
$$\int_{0}^{\infty}\frac{dx}{xe^{x}(e^{x}-1)}$$
the integral comes from trying to find an analytic continuation of
$$I(s)=s\int_{0}^{\infty} \frac{dx}{2x}\left(E_{s/2}((\pi x)^{s/2})-1\right)\omega(x)-\left(E_{s/2}(( x)^{s/2})-1\right)e^{-x}$$
$E_{s}(x)$ is the mittag-leffler function . and admits the beautiful continuation :
$E_{s}(x^{-1})=1-E_{-s}(x)$
and $$\omega(x)=\sum_{n=1}^{\infty}e^{-n^{2}\pi x}$$
which is the jacobi theta function in disguise
another representation of the integral above is - for a minute assume i'm correct about this equivalence -  :
$$I(s)=s\int_{0}^{\infty}\frac{E_{s}(x^{s})-1}{xe^{x}(e^{x}-1)}dx$$
i was wondering if we can apply Riemann's trick, and replace this integral with a contour integral to obtain a meromorphic integral !?

following Riemann's trick, here is what i did :
start with contour integral :
$$I(s)=-s\oint_{c}\frac{E_{s}((-x)^{s})-1}{xe^{x}(e^{x}-1)}dx$$
the contour is the usual Hankel contour. consider $I(-s)$ :
$$I(-s)=s\oint_{c}\frac{E_{-s}((-x)^{-s})-1}{xe^{x}(e^{x}-1)}dx=-s\oint_{c}\frac{E_{s}((-x)^{s})}{xe^{x}(e^{x}-1)}dx$$
or
$$I(s)-I(-s)=s\oint_{c}\frac{dx}{xe^{x}(e^{x}-1)}=s\oint_{c}(-x)^{-1}e^{-x}dx-s\oint_{c}\frac{(-x)^{-1}dx}{e^{x}-1}$$
now :$$\oint_{c}(-x)^{-1}e^{-x}dx=\frac{-2\pi i}{\Gamma(1)}=-2\pi i$$
and the second integral could be thought of as:
$$\oint_{c}\frac{(-x)^{-1}dx}{e^{x}-1}=\lim_{z\rightarrow 0}\oint_{c}\frac{(-x)^{z-1}dx}{e^{x}-1}=-2i\lim_{z\rightarrow 0}\sin(\pi z)\Gamma(z)\zeta(z)=i\pi$$
or :
$$I(s)-I(-s)=-3\pi is$$
lets go back to the 1st integral, and expand the Mittag-leffler function :
$$I(s)=-s\oint_{c}\frac{E_{s}((-x)^{s})-1}{xe^{x}(e^{x}-1)}dx=-s\sum_{n=1}^{\infty}\frac{1}{\Gamma(1+ns)}\oint_{c}\frac{(-x)^{sk-1}dx}{e^{x}(e^{x}-1)}$$ $$=s\sum_{n=1}^{\infty}\frac{2i \sin(k\pi s)\Gamma(ks)}{\Gamma(1+ns)}\left(\zeta(ks)-1\right)=2i\sum_{n=1}^{\infty}\sin(k\pi s)\frac{\zeta(ks)-1}{k}$$
now the problem becomes finding a function of the variable s -lets call it $A(s)$- such that:
$$\sum_{n=1}^{\infty}\sin(k\pi s)\frac{\zeta(ks)-1}{k}=A(s)\sum_{n=1}^{\infty}\frac{\zeta(ks)-1}{k}$$
if we define :
$$k(s)=\sum_{n=1}^{\infty}\frac{\zeta(ks)-1}{k}$$
then :
$$A(s)k(s)-A(-s)k(-s)=-\frac{3}{2}\pi  s$$
and the problem becomes proving the existence of $A(s)$ for all s, and of course, finding it !!

A: Edit
The formula proposed by Tom may be obtained with the classical method :
$$\int_0^{\infty} \frac{x^{s-1}}{e^x(e^x-1)} dx $$
$$    = \sum_{n=2}^{\infty} \int_0^{\infty} x^{s-1} e^{-nx} dx$$
$$    = \sum_{n=2}^{\infty} \int_0^{\infty} \left(\frac{t}{n}\right)^{s-1} e^{-t} dt/n $$
$$    = \left(\sum_{n=2}^{\infty} \frac{1}{n^s}\right) \int_0^{\infty} t^{s-1} e^{-t} dt $$
$$    = \left(\zeta(s)-1\right) \Gamma(s)$$
As $s\to 0$ we have $\displaystyle \left(\zeta(s)-1\right) \Gamma(s)\sim -\frac 3{2s} +\frac 32\gamma-\frac{\ln(2\pi)}2$ (with $\gamma=\lim_{s\to 0} \frac 1s-\Gamma(s)$ the Euler constant $0.5772156649\cdots$ and $\frac{-\ln(2\pi)}2=\zeta'(0)$) which is still divergent...
I am not sure that we can avoid this because of the equivalence $\int \frac 1{x^2} dx$ near $0$ but perhaps that your 'regularization' could consist in the limit : $\displaystyle \lim_{\epsilon\to 0} \frac{\left(\zeta(-\epsilon)-1\right) \Gamma(-\epsilon)+\left(\zeta(\epsilon)-1\right) \Gamma(\epsilon)}2=\frac 32\gamma-\frac{\ln(2\pi)}2$
or a symmetrical pondered integral centered at $0$ with the same result...

Corresponding series
Let's substitute this definition of the Mittag-Leffler function :
$$\operatorname{E}_s(z)=\sum_{k=0}^\infty \frac{z^k}{\Gamma(sk+1)}$$
in your integral :
$$I(s)=s\int_{0}^{\infty}\frac{E_{s}(x^{s})-1}{xe^{x}(e^{x}-1)}dx$$
$$I(s)=s\int_{0}^{\infty}\frac{\sum_{k=1}^\infty \frac{x^{sk}}{\Gamma(sk+1)}}{xe^{x}(e^{x}-1)}dx=s\sum_{k=1}^\infty \frac 1{\Gamma(sk+1)}\int_{0}^{\infty}\frac{ x^{sk-1}}{e^{x}(e^{x}-1)}dx$$
We may use our previous result to rewrite $I(s)$ as :
$$I(s)=s\sum_{k=1}^\infty \frac {\left(\zeta(sk)-1\right) \Gamma(sk)}{\Gamma(sk+1)}$$
this becomes the simple (and convergent if $\Re(s)>0$ and $\frac 1s$  not integer at least...) :
$$\boxed{I(s)=\sum_{k=1}^\infty \frac {\zeta(sk)-1}{k}}$$

Approximation and Visualization
We may remove the simple poles in this series at $s=\frac 1n$ for $n\in \mathbb{N}^*$ by subtracting $\frac 1{sk-1}$ at the numerator (Stieltjes expansion) with the hope of getting something simpler.
Indeed for $s$ small I found this rather accurate numerical approximation (relative error $<\frac 1{20000}$ in the rectangle $\Re(s)\in (10^{-5},2),\ |\Im(s)|<1$) :
$$\sum_{k=1}^\infty \frac {\zeta(sk)-1-\frac 1{sk-1}}{k}\approx\frac {\ln(s)}2-0.95973512990747991661376-0.04053073339766363s+0.00026485233s^2$$
I think that the $\frac {\ln(s)}2$ term is exact (from numerical 'scaling properties') as well as the provided digits of the last three constants.
Since $\displaystyle \sum_{k=1}^\infty \frac 1{k(sk-1)}=-\gamma -\psi\left(1-\frac 1s\right)$ with $\psi$ the digamma function we have the approximation :
$$I(s)\approx -\gamma -\psi\left(1-\frac 1s\right) +\frac {\ln(s)}2-0.95973512990747991661376-0.04053073339766363s+0.00026485233s^2$$
that allows us to produce this picture of the imaginary part of $I(s)$ for $\Re(s)>0$ :

We observe a jump of amplitude $\pi$ (near the line $x=\Re(s)=0$ at the right around the line $y=\Im(s)=0$) coming from $\frac {\ln(s)}2$.
Let's observe that our expansion is analytic everywhere except at the simple poles $s=\frac 1n$ and the branch point $s=0$ allowing us to show the pictures of the imaginary and real part of the (possible) Riemann surface of the analytic extension of $I(s)$ on the whole plane (imagine an infinity of sheets superposed around the branch point $0$ generated by the $\frac {\ln(s)}2$ term) :

