Expectation of Log of a general Cauchy-distributed Random Variable:$\int^{\infty}_{-\infty} {\log(|\alpha x + \beta|) \over \pi(x^{2} + 1)}{\rm d}x$

Question

The expectation of the log of the absolute value of a central Cauchy-distributed random variable is given here.

How can this be extended for a non-central case. It is equivalent to finding the following expectation:

$$\mathbb E [\log|\alpha X + \beta|]=\int^{\infty}_{-\infty} {\log\left(\left\vert \alpha x + \beta\right\vert\right) \over \pi(x^{2} + 1)}\,{\rm d}x$$

where $X$ has the standard Cauchy distribution.

For $\beta=0$, we have the nice formula of

$$\mathbb E [\log|\alpha X|]=\log|\alpha|. $$

Answer 1

Using Feymann's trick: $$\int^{\infty}_{-\infty} {\log\left(\left\vert \alpha x + \beta\right\vert\right) \over (x^{2} + 1)}\,{\rm d}x$$ $$\pi\ln\left(\left|\alpha\right|\right)+\int_{-\infty}^{\infty}\frac{\ln\left(\left|x+\frac{\beta}{\alpha}\right|\right)}{x^{2}+1}dx$$ $$s:=\frac{\beta}{\alpha}$$ $$F(s):=\pi\ln\left(\left|\alpha\right|\right)+\int_{-\infty}^{\infty}\frac{\ln\left(\left|x+s\right|\right)}{x^{2}+1}dx$$ $$f(s)=F'(s)=P.V.\int_{-\infty}^{\infty}\frac{1}{(x^{2}+1)(x+s)}dx=-\frac{\pi s}{s^2+1}$$

So

$$\int^{\infty}_{-\infty} {\log\left(\left\vert \alpha x + \beta\right\vert\right) \over \pi(x^{2} + 1)}\,{\rm d}x=\ln\left(\left|\alpha\right|\right)+\frac{1}{2}\ln\left(\left(\frac{\beta}{\alpha}\right)^{2}+1\right)$$ $$\color{blue}{\int^{\infty}_{-\infty} {\log\left(\left\vert \alpha x + \beta\right\vert\right) \over \pi(x^{2} + 1)}\,{\rm d}x=\ln\left(\sqrt{\alpha^{2}+\beta^{2}}\right)}$$

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For $\beta=0$ you have your known formula

Answer 2

Assume that $\alpha >0$, $\beta \ge 0$, and $\ln$ is the principal branch of the logarithm.

The function $$\frac{\ln \left(\alpha z+ \beta\right)}{\left(z^{2}+1\right)}$$ has a branch cut on $(-\infty, -\frac{\beta}{\alpha}]$.

On the upper side of the branch cut, $$\frac{\ln \left(\alpha z+ \beta\right)}{z^{2}+1}= \frac{\ln \left(|\alpha z+\beta|\right)}{z^{2}+1} + i \pi , \quad z < - \frac{\beta}{\alpha}.$$

And integrating on the upper half of the large semicircle $|z|=R$, we have $$\begin{align} \left| \int_{0}^{\pi} \frac{\ln \left(\alpha Re^{it}+ \beta\right)}{R^{2} e^{2it} +1} \, i R e^{it} \right| &\le \small R \int_{0}^{\pi} \frac{\sqrt{\frac{1}{4} \, \ln^{2} \left(\alpha^{2}R^{2} +2\ \alpha \beta R \cos (t) + \beta^{2}\right)+ \left(\arg\left(\alpha Re^{it}+ \beta \right)\right)^{2}}}{R^{2}-1} \, \mathrm dt \\ &\le \pi R \, \frac{\sqrt{\frac{1}{4} \, \ln^{2} \left(\alpha^{2}R^{2} +2 \alpha \beta R + \beta^{2}\right)+ \pi^{2}}}{R^{2}-1} \\& \sim \frac{\pi \ln (\alpha R)}{R}, \end{align} $$ which goes to zero as $ R \to \infty$.

Also, $(z+ \frac{\beta}{\alpha}) \frac{\ln\left(\alpha z+\beta\right)}{z^{2}+1}$ goes to zero as $z \to -\frac{ \beta}{\alpha}.$

So by integrating $\frac{\ln \left(\alpha z+\beta\right)}{z^{2}+1}$ around a closed semicircular contour in the upper half-plane that is indented as $z = -\frac{\beta}{\alpha}$, we get

$\begin{align} \int_{-\infty}^{- \beta/\alpha} \frac{\ln\left(|\alpha x + \beta|\right) + i \pi}{1+x^{2}}\, \mathrm dx + \int_{-\beta/\alpha}^{\infty} \frac{\ln\left(|\alpha x + \beta| \right)}{1+x^{2}}\, \mathrm dx &= 2 \pi i \operatorname*{Res}_{z= i} \frac{\ln \left(\alpha z + \beta \right)}{z^{2}+1} \\ &= 2 \pi i \, \frac{\ln \left(\alpha i + \beta \right)}{2i} \\ &= \pi \left(\frac{1}{2} \, \ln \left(\alpha^{2} + \beta^{2} \right) + i \arctan \left(\frac{\alpha}{\beta} \right) \right). \end{align}$

Equating the real parts on both sides of the equation, we have $$\int_{-\infty}^{\infty} \frac{\ln \left(|\alpha x + \beta |\right)}{1+x^{2}} \, \mathrm dx = \frac{\pi}{2} \, \ln \left(\alpha^{2}+\beta^{2} \right).$$

And equating the imaginary parts on both sides of the equation, we have the trivial result $$\int_{-\infty}^{- \beta/\alpha} \frac{\mathrm dx }{1+x^{2}} = \arctan \left(\frac{\alpha}{\beta} \right) = \frac{\pi}{2} - \arctan \left(\frac{\beta}{\alpha} \right). $$