Expectation of Log of a Cauchy-distributed Random Variable I found this in an article, but I cannot follow the step to get $\mathbb E[\log |a_{N,k}|]$. I'm quoting the paper:
Let $a_{N,k}$  be Cauchy-distributed random variables with parameter $N(k+1)$. 
The first moment does not exist, but there are some partial moments, for example
for $0 \leq s < 1$ we get:
$\mathbb E[|a_{N,k}|^s] = \frac{N(k+1)}{\pi}\int\limits_{-\infty}^{\infty}
\frac{|x|^s}{x^2+N^2(k+1)^2} dx$
$= \frac{1}{\pi}N^s(k+1)^s\Gamma\left( \frac{1}{2}+\frac{s}{2} \right)
\Gamma \left( \frac{1}{2} - \frac{s}{2} \right).$
Further:
$\mathbb E[\log|a_{N,k}|]=\log(N(k+1)).$
but I need a clue how to get to this.
 A: The idea is that $\dfrac{\mathrm d}{\mathrm ds} x^s=x^s\log x$ for every positive $x$ hence
$$
E[\log |a|]=\left.\frac{\mathrm d }{\mathrm ds}E[|a|^s]\right|_{s=0}.
$$
In your setting, there exists some positive $\alpha$, namely $\alpha=N(k+1)$, such that
$$
E[|a|^s]=\frac{\alpha^s}{\cos(\pi s)}.
$$
The derivative of the denominator at $s=0$ is $-\pi\sin(0)=0$ hence
$$
\left.\frac{\mathrm d }{\mathrm ds}E[|a|^s]\right|_{s=0}=\frac1{\cos(0)}\,\left.\frac{\mathrm d }{\mathrm ds}\alpha^s\right|_{s=0}=\log\alpha.
$$
A: $\displaystyle{\large%
? =\int_{0}^{\infty}{x^{s} \over x^{2} +  a^{2}}\,{\rm d}x}$
\begin{align}
2\pi{\rm i}\left(%
{\left\vert a\right\vert^{s}{\rm e}^{{\rm i}\pi s/2}
 \over
 2{\rm i}\left\vert a\right\vert}
+
{\left\vert a\right\vert^{s}{\rm e}^{-{\rm i}\pi s/2}
 \over
 -2{\rm i}\left\vert a\right\vert}
\right)
&=
\int_{-\infty}^{0}
{\left(-x\right)^{s}{\rm e}^{{\rm i}\pi s} \over x^{2} +  a^{2}}\,{\rm d}x
+
\int^{-\infty}_{0}
{\left(-x\right)^{s}{\rm e}^{-{\rm i}\pi s} \over x^{2} +  a^{2}}\,{\rm d}x
\\[3mm]&=
\int^{\infty}_{0}
{x^{s}{\rm e}^{{\rm i}\pi s} \over x^{2} +  a^{2}}\,{\rm d}x
-
\int^{\infty}_{0}
{x^{s}{\rm e}^{-{\rm i}\pi s} \over x^{2} +  a^{2}}\,{\rm d}x
\end{align}
$$
2\pi{\rm i}\left\vert a\right\vert^{s - 1}\sin\left(\pi s \over 2\right)
=
2{\rm i}\sin\left(\pi s\right)
\int^{\infty}_{0}{x^{s} \over x^{2} +  a^{2}}\,{\rm d}x
$$
$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{\large\quad%
\int^{\infty}_{-\infty}
{\left\vert x\right\vert^{s} \over x^{2} +  a^{2}}\,{\rm d}x
=
{\pi\left\vert a\right\vert^{s - 1} \over \cos\left(\pi s/2\right)}\quad}
\\ \\ \hline
\end{array}
$$
Notice that
$\displaystyle{%
{\pi \over \cos\left(\pi s/2\right)}
=
\Gamma\left({1 \over 2} + {s \over 2}\right)
\Gamma\left({1 \over 2} - {s \over 2}\right)
}$.
Derive respect $s$ in both members of the result:
$$
\int^{\infty}_{-\infty}
{\left\vert x\right\vert^{s}\ln\left(\left\vert x\right\vert\right) \over x^{2} +  a^{2}}\,{\rm d}x
=
\pi\,
{\left\vert a\right\vert^{s - 1}\ln\left(\left\vert a\right\vert\right)
 \cos\left(\pi s/2\right)
 +
 \sin\left(\pi s/2\right)\left(\pi/2\right)\left\vert a\right\vert^{s - 1}
 \over
 \cos^{2}\left(\pi s/2\right)}
$$
Set $s = 0$:$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{\large\quad%
\int^{\infty}_{-\infty}
{\ln\left(\left\vert x\right\vert\right) \over x^{2} +  a^{2}}\,{\rm d}x
=
\pi\,{\ln\left(\left\vert a\right\vert\right) \over \left\vert a\right\vert}\quad}
\\ \\ \hline
\end{array}
$$
That is all we need to derive the main results.