Asymptotic Expansion (direct method)
We want to rewrite following variant of $I(s)$ as a Maclaurin series $P(s)$ :
$$J(s):=\sum_{k=1}^\infty \frac {\zeta(sk)-1-\frac 1{sk-1}}{k}-\dfrac{\ln(s)}2$$
Let's split the $-1$ at the numerator in two $-\frac 12$ terms to study the asymptotic behavior of this series (and obtain convergence of the individual 'power $\log$ integrals' we will see later...) :
$$J(s)= \lim_{N\to\infty} J_N(s)\ \ \text{with}\ \ J_N(s)=\sum_{k=1}^N \frac {\zeta(sk)-\frac 12-\frac 1{sk-1}}{k}-\frac 12 \sum_{k=1}^N \frac 1k -\frac{\ln(s)}2$$
Let's first show that $\ \lim_{N\to\infty} J_N(s)$ exists for any $s>0$ such that $\frac 1s \notin \mathbb{N}$ (so that all the terms of the sum are defined).
Since $\ \sum_{k=1}^N \frac 1k= H_n=\gamma +\ln\bigl(N+\frac 12\bigr) + \operatorname{O}\bigl(\frac 1N\bigr)\ $ we have $\frac 12 \sum_{k=1}^N \frac 1k +\frac{\ln(s)}2=\frac {\gamma}2+\frac 12 \ln\bigl(s\bigl(N+\frac 12\bigr)\bigr)+\operatorname{O}\bigl(\frac 1N\bigr)=\frac {\gamma}2+\frac {\ln(sN)}2+\operatorname{O}\bigl(\frac 1N\bigr)$
For $sN \gg 1$ we have (setting $x:=sk$) :
$$s\sum_{k=N}^M \frac {\zeta(sk)-\frac 12-\frac 1{sk-1}}{sk}\sim \int_{sN}^{sM} \frac {\zeta(x)-\frac 12-\frac 1{x-1}}{x}\,dx$$
the error is of order $\frac 1N$ because of the $H_M-H_N$ terms and because $\zeta(x)-1 \sim 2^{-x}$ for $x \gg 1$.
A precise evaluation of the last integral is :
$\left[\frac 32 \ln(x) -\ln(x-1)+\sum_{k>1} \operatorname{Ei}(-x\ln(k))\right]^{sM}_{sN}$ (with $\operatorname{Ei}$ the exponential integral and $\operatorname{Ei}(-u)\sim \frac {e^{-u}}{-u}$ for $u\gg 1$) but this isn't needed since the integral is clearly equivalent to $\frac 12 \left[\ln(x)\right]^{sM}_{sN}\ $ with again an error of order $\frac 1N$ for $sN\gg 1$.
Both $\frac {\ln(sN)}2$ terms cancel in $J_N(s)$ for large $sN$ so that the limit $J(s)$ exists for $s>0$ (for $\frac 1s \in \mathbb{N}$ $J(s)$ may be defined as a limit since we removed the poles).
Let's use the Euler Maclaurin asymptotic formula with $f(k)=\frac {\zeta(sk)-\frac 12-\frac 1{sk-1}}{k}$ to expand directly $J_N(s)$ :
$$\sum_{k=0}^N f(k) \sim \int_0^N f(x) dx +\frac {f(0)+f(N)}2+\sum_{k=1}^\infty \frac{B_{2k}}{(2k)!} \bigl(f^{(2k-1)}(N)-f^{(2k-1)}(0)\bigr)$$
becomes :
$$ s\sum_{k=0}^N \frac {\zeta(sk)-\frac 12-\frac 1{sk-1}}{sk}\sim \int_0^{sN} \frac {\zeta(x)-\frac 12-\frac 1{x-1}}{x}\,dx + \frac {f(0)+f(N)}2+\sum_{k=1}^\infty \frac{B_{2k}}{(2k)!} \left(f^{(2k-1)}(N)-f^{(2k-1)}(0)\right)$$
Note that we started at $k=0$ with $\displaystyle\ f(0)=\lim_{k\to 0}\ s\frac {\zeta(sk)-\frac 12-\frac 1{sk-1}}{sk}=(1+\zeta'(0))s\ $
and that, for any $n\in \mathbb{N}^*$, we have :
$f^{(n-1)}(0)=\lim_{N\to 0}\ s\left(\frac d{dN}\right)^{n-1}\ \left(\frac {\zeta(sN)-\frac 12-\frac 1{sN-1}}{sN}\right)=\lim_{N\to 0}\ s\left(\frac {s\cdot d}{d(sN)}\right)^{n-1} (\cdots)=\lim_{x\to 0}\ s^n\left(\frac {\zeta(x)-\frac 12-\frac 1{x-1}}x\right)^{(n-1)}=s^n\lim_{x\to 0} \frac{\left(\zeta(x)-\frac 1{x-1}\right)^{(n)}}n$
or simply $\ \displaystyle f^{(n-1)}(0)=\frac{n!+\zeta^{(n)}(0)}n s^n\ $ for $n>0$
For $n\ge 0$ and $N \gg 1$ we may show that $f^{(n)}(N)\sim \frac {(-1)^n n!}{2 N^{n+1}}\ $ so that the Euler Maclaurin asymptotic formula becomes (after removing $\frac{H_N+\ln(s)}2$ on both sides to get $J_N(s)$ at the left) :
$$J_N(s)= s\sum_{k=0}^N \frac {\zeta(sk)-1-\frac 1{sk-1}}{sk}-\ln(s)\sim \int_0^{sN} \frac {\zeta(x)-\frac 12-\frac 1{x-1}}x\,du - \frac {\gamma}2-\frac {\ln(sN)}2+\operatorname{O}\bigl(\frac 1N\bigr)+ \frac {f(0)}2-\sum_{k=1}^\infty \frac{B_{2k}}{(2k)!} f^{(2k-1)}(0)$$
At this point let's rewrite $\ln(sN)=\int_0^{sN} \frac 1{x+1}\,du+\operatorname{O}\bigl(\frac 1N\bigr)$ and go at the limit $N\to \infty$ :
$$J(s)\sim - \frac {\gamma}2+\int_0^\infty \frac {\zeta(x)-\frac 12-\frac 1{x-1}}x-\frac 1{2(x+1)}\,dx + \frac {f(0)}2-\sum_{k=1}^\infty \frac{B_{2k}}{(2k)!} f^{(2k-1)}(0)$$
We recognize a Maclaurin series in $s$ :
$$J(s)\sim K_0 -K_1 \frac s2-\sum_{k=1}^\infty K_{2k} \frac{B_{2k}}{(2k)!} s^{2k} $$
with the constant term given by $\displaystyle\ \boxed{K_0:= -\frac{\gamma}2 +\int_0^\infty \frac {\zeta(x)-\frac 1{x-1} -\frac 12 }x -\frac 1{2(1+x)}\,dx}$
(the numerical value is $K_0=-0.95973512990\cdots$ as wished ; I don't know a closed form...)
and the $K_n$ coefficients of $s^n$ by $\boxed{K_n:=\frac{n!+\zeta^{(n)}(0)}n}\ \text{for} \ n>0$
The wished analytic extension of $I(s)$ will be :
$$\boxed{I(s)\sim -\gamma -\psi\left(1-\frac 1s\right) +\frac {\ln(s)}2+K_0 -K_1 \frac s2-\sum_{k=1}^\infty K_{2k} \frac{B_{2k}}{(2k)!} s^{2k}}$$
Short table of $K_n$ values :
($\gamma_n$ is the nth Stieltjes constant : $\gamma_0= \gamma,\ \gamma_1= 0.07281584548367672486058\cdots$)
$\displaystyle K_1=1+\zeta'(0)=1-\frac{\ln(2\pi)}2=0.0810614667953272582\cdots$
$\displaystyle K_2=1+\frac{\zeta''(0)}2=1-\frac{\ln(2\pi)^2}4+\frac{\gamma^2}4-\frac{\pi^2}{48}-\frac{\gamma_1}2=
-0.0031782279542924256\cdots$
$K_3=\frac{3!+\zeta'''(0)}3=-0.00157038895408481592\cdots$
$K_4=\frac{4!+\zeta^{(4)}(0)}4=0.0007242029965730102531\cdots$
($-\frac {K_1}2$ and $-\frac {K_2}{12}$ may be compared to the coefficients for $s$ and $s^2$ obtained previously : $-0.040530733397\cdots$ and $0.00026485233\cdots$).
(it is interesting to observe that $\zeta^{(n)}(0)\approx -n!$ as well as $\zeta^{(n)}\left(\frac 12\right)\approx -2^{n+1}n!$)

Asymptotic Expansions (the same result after an excursion with Mittag-Leffler, another expression for $K_0$)
To get the exact terms of the expansion of $I(s)$ let's provide the Maclaurin expansion in $s$ of $s\left(E_s(x^s)-1\right)$ (the coefficients are functions of $x$ that we will have to integrate later over $\mathbb{R}^+$) :
$$s\left(E_s(x^s)-1\right)=C_0(x) e^x-\frac s2-\frac{s^2}{12}\left[\ln(x)+\gamma\right]+\frac{s^4}{720}\left[\ln(x)^3+3\gamma\ln(x)^2+3\left(\gamma^2-\zeta(2)\right)\ln(x)+\gamma^3-3\gamma\zeta(2)+2\zeta(3)\right]-\frac{s^6}{30240}\left[\ln(x)^5+5\gamma\ln(x)^4+10\left(\gamma^2-\zeta(2)\right)\ln(x)^3+10\left(\gamma^3-3\gamma \zeta(2)+2\zeta(3)\right)\ln(x)^2+5\left(\gamma^4-6\gamma^2\zeta(2)+8\gamma \zeta(3)+\pi^4/60\right)\ln(x)+\gamma^5-10\gamma^3\zeta(2)+20\gamma^2\zeta(3)+\gamma\pi^4/12-20\zeta(2)\zeta(3)+24\zeta(5) \right]+P_8(\ln(x))\operatorname{O}\bigl(s^8\bigr)$$
The term independent of $s$ : $C_0(x) e^x$ with $\displaystyle C_0(x)=\int_0^1 \frac{\gamma(t,x)}{\Gamma(t)}\,dt\ $ (and $\gamma(t,x)$ the lower 'incomplete gamma function') was obtained by considering the limit as $n\to \infty$ of formula (3.4) from Haubold's paper (with $s=\frac 1n$) :
$$E_{\frac 1n}\left(x^{\frac 1n}\right)=e^x\left[1+\sum_{r=1}^{n-1}\frac{\gamma\left(1-\frac rn ,x\right)}{\Gamma\left(1-\frac rn\right)}\right]$$
A shorter expression for the Maclaurin expansion (that could again be obtained with Euler-Maclaurin) should be :
$$\boxed{s\left(E_s(x^s)-1\right)\sim C_0(x) e^x-\frac s2-\sum_{n=2}^\infty s^n\frac{B_n}{n!}\ \sum_{k=0}^{n-1} \binom{n-1}{k} S_k \ln(x)^{n-1-k}} $$
with $B_n$ the 'Bernoulli numbers' ($B_{2n+1}=0$ for $n>0$) and $S_k$ the coefficients appearing in the expansion of the reciprocal gamma function at $z=0$ (and at $z=1$) :
$$\frac 1{\Gamma(s)}=\sum_{k=0}^\infty \frac {S_k}{k!} s^{k+1}\ \ \text{so that}\ \ S_k=\lim_{s\to 0} \left(\frac 1{\Gamma(s+1)}\right)^{(k)}$$
$S_0=1$
$S_1=\gamma$
$S_2=\gamma^2-\zeta(2)$
$S_3=\gamma^3-3\gamma \zeta(2)+2\zeta(3)$
$S_4=\gamma^4-6\gamma^2\zeta(2)+8\gamma \zeta(3)+\pi^4/60$
$S_5=\gamma^5-10\gamma^3\zeta(2)+20\gamma^2\zeta(3)+\gamma\pi^4/12-20\zeta(2)\zeta(3)+24\zeta(5)$
$C_0$ too may be rewritten with the $S_k$ terms as $\displaystyle C_0(x)=\sum_{k=0}^\infty \frac{S_k}{(-\ln(x))^{k+1}}$.

We will again start with :
$\displaystyle J(s)= \lim_{N\to\infty} J_N(s)$ with
$\ \displaystyle J_N(s)=\sum_{k=1}^N \frac {\zeta(sk)-\frac 12-\frac 1{sk-1}}{k}-\frac 12 \sum_{k=1}^N \frac 1k -\frac{\ln(s)}2$
but will use Mittag-Leffler after that :
$$J_N(s)=s\sum_{k=1}^N \frac {\zeta(sk)\Gamma(sk)-\frac {\Gamma(sk)}2-\frac {(sk-1)\Gamma(sk-1)}{sk-1}}{(sk)\Gamma(sk)}-\frac 12 \sum_{k=1}^N \frac 1k -\frac{\ln(s)}2$$
$$J_N(s)=s\sum_{k=1}^N \frac {\zeta(sk)\Gamma(sk)-\frac {\Gamma(sk)}2-\Gamma(sk-1)}{\Gamma(sk+1)}-\frac 12 \sum_{k=1}^N \frac 1k -\frac{\ln(s)}2$$
Let's rewrite everything in integral form (using $\ \sum_{k=1}^N \frac 1k= \gamma +\int_0^N \frac 1{1+x} + \operatorname{O}\bigl(\frac 1N\bigr)\ $, $\Gamma(u)=\int_0^\infty t^{u-1} e^{-t} dt\ $ and considering the limit as $N\to \infty$) :
$$J_N(s)=\sum_{k=1}^N \left( \frac s{\Gamma(sk+1)}\int_{0}^{\infty}\frac{x^{sk-1}}{e^x-1}-\frac {x^{sk-1}}{2e^x}-\frac {x^{sk-1}}{x e^x}\,dx\right)-\frac{\gamma}2 -\frac{\ln(N s)}2+\operatorname{O}\bigl(\frac 1N\bigr)$$
$$J_N(s)=-\frac{\gamma}2+\int_{0}^{\infty}\frac{\sum_{k=1}^N \frac {sx^{sk}}{\Gamma(sk+1)}}{xe^{x}}\left(\frac {e^x}{e^{x}-1}-\frac 12-\frac 1x\right)\,dx -\frac{\ln(N s)}2+\operatorname{O}\bigl(\frac 1N\bigr)$$
$$J_N(s)=-\frac{\gamma}2 +\int_{0}^{\infty}\frac{\sum_{k=1}^N \frac {sx^{sk}}{\Gamma(sk+1)}}{xe^{x}}\left(\frac {e^x}{e^{x}-1}-\frac 12-\frac 1x\right)+\frac{e^{-N s x}-e^{-x}}{2x}\,dx +\operatorname{O}\bigl(\frac 1N\bigr)$$
that becomes at the limit $N \to \infty$ :
$$J(s)\sim -\frac{\gamma}2 +\int_{0}^{\infty}\frac{\sum_{k=1}^\infty\frac {sx^{sk}}{\Gamma(sk+1)}}{xe^{x}}\left(\frac {e^x}{e^{x}-1}-\frac 12-\frac 1x\right)-\frac 1{2(1+x)}\,dx$$
$$\boxed{J(s)\sim -\frac{\gamma}2 +\int_{0}^{\infty}\frac{s \left(E_{s}(x^{s})-1\right)}{xe^{x}}\left(\frac 1{e^{x}-1}-\frac 1x+\frac 12\right)-\frac 1{2(1+x)}\,dx}$$
From the MacLaurin expansion of $s \left(E_{s}(x^{s})-1\right)$ in powers of $s$, we get :
$$J(s)\sim K_0 -K_1 \frac s2-\sum_{n=2}^\infty K_n \frac{B_n}{n!} s^n $$
and the wished analytic extension of $I(s)$ :
$$\boxed{I(s)\sim -\gamma -\psi\left(1-\frac 1s\right) +\frac {\ln(s)}2+K_0 -K_1 \frac s2-\sum_{n=2}^\infty K_n \frac{B_n}{n!} s^n}$$
with the constant term given by :
$$K_0:= -\frac{\gamma}2 +\int_0^\infty \frac{C_0(x)}x\left(\frac 1{e^{x}-1}-\frac 1x+\frac 12\right)-\frac 1{2(1+x)}\,dx$$
with the same numerical value and the $K_n$ coefficients of $s^n$ for $n>0$ by :
$$K_n:= \sum_{m=0}^{n-1} f_m \binom{n-1}{m} S_{n-1-m}$$
with :
$$f_m:=\int_{0}^{\infty}\frac{\ln(x)^m}{xe^{x}}\left(\frac 1{e^{x}-1}-\frac 1x+\frac 12\right)\,dx$$
Let's use (with a derivation similar to the previous one!) :
$$Z(s):=\left(\zeta(s)-\frac 1{s-1}-\frac 12\right) \Gamma(s)=\int_0^{\infty} \frac{x^s}{xe^{x}}\left(\frac 1{e^{x}-1}-\frac 1x+\frac 12\right)\,dx$$
(the idea of the additional term $-\frac 1x+\frac 12$ appeared in the Hermite-Stieltjes letters of 1885 allowing to extend the integral giving zeta near $x=0$ by removing the $\frac 1{x^2}$ and $\frac 1x$ singularities ; just some pages later we find the famous claim of Stieltjes of a proof of R.H. !)
since $x^s=e^{s\ln(x)}$ we get :
$$f_m= Z^{(m)}(0)=\lim_{s\to 0}\left(\frac {d}{ds}\right)^m \left[\left(\zeta(s)-\frac 1{s-1}-\frac 12\right)\Gamma(s)\right]$$
From the binomial theorem it is clear that :
$$K_{n+1}= \sum_{m=0}^n \binom{n}{m} f_m S_{n-m}=  \lim_{s\to 0}\sum_{m=0}^n \binom{n}{m} Z^{(m)}\left(\frac 1{\Gamma(s+1)}\right)^{(n-m)}= \lim_{s\to 0}  \left(\frac {Z(s)}{\Gamma(s+1)}\right)^{(n)}$$
allowing some simplifications :
$$K_n=\lim_{s\to 0} \Bigl(\frac{\zeta(s)-\frac 1{s-1}-\frac 12}s\Bigr)^{(n-1)}=\lim_{s\to 0} \frac{\left(\zeta(s)-\frac 1{s-1}\right)^{(n)}}n\ \text{for} \ n>0$$
$$\text{i.e.}\ \ \boxed{K_n=\frac{n!+\zeta^{(n)}(0)}n}\ \text{for} \ n>0$$
i.e. the same result as previously

