Evaluating $\int_{0}^{\infty} \frac{\ln(x)}{(x^2-1)}\:dx$ I am asked to show that 

$$\displaystyle \int_{0}^{\infty} \frac{\ln(x)}{(x^2-1)} = \frac{\pi^2}{4}.$$

So I think I'm supposed to use residues to solve this integral and integrate around a contour on the real line. The thing is that I don't know how to choose this contour since the poles will lie on the path and I would also have to bent around zero since $\ln(x)$ has a non essential singularity there. I tried to do this with the methods that I've learned but I got as an answer $-\pi^2$ which is wrong. Any help on how to proceed is appreciated. 
 A: Hint. One may observe that
$$
\begin{align}
\int_0^\infty \frac{\ln x}{x^2-1}\:dx&=\int_0^1 \frac{\ln x}{x^2-1}\:dx+\int_1^\infty \frac{\ln x}{x^2-1}\:dx
\\&=-\int_0^1 \frac{\ln x}{1-x^2}\:dx+\int_1^\infty \frac{\ln \frac1x}{1-\large\frac1{x^2}}\left(-\frac{dx}{x^2}\right)
\\&=-\int_0^1 \frac{\ln x}{1-x^2}\:dx-\int_0^1 \frac{\ln u}{1-u^2}\:du
\\&=-2\int_0^1 \frac{\ln x}{1-x^2}\:dx
\\&=-2\int_0^1 \sum_{n=0}^\infty x^{2n}\ln x\:dx
\\&=-2\sum_{n=0}^\infty \int_0^1 x^{2n}\ln x\:dx
\\&=2\sum_{n=0}^\infty \frac1{(2n+1)^2}
\\&=2\cdot \frac34\sum_{n=1}^\infty \frac1{n^2}
\\&=\frac{\pi^2}4.
\end{align}
$$
A: Let 
$$f(\delta,x) = \frac{(\ln x)^2}{(x-i\delta)^2-1},\ \delta>0.$$
Integrate $f(\epsilon,x)$ over the contour $C$

It can be shown by comparing the magnitude of functions on the larger circle with radius $R$ and on the smaller circle with radius $\epsilon$ that as $R\to\infty$ and $\epsilon\to0^+$ the contour integral approaches
\begin{align}
\frac{\pi^2}2 + o(\delta)&=\text{Res}\big(f(\delta,\cdot),-1\big) \\
&=-\frac1{i2\pi}\oint_Cf(\delta,x)dx \\
&= -\frac1{i2\pi}\int_0^\infty \left(\frac{(\ln x+i\pi)^2}{(x-i\delta)^2-1}-\frac{(\ln x-i\pi)^2}{(x-i\delta)^2-1}\right)\,\mathrm dx \\
&= -2\int_0^\infty\frac{\ln x}{(x-i\delta)^2-1}dx. \\
\end{align}
So
$$\int_0^\infty f(\delta,x)dx \to-\frac{\pi^2}4$$
as $\delta\to0^+$.
A: Since $\displaystyle\frac{\partial x^u}{\partial u}=x^u\ln x,$ let 
$$f(u,\delta,x) = \frac{x^u}{(x-i\delta)^2-1},\ u\in[0,1),\,\delta>0.$$
Integrate $f(\epsilon,x)$ over the contour $C$

It can be shown by comparing the magnitude of functions of the radii $R$ and $\epsilon$ that as $R\to\infty$ and $\epsilon\to0^+$ the contour integral approaches
\begin{align}
\frac{1-e^{i\pi u}}2 &= \text{Res}\big(f(u,\delta,\cdot)\big) \\
&= -\frac1{2\pi i}\oint_C f(u,\delta,z) dz \\
&= -\frac{e^{i\pi u}-e^{-i\pi u}}{2\pi i} \int_0^\infty f(u,\delta,x)dx
\end{align}
$$\int_0^\infty \frac{\ln x}{x^2-1}dx = \lim_{\delta\to0^+}\frac{\partial}{\partial u}\int_0^\infty f(u,\delta,x)dx \bigg|_{u=0} = \frac{\pi^2}4.$$
