Proving: $\operatorname{P.V.} \int^\infty_{-\infty}\frac{\ln(t^2+1)}{t^2-1}dt =\frac{\pi^2}{2}$ How to prove that 
$$\operatorname{P.V.}\int^\infty_{-\infty}\frac{\ln(t^2+1)}{t^2-1}dt =\frac{\pi^2}{2}\: ?$$
 A: 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 $t=is$)
$$\int_1^\infty 2\pi/(1+s^2)\, ds=\pi^2/2$$ 
A: Since the integrand is even,
$$ \begin{align*}
PV \int_{-\infty}^{\infty} \frac{\log(t^2+1)}{t^2 - 1}\,dt
&= 2PV\int_{0}^{\infty} \frac{\log(t^2+1)}{t^2-1} \, dt \\
&= 2PV \lim_{\epsilon \to 0+} \left( \int_{0}^{1-\epsilon} \frac{\log(t^2+1)}{t^2-1} \, dt + \int_{1+\epsilon}^{\infty} \frac{\log(t^2+1)}{t^2-1} \, dt \right)
\end{align*} $$
By the substitution $t \mapsto t^{-1}$, we have
$$\int_{1+\epsilon}^{\infty} \frac{\log(t^2+1)}{t^2-1} \, dt
= \int_{0}^{\frac{1}{1+\epsilon}} \frac{\log(t^2+1) - 2\log t}{1-t^2} \, dt.$$
Thus we obtain
$$PV \int_{-\infty}^{\infty} \frac{\log(t^2+1)}{t^2 - 1}\,dt
= 2PV \lim_{\epsilon \to 0+} \left( \int_{1-\epsilon}^{\frac{1}{1+\epsilon}} \frac{\log(t^2+1)}{1-t^2} \, dt - 2\int_{0}^{\frac{1}{1+\epsilon}} \frac{\log t}{1-t^2} \, dt \right).$$
One the one hand, the function $\frac{1}{t+1}\log(t^2+1)$ is continuous on $[0, 1]$, hence is bounded on this interval by some constant $C > 0$. Also, if $\epsilon > $ is sufficiently small, we have $1 - \epsilon < \frac{1}{1+\epsilon}$. Thus we obtain an estimate
$$ \left| \int_{1-\epsilon}^{\frac{1}{1+\epsilon}} \frac{\log(t^2+1)}{1-t^2} \, dt \right|
\leq C \int_{1-\epsilon}^{\frac{1}{1+\epsilon}} \frac{dt}{1-t}
= C \log \left(1+\epsilon\right).$$
This proves that the left-hand side converges to 0 as $\epsilon \to 0+$. On the other hand, by Tonell's theorem
$$ \begin{align*}
\lim_{\epsilon \to 0+} \int_{0}^{\frac{1}{1+\epsilon}} \frac{\log t}{1-t^2} \, dt
&= \int_{0}^{1} \frac{\log t}{1-t^2} \, dt
= \sum_{n=0}^{\infty} \int_{0}^{1} t^{2n} \log t \, dt \\
&= -\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}
= -\frac{\pi^2}{8},
\end{align*}$$
where at the last equality we used the famous Euler's series. Therefore the desired result follows.