We could continue this very interesting subject and study the multiplicative partition ("Factorisatio Numerorum") obtained by rewriting $\displaystyle I(s)=\sum_{k=1}^\infty \frac {\zeta(sk)-1}{k}$ as the logarithm of $\displaystyle \prod_{n>1} \frac 1{1-n^{-s}}$ getting this Dirichlet series or... but I'll have to make all this visible one day so...
Sorry if all this is not exactly what you hoped...


text:
Top Chef Alum Jennifer Carroll Returns to Philadelphia in a Big Way
The Philly-born chef opens Spice Finch in Rittenhouse Square on Thursday.
By Regan Stephens
Credit: Eric Ashleigh
After working in some of the country's most celebrated restaurants and competing on multiple seasons of Top Chef, Jennifer Carroll is returning to her hometown. The Philadelphia-born chef and and her fiancé, chef Billy Riddle, are opening Spice Finch on Thursday in Rittenhouse Square after months of anticipation.
The contemporary Mediterranean restaurant is a fitting amalgam of Carroll and Riddle's collective pasts. Riddle's resume includes Kapnos Taverna in Washington, D.C., as well as Townsend, Ela, and Volver in Philadelphia, while Carroll worked under chefs Marcus Samuelsson and Eric Ripert, as well as at D.C. French-Mediterranean restaurant Requin. The pair return to the City of Brotherly Love after their stint in D.C., where they still operate catering company Carroll Couture Cuisine, and have noted some changes in Philly since they last worked here.
"I think there's always been diversity in Philadelphia, we've always had that," says Carroll. "But now, I think the Philadelphian palate itself has changed. The guests are willing to try more things."
Something that's remained constant, however, is the collaborative spirit within the city's culinary scene.
"Just the level of respect is something special," says Carroll. "I asked Mike Solomonov where we could get metal skewers for my merguez kebabs, and he was like, 'Ah, skewers are tough. But I'm going to Israel in August, want me to buy them for you?'" So the award-winning chef and co-owner of modern Israeli restaurant Zahav is leaving room in his suitcase for Carroll's skewers.
The restaurant's plant-focused menu includes starters like charred carrot hummus with housemade sourdough flatbread, and date truffles with pomegranate and cashew streusel. Vegetables and grains each get their own section, and plates for the table feature both fish and meat options like date braised lamb shank and whole roasted fish, as well as a salt-baked eggplant with mushroom and bechamel.
"We're doing a lot of stuff in-house," says Carroll. "We're trying to be as zero-waste as possible." Harissa that's fermented in-house, for example, uses onion and pepper ends that would otherwise wind up in the trash. "These things are being repurposed, and put into our sauces."
The thoughtful bar program is the work of beverage director Michael Haggerty, who comes to Spice Finch after stints at Wm Mulherin's Sons and a.bar.
"He's taking the concept and the flavor profile of what we've done in the kitchen, and bringing it to the bar," says Carroll.
Named for one of the busiest ports in the Mediterranean, where the spice trade took place, the Cap Spartel is made with lemon, carrot, honey, and mace—all the same flavors liberally used throughout the menu. For the cocktail's yogurt-washed gin, Haggerty marinates the gin in the same cow and sheep's milk Greek yogurt the chefs use in the kitchen, before straining the liquid and adding dill. Also, inspired by the fact that Carroll and Riddle don't drink, Haggerty has included a selection of non-alcoholic cocktails, including the N/A groni, made with sanbitter and chinotto.
"We still love the flavors of cocktails and going out and having specialty drinks, but we don't want the alcohol, so it's nice that we have a curated list," she says. "He's been having fun with that."
Haggerty has also been having fun with the series of "redemption cocktails," in which forgotten spirits like Jägermeister star in meticulously made, grown-up versions of their less mature iterations. "Everyone loves a comeback," reads the menu.
The 4,000-square-foot restaurant seats 158, and has a small outdoor area, too. Inside, there are art deco-inspired light fixtures and mirrors, wicker touches, and a wood-accented, granite-topped bar, which is a centerpiece for the bright, polished space.
When choosing their first restaurant's name, the chefs wanted something fun. "The bird is native to the Mediterranean and eats many of the spices that we're incorporating into the menu," says Carroll. "Have you ever seen a finch?" asks Riddle. "They're so adorable."
Today, Spice Finch opens its doors for dinner at 5 p.m., with plans in the works for breakfast, brunch, and lunch, too.

redpajama_set_name: RedPajamaCommonCrawl


text:
Michigan Wolverines NCAA FB
Michigan Wolverines NCAA BK
Michigan State Spartans NCAA FB
Michigan State Spartans NCAA BK
Prep Zone
Wings fall to Canucks
foxsports Jan 4, 2015 at 1:17a ET
Henrik Sedin's two goals gave him 201 for his career.
Anne-Marie Sorvin
VANCOUVER, British Columbia (AP) — Ryan Miller made 29 saves, Henrik Sedin scored two empty-netters to pass 200 goals for his career and the Vancouver Canucks beat the Detroit Red Wings 4-1 Saturday night.
Alexander Edler and Radim Vrbata scored on the power play for Vancouver (22-12-3), which bounced back with a solid effort after being outplayed in a 3-2 loss the Los Angeles Kings on Thursday.
Sedin's two goals gave him 201 for his career. He also had up an assist.
Tomas Tatar scored for Detroit (20-10-9), which got 20 stops from Jimmy Howard.
The Canucks talked a lot about consistency in the wake of the loss to the Kings — a game in which they were outshot 40-16 — that came on the heels of a hard-fought 3-1 road win over the San Jose Sharks.
Vancouver was much better defensively on this night and grabbed the lead against the league's No. 2 penalty-killing unit 1:48 into the second when Edler blasted a one-timer from the top of the face-off circle past a screened Howard.
All four of Edler's goals this season have come on the power play and Saturday's was the defenseman's first in nearly two months.
The Red Wings started to push back as the period wore on, with Miller — who came in with a career 2-10-2 mark against Detroit — stopping a couple chances from Pavel Datsyuk, including a nice glove save on a redirection off the rush.
Detroit got a power play early in the third, and Miller made two more big stops on Datsyuk that had fans at Rogers Arena chanting "Miller! Miller!"
Detroit got a brief 5-on-3 advantage when Daniel Sedin went off for hooking and Miller made two more big saves on Stephen Weiss toward then end of the second penalty with his team still down a man.
The Canucks then put the game away on another power play at 9:11 when Vrbata buried his own rebound past a down-and-out Howard for his 16th of the season and third in as many games.
Vancouver gave up two late goals in Thursday's loss, and Tatar made the Canucks sweat when he buried his 16th of the season with 3:30 to go in the third and Howard on the bench for an extra attacker.
But the Canucks sealed it when Henrik Sedin scored his seventh of the season into an empty net for his milestone goal with 1:51 left on the clock. He added another with 31.2 seconds remaining.
NOTES: Edler picked up an assist on Sedin's first empty-net goal. … Drew Miller was 7-1-0 in NHL games against his sibling prior to Saturday, including Detroit's 5-3 win over the Canucks on Nov. 30. … Canucks forward Zack Kassian suited up for the first time since Nov. 25. The 23-year-old missed 13 games with a broken finger before sitting out one more as a healthy scratch on Thursday. Kassian took the place of Linden Vey, who has just two assists in the last 12 games. … Canucks tough guy Tom Sestito cleared waivers and has been assigned to the AHL's Utica Comets. … The Canucks continue their five-game homestand on Tuesday against the New York Islanders. … The Red Wings opened a six-game road trip with just their second game outside the Eastern time zone in 2014-15. Detroit visits the Edmonton Oilers on Tuesday.


