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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.

Davide Giraudo
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user95166
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2 Answers2

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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 $$ \left.\frac{\mathrm d }{\mathrm ds}E[|a|^s]\right|_{s=0}=\frac1{\cos(0)}\,\left.\frac{\mathrm d }{\mathrm ds}\alpha^s\right|_{s=0}=\log\alpha. $$

Did
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$\displaystyle{\large% ? =\int_{0}^{\infty}{x^{s} \over x^{2} + a^{2}}\,{\rm d}x}$

\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.

Felix Marin
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