I was trying to evaluate an integral related to the product of two cauchy distributions and in one of the steps got stuck in the integral
$$\int_0^{\infty} \frac{\ln(x)}{\sqrt{x}(x-1)} dx. $$
I tried to evaluate the integral using Mathematica, and it seems that the answer is $\pi^2$. Furthermore, if I restrict the integral to $(0,1)$, the answer is just $\pi^2/2$, i.e. the integral over $(0,1)$ and the integral over $(1,\infty)$ are equal.
I was wondering if anyone could help me verify/disprove this identity? I apologize in advance if this seems an ill-posed question.
Best Answer
Another (low-tech) way to prove your claim is to consider that: $$\begin{eqnarray*} I &=& 4\int_{0}^{+\infty}\frac{\log x}{x^2-1}\,dx =4\left(\int_{0}^{1}\frac{\log x}{x^2-1}\,dx + \int_{1}^{+\infty}\frac{\log x}{x^2-1}\,dx \right)\\&=&8\int_{0}^{1}\frac{-\log x}{1-x^2}\,dx=8\sum_{k=0}^{+\infty}\int_{0}^{1}(-\log x)\,x^{2k}\,dx=8\sum_{k=0}^{+\infty}\frac{1}{(2k+1)^2}\\&=&8\left(\zeta(2)-\frac{1}{4}\zeta(2)\right)=8\cdot\frac{3}{4}\cdot\frac{\pi^2}{6}=\color{red}{\pi^2}.\end{eqnarray*}$$