If you are looking for an arborist who is qualified the first thing that you need to do is to make sure that your research thoroughly on different search engines that are there for an arborist in your area who is popular. If you find an arborist who so many people in your local area know about then know that the service has to be the best and that is why they are quite popular. One thing that you need to know is that when choosing an arborist make sure that you first check what people are saying about their service. If you find that there are a lot of negative testimonies from their previous clients it is usually advisable for you to find another specialist instead of you risking your money. The service cost is also something that you cannot ignore, and you need to check how much tree care cypress different specialist charge if you hire them for the job. The good thing is that different specialists charge tree care the woodlands differently there for you can be able to find one whose charges you can easily afford without going broke. At the end of the day how much an arborist will charge you does not reflect on the type of service that you will get therefore tree care atascocita choose one that you can afford.


text:
Costume Idea?
How to Dress Like Eazy-E
Straight Out of Compton Male Music
Cosplay Wig
Rapper Chain Necklace
Compton Baseball Cap
Front-Zip Jacket
Motorcycle Biker Glove
Pro Club Men's T-Shirt
Levi's Men's Jean
Dark Lens Sunglasses
Items You'll Need
Cosplay Wig See on Amazon
Rapper Chain Necklace See on Amazon
Compton Baseball Cap See on Amazon
Front-Zip Jacket See on Amazon
Motorcycle Biker Glove See on Amazon
Pro Club Men's T-Shirt See on Amazon
Levi's Men's Jean See on Amazon
Dark Lens Sunglasses See on Amazon
Nike Classic Cortez See on Amazon
Best Eazy-E Costume Guide
Eazy-E was the stage name of the American rapper, record producer and entrepreneur, Eric Lynn Wright (September 7, 1964 – March 26, 1995). This highly-revered "Godfather of Gangsta Rap" notoriety was one of the forerunners in extending the boundaries of both musical lyrics and visual imagery way beyond conventional limits and acceptance in mainstream popular music. Eazy-E derived his inspiration largely from musical genres like funk, contemporary hip-hop, and, surprisingly, comedians. He is also known for his close collaboration with Ice Cube and Dr. Dre. As a result, the trio formed the group N.W.A. and released the hugely controversial album, Straight Outta Compton in 1988. This album is regarded as one of the greatest albums of all-time, and undoubtedly one of the most influential in the genre, too.
If you really wanna be hip and make the jaws drop at the next Halloween costume party, just get into costume as Eazy-E. Start with a Compton hat, and add on a Cosplay Wig, Dark Lens Sunglasses, Pro Club Men's T-Shirt, Front-Zip Jacket, Levi's Men's Jeans, Motorcycle Biker Gloves, Rapper Chain Necklace and the Nike Classic Cortez!
Eazy-E Cosplay Costumes
It's not every day that you get the chance to dress up as one of the meanest gangsta rappers of all time dubbed as the "Godfather of Gangsta Rap" and be as hip and tough as you can look. It's also pretty rare an opportunity to strut out into the spotlight in equally glittering company with the likes of Ice Cube and Dr. Dre, not to mention the numerous others who have made it to the big league in rap and hip-hop music. This highly-controversial figure was also the founder of Ruthless Records in 1986. In subsequent years, Eazy-E was caught in a high-profile legal tussle with Dr. Dre. To add to the headlines, he died finally of AIDS in 1995.
You will definitely bring awe, shock and lots of fond memories to any costume party with your own group of "gangsta" rappers and hip-hop friends in all kinds of apparel popularly associated with this musical culture and revolution!
About Eazy-E
Eazy-E was born and raised in Compton. In 1986, even at the age of 22, he was reputed to have earned as much as $250,000 just from dealing drugs. It was only after the tragic demise of his cousin who was killed in a gunfight that he decided to withdraw from the drug trade and make his entry into the Los Angeles hip-hop scene. Capitalizing on the new genre that was beginning to grow in popularity, Eazy-E began recording his own songs in his parents' garage. Not long afterward, he founded his own record label, Ruthless Records, in 1986.
After a short solo stint, Eazy got together with Ice Cube and Dr. Dre to form the group N.W.A. and released their controversial, landmark album, Straight Outta Compton. This highly-acclaimed album dealt with various socio-political issues and was lauded as one of the most influential ever within its genre. Eazy-E died in 1995 of AIDS.
Other Costume Ideas and Cosplay Guides
Bring some joy to your costume party by dressing up like Joy, one of the five emotions from Disney/Pixar's Inside Out.
Become a Supreme Being of the universe, dressed as Leeloo from the 1997 movie The Fifth Element.
Crimson Viper
Get ready to take down your opponents when you dress as Crimson Viper, the CIA secret agent from Street Fighter.
Cosplay the look of N, a Pokemon trainer, and leader of Team Plasma that appears in Pokemon Black, White.
Prepare for Star Wars mission dressed up to look like Cara Dune from the popular Disney+ series, The Mandalorian.
Suit up like one of the most popular Internet memes, Pepe the Frog, from the comic series Boy's Club by Matt Furie.
Get the guide to dress like just like Sherlock Holmes played by Benedict Cumberbatch on the BCC television show.
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Emma Clayton is a British children's novelist and author of dystopian thriller, The Roar. The Roar was nominated for the Carnegie Medal in 2008.
The book is set sometime in the future where an animal plague turns all animals into infected, vicious killers. The plague is televised and the population of the world are terrified. The government build a large wall circling the northern hemisphere to protect the population from the dangers that lurk outside. As the population grows, conditions inside the wall deteriorate and this causes unrest amongst the poorest of the population. A time after the plague, one of the main characters, Ellie, goes missing and is presumed dead. Her twin brother Mika, the second of the main characters knows she is still alive as a connection between the two leads him to believe so. The twins are rare mutants as a result of evolution. The story follows Mika's desperate quest to retrieve Ellie and along the way the plot thickens as he uncovers secrets of deceit and betrayal that the government have kept hidden. These secrets, if discovered, could change life as the children of the northern hemisphere know it.
This is an extremely interesting, entertaining and fast paced read. The story jumps from being mostly narrated by Mika to then also being narrated by Ellie. This aspect of the novel is great, as we get to read two different tales that intertwine and come together, from two different perspectives. Each page is intriguing and entices you to read on, it is hard not to read it all in one sitting and will leave you anxiously awaiting the release of the sequel. Since the book is set in a post-apocalyptic, dismal world, the reader feels that the main characters are brave and determined to persevere through these conditions to take a stand against injustice. This futuristic novel with weapons and fighting scenes abound will appeal to fans of science fiction with a wartime twist. The Roar is a definite must-read.


Four-wheel drive is helpful for driving in adverse conditions such as the snow, but don't let it lull you into a false sense of security. When driving in the snow, you still have to take care to optimize your vehicle so the four-wheel drive can perform at its best. Here are some tips for doing so.
This applies to driving in any situation, not just snow. Don't' drive on tires of different circumference as it can damage the driveline of the truck and create handling problems.
The slower you go, the better, especially when there's a lot of snow on the road. Going into a curve or turn with too much speed can cause you to spin out.
If you're going down a hill in low gear, shift to two-wheel drive. This will prevent the front wheels from sliding and loss of control of the truck.
Turn off traction control systems so they don't stop the truck if the tires start spinning when you're trying to make it up a snowy hill. Try to get a little more momentum before going up a hill, but not so much that you lose control.
The more experience you have driving in the snow, the better you'll be. Practice in an empty snow-covered parking lot if possible. Remember that while four-wheel drive can help you get going in the snow, it won't help you stop.
This entry was posted on Friday, January 3rd, 2014 at 7:11 pm. Both comments and pings are currently closed.


text:
\section{Introduction}

Let $A$ be an algebra over a field $\Bbbk $, i.e., a linear space equipped with a
binary linear operation (multiplication) $\cdot : A\otimes A \to A$.
Suppose $T$ is a linear operator on $A$, and let $\prec$ and $ \succ $ be two
new linear maps $A\otimes A\to A$ defined by
\[
a\prec b = a\cdot T(b), \quad a\succ b= T(a)\cdot b, \quad a,b\in A.
\]
Denote the system $(A,\prec, \succ)$ by $A^{(T)}$.
If the initial algebra $A$ satisfies a polynomial identity then what could be said about $A^{(T)}$?
The answer is known if $T$ is a Rota---Baxter operator \cite{BGuoNi13}, \cite{GubKol13}, \cite{FardGuo07} or
averaging operator \cite{GubKol14}. In these cases, the identities of $A^{(T)}$ may be obtained by
means of categoric procedures (black and white Manin products of operads \cite{GK94}).

The purpose of this note is to consider the case when $T$ is a derivation (or generalized derivation).
It is well-known \cite{DzhLowf09}
that a commutative (and  associative) algebra $A$ with a derivation $d$ induces Novikov
algebra structure on $A^{(d)}$,
assuming $(a\succ b)=(b\prec a)$. Conversely, if an identity holds on $A^{(d)}$ for an arbitrary
commutative algebra $A$ with a derivation $d$ then this identity is a consequence of Novikov
identities.

In this paper, we generalize this observation for an arbitrary variety $\mathop {\fam0 Var}\nolimits $ of algebras. Namely,
if an identity $f$ holds on $A^{(d)}$ for every $\mathop {\fam0 Var}\nolimits $-algebra $A$ with a derivation $d$
then we say $f$ is a derived identity of $\mathop {\fam0 Var}\nolimits $. For $\mathop {\fam0 Var}\nolimits = \mathop {\fam0 As }\nolimits$, some derived identities were
found in \cite{Lod10}. We show that for a multi-linear variety $\mathop {\fam0 Var}\nolimits $ the set of derived identities
coincides with the set of relations on the operad $\mathop {\fam0 Var}\nolimits\circ \mathop {\fam0 Nov}\nolimits$, where
$\mathop {\fam0 Var}\nolimits $ and $\mathop {\fam0 Nov}\nolimits $ are the operads governing the varieties $\mathop {\fam0 Var}\nolimits $ and $\mathop {\fam0 Nov}\nolimits$, respectively,
$\circ $ is the Manin white product of operads.

Calculation of Manin products is a relatively simple linear algebra problem, it is based on finding
intersections of vector spaces. Therefore, our result provides an easy way for finding a complete list
of derived identities for an arbitrary binary operad.

\section{White product of operads}

Suppose $\mathop {\fam0 Var}\nolimits $ is a multi-linear variety of algebras, i.e.,
a class of all algebras that satisfy a given family of multi-linear
identities (over a field $\Bbbk $ of characteristic zero, every variety is multi-linear).
Fix a countable set of variables $X=\{x_1,x_2,\dots \}$ and denote
\[
 \mathop {\fam0 Var}\nolimits(n) = M_n(X)/M_n(X)\cap T_{\mathop {\fam0 Var}\nolimits}(X),
\]
where $M_n(X)$ is the space of all (non-associative) multi-linear polynomials of degree $n$
in $x_1,\dots, x_n$, $T_{\mathop {\fam0 Var}\nolimits}(X)$ is the T-ideal of all identities that hold in $\mathop {\fam0 Var}\nolimits $.

The collection of spaces $(\mathop {\fam0 Var}\nolimits(n))_{n\ge 1}$ forms a symmetric operad relative to the natural
composition rule and symmetric group action (see, e.g., \cite{LodVal08}). We will denote this operad by the same symbol $\mathop {\fam0 Var}\nolimits $.
Every algebra $A\in \mathop {\fam0 Var}\nolimits $ may be considered as a morphism of multi-categories
$\mathop {\fam0 Var}\nolimits \to \mathop {\fam0 Hom}\nolimits_\Bbbk$, where $\mathop {\fam0 Hom}\nolimits_\Bbbk $ stands for the multi-category of linear spaces
over $\Bbbk $ and multi-linear maps.
The operad $\mathop {\fam0 Var}\nolimits $ is a {\em binary} one, i.e., it is generated (as a symmetric operad)
by the elements of $\mathop {\fam0 Var}\nolimits(2)$.

\begin{example}\label{exmp:Lie}
The operad $\mathop {\fam0 Lie}\nolimits $ governing the variety of Lie algebras
is generated by 1-dimensional space $\mathop {\fam0 Lie}\nolimits(2)=\Bbbk\mu $,
$\mu^{(12)}=-\mu$. The space $\mathop {\fam0 Lie}\nolimits(1)$ is also 1-dimensional, it is spanned by
the identity $\mathop {\fam0 id}\nolimits $ of the operad.
If we identify $\mathop {\fam0 id}\nolimits $ with $x_1$ and $\mu $ with $[x_1x_2]$ then
\[
 \mu(\mathop {\fam0 id}\nolimits, \mu) = [x_1[x_2x_3]], \quad \mu(\mu,\mathop {\fam0 id}\nolimits ) = [[x_1x_2]x_3],
\]
so the Jacobi identity may be expressed as
\[
 \mu(\mathop {\fam0 id}\nolimits,\mu) - \mu(\mathop {\fam0 id}\nolimits,\mu)^{(12)} = \mu(\mu, \mathop {\fam0 id}\nolimits).
\]
\end{example}

\begin{example}\label{exmp:As}
 The operad $\mathop {\fam0 As }\nolimits $ governing the variety of associative algebras
 is generated by 2-dimensional space $\mathop {\fam0 As }\nolimits(2)$ spanned by $\nu=(x_1x_2)$
 and $\nu^{(12)}=(x_2x_1)$. Associativity relations form an $S_3$-submodule
 in $M_3(X)$ spanned by
 \[
  \nu(\nu,\mathop {\fam0 id}\nolimits) = \nu(\mathop {\fam0 id}\nolimits,\nu).
 \]
\end{example}

\begin{example}\label{exmp:Nov}
{\em Novikov algebra} is a linear space with a multiplication
satisfying the following axioms:
\begin{gather}
 (x_1x_2)x_3 - x_1(x_2x_3) = (x_2x_1)x_3 - x_2(x_1x_3)   \label{eq:LSymm},\\
 (x_1x_2)x_3 = (x_1x_3)x_2. \label{eq:RComm}
\end{gather}
The corresponding operad is generated by 2-dimensional $\mathop {\fam0 Nov}\nolimits(2)$
spanned by $\nu = (x_1x_2)$ and $\nu^{(12)}=(x_2x_1)$. Defining
identities of the variety $\mathop {\fam0 Nov}\nolimits $ may be expressed as
\[
\begin{gathered}
\nu(\nu, \mathop {\fam0 id}\nolimits) - \nu(\mathop {\fam0 id}\nolimits ,\nu) = \nu(\nu^{(12)}, \mathop {\fam0 id}\nolimits ) - \nu^{(12)}(\nu, \mathop {\fam0 id}\nolimits)^{(23)}, \\
\nu(\nu, \mathop {\fam0 id}\nolimits) = \nu(\nu, \mathop {\fam0 id}\nolimits)^{(23)}.
\end{gathered}
\]
\end{example}

