TITLE: Proving: $\operatorname{P.V.} \int^\infty_{-\infty}\frac{\ln(t^2+1)}{t^2-1}dt =\frac{\pi^2}{2}$
QUESTION [6 upvotes]: How to prove that 
$$\operatorname{P.V.}\int^\infty_{-\infty}\frac{\ln(t^2+1)}{t^2-1}dt =\frac{\pi^2}{2}\: ?$$

REPLY [6 votes]: If you integrate along this closed curve in $\mathbb{C}$

then you get $0$ by Cauchy theorem. The integral you want is along the horizontal parts of the contour, as the small circles vanish and the big one goes to $\infty$. The contributions of the small circles cancel each other (the two residues have opposite signs), the integral along to big semicircle goes to $0$, so your integral is equal to the integral along the vertical part of the contour. Now notice that $\log(t^2+1)$ have different values on the two sides of the vertical path (that's why the integral is not $0$), but they differ just by $2\pi i$. What remains is (setting $t=is$)
$$\int_1^\infty 2\pi/(1+s^2)\, ds=\pi^2/2$$