TITLE: Integral: $\int^\infty_{-\infty}\frac{\ln(t+1)}{t^2+1}\mathrm{d}t$
QUESTION [8 upvotes]: How to prove the following:
$$\int^\infty_{-\infty}\frac{\ln(t+1)}{t^2+1}\mathrm{d}t=\frac{\pi}{2}\left(\ln(2)+\frac{\pi}{2} i\right)$$

REPLY [4 votes]: It would be easier if you define
$$I(\alpha)=\int_{-\infty}^\infty\frac{\ln(\alpha t+1)}{t^2+1}dt.$$
Then
\begin{eqnarray*}
I'(\alpha)&=&\int_{-\infty}^\infty\frac{t}{(\alpha t+1)(t^2+1)}dt\\
&=&\frac{1}{\alpha^2+1}\int_{-\infty}^\infty\left(-\frac{1}{t+\frac{1}{\alpha}}+\frac{t}{t^2+1}+\frac{\alpha}{t^2+1}\right)dt\\
&=&\frac{1}{\alpha^2+1}\left(-\ln\frac{t+\frac{1}{\alpha}}{\sqrt{t^2+1}}+\alpha\arctan t\right)\big|_{-\infty}^\infty\\
&=&\frac{1}{\alpha^2+1}\left[\ln(-1)+\alpha\pi\right]\\
&=&\frac{\pi}{\alpha^2+1}(i+\alpha),
\end{eqnarray*}
and hence $I(\alpha)=\pi i\arctan\alpha+\frac{\pi}{2}\ln(\alpha^2+1)+C$. But $I(0)=0$  implies $C=0$. Thus
$$I(1)=\pi i\frac{\pi}{4}+\frac{\pi}{2}\ln 2=\frac{\pi}{2}\left(\ln 2+i\frac{\pi}{2}\right).$$