Let $\mathop {\fam0 Var}\nolimits_1$ and $\mathop {\fam0 Var}\nolimits_2$ be two  operads. Then the family of spaces
$(\mathop {\fam0 Var}\nolimits_1(n)\otimes \mathop {\fam0 Var}\nolimits_2(n))_{n\ge 1}$ is an operad relative to
the natural (componentwise) composition and symmetric group action
(known as the {\em Hadamard product} of $\mathop {\fam0 Var}\nolimits_1$ and $\mathop {\fam0 Var}\nolimits_2$).
Even if $\mathop {\fam0 Var}\nolimits_1$ and $\mathop {\fam0 Var}\nolimits_2$ were binary operads, their Hadamard product
$\mathop {\fam0 Var}\nolimits_1\otimes \mathop {\fam0 Var}\nolimits_2$
 may  be non-binary. The sub-operad of $\mathop {\fam0 Var}\nolimits_1\otimes \mathop {\fam0 Var}\nolimits_2$ generated by
 $\mathop {\fam0 Var}\nolimits_1(2)\otimes \mathop {\fam0 Var}\nolimits_2(2)$ is known as {\em Manin white product} of $\mathop {\fam0 Var}\nolimits_1$ and $\mathop {\fam0 Var}\nolimits_2$,
 it is denoted $\mathop {\fam0 Var}\nolimits_1\circ \mathop {\fam0 Var}\nolimits_2$ \cite{GK94}.

\begin{example}\label{exmp:Lie-Nov}
 The operad $\mathop {\fam0 Lie}\nolimits \circ \mathop {\fam0 Nov}\nolimits$ is isomorphic to the operad governing the class of all
 algebras ({\em magmatic} operad).
\end{example}

Indeed, both $\mathop {\fam0 Lie}\nolimits $ and $\mathop {\fam0 Nov}\nolimits $ are quadratic operad, and so is $\mathop {\fam0 Lie}\nolimits\circ \mathop {\fam0 Nov}\nolimits$ \cite{GK94}.
Let $\mu $ and $\nu $ be the generators of $\mathop {\fam0 Lie}\nolimits$ and $\mathop {\fam0 Nov}\nolimits$.
It is enough to find the defining identities of $\mathop {\fam0 Lie}\nolimits \circ \mathop {\fam0 Nov}\nolimits$ that are quadratic with
respect to $\mu $ and $\nu $.

Identify $\mu\otimes \nu $ with $[x_1\prec x_2]$, then $\mu\otimes \nu^{(12)} = -(\mu\otimes \nu)^{(12)}$
corresponds to $-[x_2\prec x_1]$. Hence, $(\mathop {\fam0 Lie}\nolimits\circ \mathop {\fam0 Nov}\nolimits)(n)$ is an image of the space $M_n(X)$
of all multi-linear non-associative polynomials of degree $n$ in $X=\{x_1,x_2,\dots \}$ relative to the
operation $[\cdot \prec \cdot ]$. Calculating the  compositions
$\mu(\mu,\mathop {\fam0 id}\nolimits)\otimes \nu(\nu,\mathop {\fam0 id}\nolimits)$ and $\mu(\mathop {\fam0 id}\nolimits,\mu)\otimes \nu(\mathop {\fam0 id}\nolimits,\nu)$, we obtain
\[
\begin{gathered}
m_1= [[x_1\prec x_2]\prec x_3] = [[x_1x_2]x_3]\otimes (x_1x_2)x_3, \\
m_2= [x_1\prec [x_2\prec x_3]] = [x_1[x_2x_3]]\otimes x_1(x_2x_3),
\end{gathered}
\]
It remains to find the intersection of the $S_3$-submodule generated by $m_1$ and $m_2$
in $M_3(X)\otimes M_3(X)$ with the kernel of the projection
$M_3(X)\otimes M_3(X)\to \mathop {\fam0 Lie}\nolimits(3)\otimes \mathop {\fam0 Nov}\nolimits(3)$. Straightforward calculation
shows the intersection is zero. Hence, the operation $[\cdot \prec \cdot]$
satisfies no identities.

\begin{example}\label{exmp:As-Nov}
 The operad $\mathop {\fam0 As }\nolimits \circ \mathop {\fam0 Nov}\nolimits$ generated by 4-dimensional space $(\mathop {\fam0 As }\nolimits\circ \mathop {\fam0 Nov}\nolimits)(2)$
 spanned by $(x_1\succ x_2)$, $(x_2\succ x_1)$, $(x_1\prec x_2)$, $(x_2\prec x_1)$
 relative to the following identities:
 %
 \begin{gather}
  (x_1\succ x_2)\prec x_3 - x_1\succ (x_2\prec x_3)=0, \label{eq:L-alg} \\
  (x_1\prec x_2)\prec x_3 -x_1\prec (x_2\succ x_3)+ (x_1\prec x_2)\succ x_3 - x_1\succ (x_2\succ x_3)=0.
								  \label{eq:LodDer}
 \end{gather}
\end{example}

This result may be checked with a straightforward computation, as in Example \ref{exmp:Lie-Nov}.
Namely, one has to find the intersection of the $S_3$-submodule in $M_3(X)\otimes M_3(X)$
generated by
\[
\begin{gathered}
 (x_1\prec x_2)\prec x_3 = (x_1x_2)x_3\otimes (x_1x_2)x_3, \\
 x_1\succ (x_2\succ x_3) = x_1(x_2x_3)\otimes (x_3x_2)x_1, \\
 (x_1\succ x_2)\prec x_3 = (x_1x_2)x_3\otimes (x_2x_1)x_3, \\
 x_1\succ (x_2\prec x_3) = x_1(x_2x_3)\otimes (x_2x_3)x_1, \\
 (x_1\prec x_2)\succ x_3 = (x_1x_2)x_3 \otimes x_3(x_1x_2), \\
 x_1\prec (x_2\succ x_3) = x_1(x_2x_3)\otimes x_1(x_3x_2), \\
 x_1\prec (x_2\prec x_3) = x_1(x_2x_3)\otimes x_1(x_2x_3), \\
 (x_1\succ x_2)\succ x_3 = (x_1x_2)x_3\otimes x_3(x_2x_1)
\end{gathered}
\]
with the kernel of the projection $M_3(X)\otimes M_3(X)\to \mathop {\fam0 As }\nolimits(3)\otimes \mathop {\fam0 Nov}\nolimits(3)$.

\section{Derived identities}

Given a (non-associative) algebra $A$ with a derivation $d$, denote by
$A^{(d)}$ the same linear space $A$ considered as a system with two binary
linear operations of multiplication  $(\cdot \prec \cdot)$
and $(\cdot \succ \cdot)$ defined by
\[
 x\prec y = xd(y), \quad x\succ y = d(x)y,\quad x,y\in A.
\]

Let $\mathop {\fam0 Var}\nolimits $ be a multi-linear variety of algebras. As above, denote by $T_{\mathop {\fam0 Var}\nolimits}(X)$
the T-ideal of all identities in a set of variables $X$ that hold on $\mathop {\fam0 Var}\nolimits $.
A non-associative polynomial $f(x_1,\dots, x_n)$ in two operations
of multiplication $\prec $ and $\succ $ is called
a {\em derived identity} of $\mathop {\fam0 Var}\nolimits $ if for every $A\in \mathop {\fam0 Var}\nolimits $ and for every derivation
$d\in \mathop {\fam0 Der}\nolimits (A)$ the algebra $A^{(d)}$ satisfies the identity $f(x_1,\dots ,x_n)=0$.
Obviously, the set of all derived identities is a T-ideal of the algebra of non-associative
polynomials in two operations.

For example,  $(x\succ y)=(y\prec x)$ is a derived identity of $\mathop {\fam0 Com}\nolimits $.
Moreover, the operation $(\cdot \prec \cdot )$ satisfies
the axioms of Novikov algebras \eqref{eq:LSymm} and \eqref{eq:RComm}.
It was actually shown in \cite{DzhLowf09} that the entire T-ideal of derived
identities of $\mathop {\fam0 Com}\nolimits $ is generated by these identities.

For the variety of associative algebras, it was mentioned in
\cite{Lod10} that
\eqref{eq:L-alg} and \eqref{eq:LodDer} are derived identities of~$\mathop {\fam0 As }\nolimits $.

\begin{remark}[c.f. \cite{GuoKeig08}]
For a multi-linear variety $\mathop {\fam0 Var}\nolimits$, the
free differential $\mathop {\fam0 Var}\nolimits$-algebra generated by a set $X$
with one derivation is nothing but
$\mathop {\fam0 Var}\nolimits\<d^\omega X\>$, where $d^\omega X = \{d^s(x) \mid s\ge 0, x\in X\}$.
\end{remark}

Indeed, consider the free magmatic algebra $F(X;d)=M\<d^\omega X\>$ and define a linear map
$d: F(X;d)\to F(X;d)$ in such a way that $d(d^s(x))=d^{s+1}(x)$, $d(uv)=d(u)v+ud(v)$.
Since $\mathop {\fam0 Var}\nolimits $ is defined by multi-linear relations, the T-ideal $T_{\mathop {\fam0 Var}\nolimits} (d^\omega X)$
is $d$-invariant. Hence, $U=F(X;d)/T_{\mathop {\fam0 Var}\nolimits}(d^\omega X)$ is a $\mathop {\fam0 Var}\nolimits $-algebra with a derivation~$d$.
Since all relations from $T_{\mathop {\fam0 Var}\nolimits}(d^\omega X)$ hold in every differential $\mathop {\fam0 Var}\nolimits$-algebra,
$U$ is free.

Let $\mathcal N_{\mathop {\fam0 Var}\nolimits }$ be the class of differential $\mathop {\fam0 Var}\nolimits $-algebras with one locally nilpotent
derivation, i.e., for every $A\in \mathcal N_{\mathop {\fam0 Var}\nolimits}$ and for every $a\in A$ there
exists $n\ge 1$ such that $d^n(a)=0$.

The following statement shows why the variety generated by $\mathcal N_{\mathop {\fam0 Var}\nolimits }$ coincides
with the class of all differential $\mathop {\fam0 Var}\nolimits $-algebras.

\begin{lemma}\label{lem:LocNilp}
Suppose $f=f(x_1,\dots, x_n)$ is a multi-linear identity that
holds on $A^{(d)}$ for all $A\in \mathcal N_{\mathop {\fam0 Var}\nolimits }$.
Then $f$ is a derived identity of $\mathop {\fam0 Var}\nolimits $.
\end{lemma}

\begin{proof}
Consider the free differential $\mathop {\fam0 Var}\nolimits$-algebra $U_n$ generated
by the set $X=\{x_1,x_2,\dots \}$
with one derivation $d$ modulo defining relations $d^{n}(x)=0$, $x\in X$.

