Evaluating $\int_{0}^{\infty} \frac{\ln(x)}{(x^2-1)}\:dx$

Question

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.

Answer 1

Hint. One may observe that $$ \begin{align} \int_0^\infty \frac{\ln x}{x^2-1}\:dx&=\int_0^1 \frac{\ln x}{x^2-1}\:dx+\int_1^\infty \frac{\ln x}{x^2-1}\:dx \\&=-\int_0^1 \frac{\ln x}{1-x^2}\:dx+\int_1^\infty \frac{\ln \frac1x}{1-\large\frac1{x^2}}\left(-\frac{dx}{x^2}\right) \\&=-\int_0^1 \frac{\ln x}{1-x^2}\:dx-\int_0^1 \frac{\ln u}{1-u^2}\:du \\&=-2\int_0^1 \frac{\ln x}{1-x^2}\:dx \\&=-2\int_0^1 \sum_{n=0}^\infty x^{2n}\ln x\:dx \\&=-2\sum_{n=0}^\infty \int_0^1 x^{2n}\ln x\:dx \\&=2\sum_{n=0}^\infty \frac1{(2n+1)^2} \\&=2\cdot \frac34\sum_{n=1}^\infty \frac1{n^2} \\&=\frac{\pi^2}4. \end{align} $$

Answer 2

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 $\delta\to0^+$.

Answer 3

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} $$\int_0^\infty \frac{\ln x}{x^2-1}dx = \lim_{\delta\to0^+}\frac{\partial}{\partial u}\int_0^\infty f(u,\delta,x)dx \bigg|_{u=0} = \frac{\pi^2}4.$$