TITLE: Integral:$ \int^\infty_{-\infty}\frac{\ln(x^{2}+1)}{x^{2}+1}dx $
QUESTION [12 upvotes]: How to evaluate:
$$ \int^\infty_{-\infty}\frac{\ln(x^{2}+1)}{x^{2}+1}dx $$
Maybe we can evaluate it using the well-known result:$\int_{0}^{\frac{\pi}{2}} \ln{\sin t} \text{d}t=\int_{0}^{\frac{\pi}{2}} \ln{\cos t} \text{d}t=-\frac{\pi}{2}\ln{2}$ 
But how do I evaluate it, using that ?

REPLY [8 votes]: Let
$$I(\alpha)=\int_{-\infty}^{\infty}\frac{\ln(\alpha x^2+1)}{x^2+1}dx.$$
Then
\begin{eqnarray*}
I'(\alpha)&=&\int_{-\infty}^{\infty}\frac{x^2}{(\alpha x^2+1)(x^2+1)}dx\\
&=&\frac{1}{\alpha-1}\int_{-\infty}^{\infty}\left[\frac{1}{x^2+1}-\frac{1}{\alpha x^2+1}\right]dx\\
&=&\frac{\pi}{\sqrt{\alpha}+\alpha}
\end{eqnarray*}
and hence
\begin{eqnarray*}
I(\alpha)&=&\int\frac{\pi}{\sqrt{\alpha}+\alpha}d\alpha\\
&=&2\pi\ln(1+\sqrt{\alpha})+C.
\end{eqnarray*}
Clearly $I(0)=0$ implies $C=0$. Thus
$$\int_{-\infty}^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx=I(1)=2\pi\ln 2.$$