Then $U_n\in \mathcal N_{\mathop {\fam0 Var}\nolimits }$. As a differential $\mathop {\fam0 Var}\nolimits$-algebra, $U_n$ is a homomorphic
image of the free magma $M\<d^{\omega }X\>$.
Denote by $I_n$ the kernel of the homomorphism $M\<d^{\omega }X\> \to U_n$.
As an ideal of $M\<d^{\omega }X\>$, $I_n$ is a sum of two ideals: $I_n=T_{\mathop {\fam0 Var}\nolimits}(d^\omega X)+N_n$,
where $N$ is generated by $d^{n+t}(x)$, $x\in X$, $t\ge 0$.
Note that the last relations form a $d$-invariant subset of $M\<d^\omega X\>$,
so the ideal $N_n$ is $d$-invariant.

If a polynomial $F\in M\<d^\omega X\>$ belongs to $I_n$ and
its degree in $d$ is less than $n$ then $F$ belongs to $T_{\mathop {\fam0 Var}\nolimits }(d^\omega X)$.
If $f=f(x_1,\dots , x_n)$ is a multi-linear identity in $U_n^{(d)}$
then the image $F$ of $f$ in $M\<d^\omega X\>$ has degree $n-1$ in $d$,
so $F\in T_{\mathop {\fam0 Var}\nolimits }(d^\omega )$.
Hence, $f$ is a derived identity of $\mathop {\fam0 Var}\nolimits $.
\end{proof}

\begin{theorem}\label{thm:WhiteNov->Derived}
If a multi-linear identity $f$ holds on the variety
governed by the operad $\mathop {\fam0 Var}\nolimits\circ \mathop {\fam0 Nov}\nolimits $ then
$f$ is a derived identity of $\mathop {\fam0 Var}\nolimits $.
\end{theorem}

\begin{proof}
For every $A\in \mathcal N_{\mathop {\fam0 Var}\nolimits }$ we may construct an algebra in the variety
$\mathop {\fam0 Var}\nolimits \circ \mathop {\fam0 Nov}\nolimits $ as follows. Consider the linear space $H$ spanned by elements $x^{(n)}$, $n\ge 0$, with
multiplication
\[
 x^{(n)}\cdot x^{(m)} = \binom{n+m-1}{n} x^{(n+m-1)}.
\]
This is a Novikov algebra (in characteristic 0, this is just the ordinary polynomial algebra).
Denote $\hat A = A\otimes \Bbbk [x]$ and define operations $\prec $ and $\succ $ on $\hat A$ by
\[
 (a\otimes f)\prec (b\otimes g) = ab\otimes f\cdot g,
 \quad
 (a\otimes f)\succ (b\otimes g) = ab\otimes g\cdot f.
\]
The algebra $\hat A$ obtained belongs to the variety governed by the operad $\mathop {\fam0 Var}\nolimits \circ \mathop {\fam0 Nov}\nolimits $.
There is an injective map
\begin{align}
 & A \to \hat A, \nonumber \\
 & a\mapsto \sum\limits_{s\ge 0} d^s(a)\otimes x^{(s)}, \label{eq:A-to-hatA}
\end{align}
which preserves operations $\prec $ and $\succ $. Indeed, apply \eqref{eq:A-to-hatA} to
$a\prec b$, $a,b\in A$:
\begin{multline}\nonumber
(a\prec b)=ad(b) \mapsto \sum\limits_{s\ge 0}  d^s(ad(b))\otimes x^{(s)}
= \sum\limits_{s,t\ge 0} \binom{s+t}{s} d^s(a)d^{t+1}(b)\otimes x^{(s+t)} \\
= \sum\limits_{s,t\ge 0}  d^s(a)d^{t+1}(b) \otimes x^{(s)}\cdot x^{(t+1)}
=
\left(\sum\limits_{s\ge 0} d^s(a)\otimes x^{(s)}\right)
\prec
\left(\sum\limits_{t\ge 0} d^t(a)\otimes x^{(t)}\right).
\end{multline}
Therefore, $A^{(d)}$ is a subalgebra of a $(\mathop {\fam0 Var}\nolimits\circ \mathop {\fam0 Nov}\nolimits )$-algebra,
so all
defining identities of $\mathop {\fam0 Var}\nolimits\circ \mathop {\fam0 Nov}\nolimits $ hold on $A^{(d)}$.
Lemma \ref{lem:LocNilp} completes the proof.
\end{proof}

\begin{theorem}\label{thm:Converse}
Let $f$ be a  multi-linear derived identity of $\mathop {\fam0 Var}\nolimits $. Then $f$ holds on all ($\mathop {\fam0 Var}\nolimits\circ\mathop {\fam0 Nov}\nolimits$)-algebras.
\end{theorem}

\begin{proof}
Let $X=\{x_1,x_2,\dots \}$, and let $f(x_1,\dots , x_n)$ be a non-associative multi-linear polynomial
in two operations of multiplication $\prec$ and $ \succ$.
By definition, $f$ is an identity of the algebra $\mathop {\fam0 Var}\nolimits\<X\>\otimes \mathop {\fam0 Nov}\nolimits\<X\>$
with operations
\[
(u\otimes f)\prec (v\otimes g)= (uv\otimes fg),
\quad
(u\otimes f)\succ (v\otimes g)= (uv\otimes gf)
\]
if and only if $f$ is a defining identity of $\mathop {\fam0 Var}\nolimits\circ \mathop {\fam0 Nov}\nolimits$.

It is well-known that $\mathop {\fam0 Nov}\nolimits\<X\>$ is a subalgebra of $C^{(d)}$ for
the free differential commutative algebra $C$ generated by $X$ with a derivation~$d$
\cite{DzhLowf09}. Therefore,
$\mathop {\fam0 Var}\nolimits\<X\>\otimes \mathop {\fam0 Nov}\nolimits\<X\>$ is a subalgebra of $(\mathop {\fam0 Var}\nolimits\<X\>\otimes C)^{(d)}$,
where $d(u\otimes f)=u\otimes d(f)$ for $u\in \mathop {\fam0 Var}\nolimits\<X\>$, $f\in C$.
Hence, $\mathop {\fam0 Var}\nolimits\<X\>\otimes \mathop {\fam0 Nov}\nolimits\<X\>$ is embedded into a differential
$\mathop {\fam0 Var}\nolimits $-algebra.
\end{proof}

\begin{remark}
In this paper, we consider algebras with one binary operation, so $\mathop {\fam0 Var}\nolimits $ has only one generator.
However, Theorems~\ref{thm:WhiteNov->Derived}, \ref{thm:Converse} are easy to generalize for algebras
with multiple binary operations, i.e., for an arbitrary binary operad $\mathop {\fam0 Var}\nolimits $.
\end{remark}

\begin{remark}
In the definition of a derived identity, a derivation $d$ of an algebra $A$
may be replaced with {\em generalized derivation},
i.e., a linear map $D: A\to A$ such that
\[
D(ab)=D(a)b+aD(b)+\lambda ab, \quad a,b\in A,
\]
where $\lambda $ is a fixed scalar in $\Bbbk $. Indeed, it is enough to note that
 $D-\lambda \mathop {\fam0 id}\nolimits$ is an ordinary derivation.
\end{remark}

redpajama_set_name: RedPajamaArXiv


text:
Francisco Franco de Sousa (Funchal, 9 de outubro de 1885 — Lisboa, 15 de fevereiro de 1955) foi um escultor português.

Pertence à primeira geração de artistas modernistas portugueses e destaca-se, segundo José-Augusto França, como o maior escultor português da década de 1920. A sua obra desse período é um marco na renovação da escultura nacional. O seu o Monumento a Gonçalves Zarco, 1927, assinala uma inflexão e tornar-se-ia numa referência para as direções da escultura pública portuguesa a partir dos anos de 1930; Francisco Franco foi, então, um dos autores mais solicitados para a realização da estatuária oficial do Estado Novo.

Um dos bustos de António Oliveira Salazar, da sua autoria que estava à guarda do Estado português, em 2020 foi dado como desaparecido.

Biografia

Nasce no Funchal a 9 de outubro de 1885. Trabalha com o seu pai, mestre do Ensino Técnico, em carpintaria e marcenaria, recebendo dele as primeiras lições de desenho. Aos 15 anos de idade parte para Lisboa com o seu irmão, Henrique Franco. Frequenta a Escola de Belas Artes de Lisboa entre 1902 e 1909, onde é aluno de Simões de Almeida (tio). Em 1909 interrompe os estudos e parte para Paris como bolseiro do Estado, através do Legado Valmor; convive com outros artistas portugueses então residentes nessa cidade, entre os quais Amadeo de Souza Cardoso, Eduardo Viana, Manuel Jardim, José Pacheko e, sobretudo, Santa-Rita Pintor, o seu principal companheiro nesses anos. Interessa-se de modo particular pela obra de Auguste Rodin.

Em 1911 viaja pelos Países Baixos. No ano imediato a sua bolsa, tal como a de outros colegas, é interrompida devido a desentendimentos de caráter político, levando-o a regressar a Lisboa, mas em breve está de novo em Paris. Em 1914, após a eclosão da I Guerra Mundial, fixa-se no Funchal, onde realiza diversas esculturas, entre as quais quatro monumentos públicos.

Em 1919 (ou 1921, segundo algumas fontes consultadas) volta para Paris como bolseiro, convivendo com Diogo de Macedo, Dordio Gomes, Heitor Cramez e Modigliani. É uma fase de trabalho intenso: executa diversas esculturas, desenha e grava em madeira e ponta-seca, realiza algumas monotipias. Expõe no Salon d'Automne em 1921 e na Société nationale des beaux-arts, da qual é nomeado «Associé» com privilégio «Hors Concours». Em 1923 desloca-se a Lisboa para organizar a exposição Cinco Independentes (com Diogo de Macedo, Dordio Gomes, Henrique Franco e Alfredo Miguéis). No ano imediato viaja a Paris, expondo no Salon a sua escultura Semeador.

Em 1925 parte para Roma como pensionista do Estado, visitando, com Dordio Gomes, diversas cidades italianas entre as quais Turim, Veneza, Florença, Pompeia, Assis. Regressa ao Funchal em 1926, começando a trabalhar na estátua de Gonçalves Zarco (completada em 1927); essa obra é exposta na Avenida da Liberdade (1928) e, uma cópia, na Feira de Sevilha (1929). Em 1930 o monumento é inaugurado na cidade do Funchal. Nesse mesmo ano Francisco Franco participa no I Salão dos Independentes (SNBA, Lisboa), onde se faz um balanço do modernismo à escala nacional.

A partir da década de 1930 Francisco Franco irá ser um dos autores mais solicitados para a realização da estatuária do Estado Novo, tornando-se numa figura de referência a nível nacional.

Em 1931 realiza a estátua do Infante D. Henrique para a Exposição Colonial de Paris (Vincennes). Em 1934 candidata-se ao lugar de professor de desenho da Escola de Belas-Artes de Lisboa, sendo preterido em favor de Leopoldo de Almeida. Nesse mesmo ano executa um primeiro retrato de Salazar, que expõe na I Exposição de Arte Moderna do S.P.N. (1935), e é agraciado com o grau de Comendador da Ordem Militar de Sant'Iago da Espada, a 5 de outubro. Realiza a estátua de Salazar que será apresentada na Exposição Internacional de Paris de 1937. Durante dois anos trabalha na estátua equestre de D. João IV para o terreiro do Paço Ducal de Vila Viçosa. Recebe a Medalha de Ouro de Mérito do Município do Funchal.

Os últimos anos revelam um abrandamento da sua obra, em grande parte resultante do atropelamento de que foi vítima junto ao seu estúdio e do qual nunca recuperou por completo. O monumento ao Cristo Rei, em Almada, de que apenas realizou o esboceto, é revelador desse esmorecimento. Viria a falecer a 15 de fevereiro de 1955 aos 69 anos de idade.

No ano de 1979, a Escola Secundária Francisco Franco, localizada no Funchal, escolheu-o como seu patrono passando assim a constar o seu nome na designação da mesma escola.

Obra

A obra de Francisco Franco pertencente ao seu primeiro período parisiense (1910-1911), é ainda tendencialmente convencional; da sua fixação na ilha da Madeira, em 1914, irão resultar esculturas "com bom valor dramático" onde se sentem ecos da obra de Rodin. No início do segundo período parisiense, ao mesmo tempo que realiza notáveis gravuras em madeira ou a ponta-seca, Franco abandona, a pouco e pouco, a influência de Rodin (ainda presente em O semeador, 1924), modelando obras como Busto de Manuel Jardim, 1921 – um poderoso retrato interior e a escultura mais significativa desse ano –, ou Torso de Mulher, 1922, duas peças peças que contam como as primeiras de uma "escultura portuguesa modernizada, libertada em valores expressionistas".

Torso de Mulher foi exposto no Salon d'Automne e assinala a opção por valores mais classizantes próprios do início da década de 1920. "A arquitetura das massas bem equilibradas, quase hierarquicamente entendidas como puros valores escultóricos, aproximam-no  de Bourdelle que será o seu mestre entre os contemporâneos, no entanto os cortes que fragmentam a peça, sobretudo o da cabeça, evocando uma memória arqueológica, têm ainda presente a lição de Rodin".

Em 1927 realiza um trabalho que irá marcar a estatuária portuguesa entre as décadas de 1930 e 1960: o monumento a Gonçalves Zarco, Funchal. Inspirada nos painéis de Nuno Gonçalves, esta obra fixa o cânone da «idade de ouro da escultura portuguesa» – assim denominada por António Ferro em 1949. Caracterizado por um «classicismo austero», esse modelo seria longamente explorado (pelo próprio e por autores mais jovens), garantindo a Franco uma carreira de escultor oficial do Estado Novo durante mais de um quarto de século. O abandono do lirismo expressivo inicial estender-se-ia à série de estátuas públicas que vai do seu Infante D. Henrique (1931) ao Cristo-Rei monumental, "que esboçou antes de morrer e seria infelizmente erguido numa colina de Almada em 59". Esse «naturalismo clássico» ficaria claramente expresso na sua D. Leonor (1935; Caldas da Rainha), D. Dinis (1943, Universidade de Coimbra), Bispo D. Miguel (1950; Lamego), Oliveira Salazar (1937), ou na notável estátua equestre de D. João IV (1940; Vila Viçosa). Realizada em 1940 no âmbito das comemorações dos centenários, uma cópia em gesso da estátua de D. João IV foi apresentada na Exposição do Mundo Português; nas palavras de José-Augusto França, trata-se de uma obra "de grande dignidade na sua cuidada elaboração, síntese de bons exemplos bem unificados, velasquenha por necessidade e funcionando como se pretendia, e com escala suficiente, no vasto terreiro do Paço Ducal".

Incompleta talvez para as exigências do próprio autor, a obra plástica de Franco será, segundo Diogo de Macedo, um reflexo dos constrangimentos do tempo em que viveu. A imposição dos temas históricos na sua vasta produção não lhe permitiu realizar plenamente o que sonhara na mocidade. Para Francisco Franco, "não houve tempo nem oportunidade para esculpir para sua satisfação a obra-prima de total intimidade de que era capaz. Escravizado às circunstâncias que o destino ordena […], só nas pequenas obras de criação afetiva pôde fazer declarações desse desejo", como acontece, por exemplo, no Busto de Manuel Jardim ou na Rapariga polaca (1921), obras parisienses que são excelentes exemplos de uma arte de penetração psicológica e de um superior domínio formal.

Algumas obras (escultura)

 Cabeça de Velho, 1909, Col. Fundação Calouste Gulbenkian, Lisboa.
 Rapariga Viloa, 1914.
 Busto simbólico de aviador (comemorando o voo de Gago Coutinho à ilha da Madeira em 1921).
 Busto monumental de Gonçalves Zarco, Funchal.
 Rapariga polaca, 1921, Col. Museu do Chiado, Lisboa.
 Busto de Manuel Jardim, 1921, Col. Museu do Chiado.
 Torso de mulher, 1922, Col. Museu do Chiado.
 Adão e Eva, 1922-23, Col. Museu do Chiado.
 Retrato do Gil, 1923, Col. Museu do Chiado.
 Rapariga francesa, 1923, Col. Museu do Chiado.
 O semeador, 1924, Funchal.
 Monumento a João Gonçalves Zarco, 1927, Funchal.
 Estátua do Infante D. Henrique, 1931 (para a exposição em Vincennes).
 Lusitânia, 1931, baixo-relevo, Tribunal do Comércio.
 Busto do Prof. Doutor Oliveira Salazar, 1934.
 Estátua da Rainha D. Leonor, 1935, Caldas da Rainha.
 Relevo para a Casa da Moeda, 1936, Lisboa.
 Estátua ao Prof. Doutor Oliveira Salazar, para a Exposição de Paris, 1937.
 Apostolado, 1938, friso sobre o pórtico de entrada, Igreja de Nossa Senhora do Rosário de Fátima, Lisboa.
 Estátua equestre de D. João IV, 1940, erguida em frente ao Paço Ducal de Vila Viçosa.
 Estátua de D. Dinis, 1943, Universidade de Coimbra.
 Estátua de D. João III, 1948, Universidade de Coimbra.
 Estátua do bispo de Lamego D. Miguel de Portugal, 1950, Lamego.
 Monumento ao Cristo Rei, Almada (inaugurado postumamente).

Ligações externas

 "Visita Guiada - Museu Henrique e Francisco Franco, Funchal", episódio 18, 15 de outubro de 2018, temporada 8, programa de Paula Moura Pinheiro, na RTP

Naturais do Funchal
Escultores de Portugal
Monárquicos de Portugal
Artistas plásticos da Madeira
Alumni da Escola Superior de Belas-Artes de Lisboa
Comendadores da Ordem Militar de Sant'Iago da Espada

redpajama_set_name: RedPajamaWikipedia


text:
Q: Android countdown timer gradually losing time I'm trying to create a countdown timer optimised for tablets, and it runs fine except for the slight detail that it gradually loses time, over time: over 5 mins it loses probably 1-2 seconds, when testing on my Xoom.
Here is my onTick code:
        public void onTick(long millisUntilFinished) {
        // TODO Auto-generated method stub
        currentTime = System.currentTimeMillis();
        currentCount = startTime - currentTime;
        //currentCount = millisUntilFinished;
        displayCorrectTime();
    }

Initially I started with just using the 'millisUntilFinished', when I noticed the time was slipping. I then read around and tried to get it to sync up better by using the difference between the start time and 'now' to inform the timer. Unfortunately, both methods seem to lose time (I don't really understand how this second method can lose time, but nevertheless).
Any suggestions?

A: That's normal, the System.currentTimeMillis() has that inprecision. You can try System.nanoTime() for a (theoretical) better precision. At any rate, you need to compensate for the time it takes to execute the  displayCorrectTime() call.
Better solution would be to use :
http://developer.android.com/reference/java/util/Timer.html

A: new CountDownTimer(60000, 1000) {
        long startTime = System.currentTimeMillis();
        public void onTick(long millisUntilFinished) {
            long timePassed = System.currentTimeMillis() - startTime;
            pb.setProgress(60000-(int)timePassed);
            t.setText(String.valueOf((60000-timePassed)/1000));
        }

        public void onFinish() {
            Intent intent = new Intent(GameActivity.this, SummaryActivity.class);
            startActivity(intent);
        }
    }.start();

This worked for me. It's a count down timer of 1 minute with a progressbar display.

A: Here's a simple function I wrote a very long time ago for Java server apps:
/// <summary>
/// A static helper class the enables a thread to sleep for a specific period of
/// time, manage drift and supports clock changes.
/// </summary>
public static class ThreadHelper
{
    /// <summary>
    /// Enables a thread to sleep for a specific period of time, manage drift and
    /// supports clock changes. This function enables the user to break from the
    /// call.
    /// </summary>
    /// <param name="startTicks">A long value that specifies the time to start sleeping.</param>
    /// <param name="intervalTicks">The long value that specifies the duration to sleep for.</param>
    /// <param name="signalStop">A reference to a bool that enables the caller to
    /// break out of this call from another thread.</param>
    /// <returns>A long value that specifies the value the caller should use for the
    /// startTicks parameters on subsequent calls into this function.</returns>
    public static long Sleep(long startTicks, long intervalTicks, ref bool signalStop)
    {
        long finalTicks = startTicks + intervalTicks;

        while (finalTicks + intervalTicks < DateTime.Now.Ticks)
        {
            finalTicks += intervalTicks;
        }

        while (!signalStop && DateTime.Now.Ticks < finalTicks)
        {
            while (finalTicks - intervalTicks > DateTime.Now.Ticks)
            {
                finalTicks -= intervalTicks;
            }

            Thread.Sleep(1000);
        }
        return finalTicks;
    }
}

redpajama_set_name: RedPajamaStackExchange


Ever since I got here, I feel like I've been on a long-term vacation. I didn't attend summer school, so I had all summer to work, relax, and prepare for my travels. When I arrived in London, I returned to the mentality of school. School was on my mind because I know myself and I can easily get distracted. Especially since I'm in a foreign country where it's even easier to get distracted where there is so much to do. And sadly that mentality of focusing on school has wore off and I have been distracted lately.
My week here consists of one class a day, Monday-Thursday, BUT they are 3 hours each class. While we don't have any physical homework to do in many classes, we do have a lot of readings each week.
On the bright side, in October we get a 10 day break from classes to travel and do whatever we please, but of course before any break means a ton of work to be done.
This week I've been swamped with work. In just one week I had two papers due and a presentation, the following week I have another presentation due and a take home midterm exam. Then the week after that I have another paper due.
While I've been attending classes, doing my readings, and understanding the subjects, for some reason I have found it difficult to sit down and do these papers and presentations. Maybe because I rather be out exploring the city; but yet again I have to remind myself, it's called STUDYING abroad for a reason.
It will definitely pay off and after all of this is done, I luckily get a break where I'll be traveling to other countries and major cities! So be on the look out for posts about those adventures coming soon!


text:
Toward making people invisible to mosquitoes
EMBARGOED FOR RELEASE | September 09, 2013
Note to journalists: Please report that this research was presented at a meeting of the American Chemical Society.
A press conference on this topic will be held on Monday, Sept. 9, at 11:30 a.m. in the ACS Press Center, Room 211, in the Indiana Convention Center. Reporters can attend in person or access live audio and video of the event and ask questions at www.ustream.tv/channel/acslive.
INDIANAPOLIS, Sept. 9, 2013 — In an advance toward providing mosquito-plagued people, pets and livestock with an invisibility cloak against these blood-sucking insects, scientists today described discovery of substances that occur naturally on human skin and block mosquitoes' ability to smell and target their victims.
The presentation was among almost 7,000 scheduled here this week at the 246th National Meeting & Exposition of the American Chemical Society, the world's largest scientific society. Thousands of scientists and others are expected to attend the sessions, being held in the Indiana Convention Center and downtown hotels.
Ulrich Bernier, Ph.D., who gave the talk, cited the pressing need for better ways to combat mosquitoes. Far from being just a nuisance, mosquitoes are more deadly to humans than any other animal. Their bites transmit malaria and other diseases that kill an estimated 1 million people around the world each year. In the United States, mosquitoes spread rare types of encephalitis, an inflammation of the brain. They also transmit heartworms to pet dogs and cats.
"Repellents have been the mainstay for preventing mosquito bites," said Bernier. "The most widely used repellant, DEET, is quite effective and has been in use for a long time. However, some people don't like the feel or the smell of DEET. We are exploring a different approach, with substances that impair the mosquito's sense of smell. If a mosquito can't sense that dinner is ready, there will be no buzzing, no landing and no bite."
Female mosquitoes, which suck blood to obtain a protein needed to produce fertile eggs, can smell people from over 100 feet away. The Mosquito and Fly Unit at the U.S. Department of Agriculture's Agricultural Research Service-Center for Medical, Agricultural, and Veterinary Entomology in Gainesville, Fla., has been doing research on mosquito repellents since the 1940s. In the 1990s, they accumulated information on substances secreted through the human skin or formed by bacteria on the skin that make some people more attractive to mosquitoes than others.
A person's scent, Bernier explained, comes from hundreds of compounds on the skin, many emitted through sweat and others produced by bacteria. To identify which of these attract mosquitoes, Bernier and colleagues used a special mosquito cage divided by a screen. They sprayed various substances into one side of the cage, and documented the effects in attracting mosquitoes. Some compounds, like lactic acid — a common component of human sweat — were definite mosquito lures, drawing 90 percent of the mosquitoes to the screen. With other compounds, however, many of the mosquitoes didn't even take flight or seemed confused.
"If you put your hand in a cage of mosquitoes where we have released some of these inhibitors, almost all just sit on the back wall and don't even recognize that the hand is in there. We call that anosmia or hyposmia, the inability to sense smells or a reduced ability to sense smells," explained Bernier.
He said that a group of chemical compounds, including 1-methylpiperzine, block mosquitoes' sense of smell. This may help explain why mosquitoes fly toward some people but not others. The substances have a molecular architecture found in ingredients in dozens of medicines and other products. They appear suitable for use in cosmetics, lotions, clothing and other products that currently incorporate mosquito repellants.
Bernier's team cited funding from the U.S. Departments of Agriculture and Defense.
His presentation was part of a meeting symposium on biopesticides, which come from natural sources such as plants, animals, minerals and bacteria.
To automatically receive news releases from the American Chemical Society, contact newsroom@acs.org.
The American Chemical Society is a nonprofit organization chartered by the U.S. Congress. With more than 163,000 members, ACS is the world's largest scientific society and a global leader in providing access to chemistry-related research through its multiple databases, peer-reviewed journals and scientific conferences. Its main offices are in Washington, D.C., and Columbus, Ohio.
During the national meeting, Sept. 6-11, the contacts can be reached at 317-262-5907.
Michael Bernstein
m_bernstein@acs.org
m_woods@acs.org
Ulrich Bernier, Ph.D., uses a hands-on approach for testing substances that block the mosquito's sense of smell and ability to target humans.
Credit:Greg Allen, USDA-ARS
High-resolution image


Odakyu Line Yoyogi-koen 10 min.
New Properties!! Luxury High Rise Apt.in Aoyama, near "Outer Garden of Meiji Shrine"
New Properties!! New Apt. in Quiet Neighborhood of Setagaya, near "Nakameguro"
Ginza Line Gaienmae 9 min.
Luxury Terrace House in shibuya area.


I love how this album starts out with a southern gospel foot stomping "My Feet Are on the Rock." It's the kind of song that gets your blood a pumping and your praises running high. That's followed up by an equally powerful praise song "To the One." In fact, after previewing the album on iTunes on a few occasions I have found that I rated every track with 4 stars.
I listen to projects like this on multiple devices and typically mark those I really like randomly, so I know what to give a good listen to as I write a review. Apparently, over time, I marked each and every track on "Trial & Triumph" with four stars. I think what made a huge difference is that they can switch around the leads quite easily. They have a nice combination of vocals that makes for a good variety of ambiances from song to song.
Therefore, finding a few stand outs is very difficult, but overall, I was drawn to the tender tones of songs like "The Water" and "Near to Me." These two speak to the "weak and weary" ways which we struggle with each day, but the lyrics speak of "hope," "justice," and being filled with the holy spirit. These two in particular are very uplifting and beautiful songs.
For more on "Trial & Triumph" check out the official I Am The website at https://www.iamtheyband.com/.
This entry was posted on Monday, April 9th, 2018 at 7:39 pm and is filed under CD Reviews . You can follow any responses to this entry through the RSS 2.0 feed. Both comments and pings are currently closed.


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Voice for African Migrants
IOM provides 40,000 surgical gloves, 4,800 surgical masks other equipment to support Coronavirus response in China
The International Organization for Migration (IOM) is donating 40,000 surgical gloves, 4,800 high quality surgical masks and 2,375 isolation suits to protect frontline health personnel working tirelessly on the novel coronavirus (2019-nCoV) response in China.
The first batch of critical personal protective equipment (PPE) is due to land in Shanghai, China, on Sunday. A second batch of PPE will follow on the next available flights.
"This donation is a small but meaningful contribution as it shows IOM's appreciation for, solidarity with and confidence in the capable steps China continues to take to protect and support citizens, both nationals and foreign, within and beyond its borders," said IOM Director General António Vitorino.
The supplies will be directly donated to the Chinese government and then transported to health facilities in Wuhan and other locations which are currently experiencing PPE shortages due to unprecedented demand.
Though figures are being updated rapidly, as of yesterday, virtually all 28,276 confirmed cases of coronavirus were in China. All but one of the 565 deaths attributed to it were within China.
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PPE ensures that those on the frontline have the tools they need to treat those affected and mitigate the further spread of the virus.
Organizations including IOM have faced procurement challenges in part due to the problems freight forwarders face in prioritizing the shipment of emergency relief supplies amidst travel restrictions introduced by several countries.
At the global level, IOM is discussing with the World Health Organization (WHO) and other UN agencies ways and means to address risks of potential market disruption and price inflation in the manufacturing and distribution of PPEs.
At the country level, to ensure that IOM's efforts support China's national priorities and response strategies, IOM is liaising closely with the Chinese government, including the Department of International Trade and Economic Affairs of the Ministry of Commerce, and the United Nations System, on this task.
Last week, IOM called on all parties to work together to prevent the undue stigmatization of international travelers, in line with international health principles, and in support of the WHO's Emergency Committee cautioning against actions that promote stigma or discrimination
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"A responsible global approach to the battle to contain this novel coronavirus, one that is grounded in solidarity, humanitarian values, and knowledge rather than fear, is not one that is discriminatory, but one that unites nations and people to fight together to control this epidemic," said IOM DG Vitorino.
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Human trafficking: PJI urges proper trauma management for returnees
The Pathfinder Justice Initiative (PJI), a Non-Governmental Organisation, has called for proper trauma care for migrant returnees to prevent them from becoming vulnerable to subsequent trafficking.
Evon Benson-Idahosa, the Executive Director, PJI, made the call at a Rehabilitation Workshop for Providers Serving Survivors of Human Trafficking held in Benin on Thursday.
The workshop was organised by PJI and funded by INSighT- Building Capacity to deal with human trafficking and transit routes to Nigeria, Italy and Sweden.
Benson-Idahosa said that a majority of returnee-migrants usually undergo different traumatic situations and needed to be properly rehabilitated before being integrated back into the society. She noted that if the migrant returnees were not properly rehabilitated, they would not be able to put into good use any form of skills acquisition or empowerment received.
"Providers serving survivors should know how to handle traumatised victims because many of them, especially females, have been raped and have gone through horrible experiences during their trafficking journey.
"The providers should know that there are best practices in terms of handling trafficked victims; they need to use a survivor centred approach to prioritise the needs of the victims," she said.
She called on the government at all levels to partner more with NGOs on providing best traumatic care for returned migrants in the country.
How Nigerian-American police officer burst human trafficking syndicate in US
A retried Nigerian American Police officer, Samuel Balogun narrated how he burst a human trafficking syndicate that specialized in using minors for prostitution.
"My biggest accomplishment was bursting a human trafficking crime," Balogun said.
Giving details of how he executed the task, the dark skinned retired police officer said: " There was a guy that was using minors for prostitution on the internet. I have an accent and when I speak people know I am an African. So, I had to go undercover and had to call the guy on the internet. I said ' hey! what is going on, I am in town. I am a truck driver and I want some girls.' I asked how old? He said the younger they are, the more money. I said about 15 to 16 years. He said ok. I asked how many he could bring and he replied two. He said which hotel was I and I gave the name to him. He told me to hang up and he called back the hotel. He subsequently called me and asked if I was there and I said yes. He said he would be there in 20 minutes.
"We were waiting for him to come but he was smart too. He dropped the girls down the street and made them walk to the room. The girls asked how much I was ready to pay and wanted to take off their clothes but I said not yet. In the next room were officers listening to our conversation. When I make a signal, that means it is time for them to come in. but before you make the signal, you have to make sure they have mentioned the price, they have given the reason why they were there, so it doesn't look like you are entrapping them. When I made the signal, the officers burst in and arrested everybody including me.
Thereafter, Balogun said the police processed the girls and after that, "they said look, you are minors and we know somebody is pushing you to do this. Now we don't want to arrest you but tell us how to get to the boss. The girls cooperated and made as if they were leaving. When the man pulled up to pick them up, and that was how we arrested him. That stopped a lot of those crimes."
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Balogun said he was in Nigeria to bring his wealth of experience to bear on the disturbing security situation in the country. " I am trying to bring back my experience as a police officer in the states to Nigeria. When you look at the #endsars period, the performance of the police was something that hurt my feelings. How can we make it better? How can we make the police job something that people will look with respect and want to join?"
He hinted that his security firm is involved in training not only police officers but " I also train private security companies. I am in touch with a lot of private security companies in Nigeria. There is another concept which Nigeria is embracing right now.
"It is called community policing. In the states it is called neighbourhood policing or community policing. It works in a way that in every street, there would be a police officer that lives in that neighbourhood. You get to know the people and the people know you. In some apartments, they will give you a discount just for the police officer to be there because they know once a police officer is living there, the police car is outside and the crime level will reduce. People are more likely to talk to that officer because they know him. They are more able to tell him' hey we know who committed that crime.' For every crime, you need people to tell you what happened. You can have all the gadgets but if people are not talking, you can't solve the crime."
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He further said: "I am training police officers, security companies and executive protection. What my security company is doing is to free the police officers from attachment to chiefs, politicians and all that. We train civilians to represent those officers so that they can go back to the street and do their normal jobs. We have what we call executive protection/training. We have people that follow the president. We can train you on how to be efficient and sometimes using less force, description tactics."
Further expatiating on what his security firm does, the soft spoken officer said: "What my company is trying to do is to bring people to the table. We are trying to train companies that there is a better way of security where we can teach you how to defend yourself, how to prepare for any emergency, and how to use less force. I have a guy, a navy seal that worked for the United States of America. You will be amazed about what he can do. He can disarm you in a minute even when you come with AK 47. I am also bringing Hostage Negotiation, people that can talk to you when ransom has to be paid. In the US, we call it Hostage Negotiation. They can talk to these people, and know their psyche. It is a full package. When you come to my firm, you can see the whole spectrum and choose."
As a vastly travelled person, Blagun said: "I travel a lot and in all the African nations is where you see officers with AK 47. They said it is more intimidating. Criminals use AK 47 in America too but we still don't carry it. Is that the right weapon for the police officers, I leave that question open. "
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On the attitude of the Nigerian authorities his plans, he said: "I have talked to a lot of people in higher positions. In some places I don't want to mention, I have got good responses. My firm has done some things with certain private firms and the police. I have dealt with some highly placed security firms. So, this is not my first time here. We are looking at having training in Sheraton around July/August this year. It is going to be a big one. I am bringing a retired FBI agent, a navy seal, a retired marine , myself and may be two other officers.
"This is my country, I am proud of it. I am sad sometimes when you look at the security aspect of it. With my experience, I am trying to make it a better place. It has always been my passion to come back home. I am retired and don't really need to work again. My benefits are okay untill I die. But why die with all this experience when I can pass it to the next person."
Hundreds of thousands of people leave Britain due to pandemic
Hundreds of thousands of people have left Britain as a fallout of the pandemic on the economy, according to a study released yesterday.
There is an "unprecedented exodus" of workers born outside Britain, researchers at London's Economic Statistics Centre of Excellence said.
"It seems that much of the burden of job losses during the pandemic has fallen on non-UK workers and has manifested itself in return migration, rather than unemployment," said the authors.
The study is based on labour market data.
The trend was particularly notable in London, where one in five residents was born abroad.
The capital's population has fallen by 700,000, the study said, adding that nationwide, the figure could be more than 1.3 million.
If these numbers are accurate, this is the largest decline in Britain's population since World War II, according to the study.
No evidence suggests that similar numbers of British people who live abroad are returning to Britain.
However, this could be a temporary trend, the researchers said, noting that workers from abroad might return after the pandemic.
The British economy depends on workers from abroad and it is not only threatened by migration due to the pandemic.
Many industries fear the loss of skilled workers due to Britain's departure from the European Union and stricter migration laws.
A further trend in 2021 is also causing concern, described as a "baby bust" by consultancy PwC, which said many couples were postponing having children due to the uncertainty caused by the pandemic.
This could lead to the lowest birth rate since 1900, PwC said in early January.
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