TITLE: How to show this integral equals $\pi^2$?
QUESTION [7 upvotes]: 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.

REPLY [5 votes]: Letting $ \displaystyle f(z) = \frac{\ln^{2}z}{\sqrt{z} (z-1)} $ and integrating around a keyhole contour where the branch cut is along the positive real axis, $$\int_{0}^{\infty} \frac{\ln^{2} x}{\sqrt{x}(x-1)} \ dx + \text{PV} \int_{\infty}^{0} \frac{(\ln x + 2 \pi i )^{2}}{\sqrt{x e^{2 \pi i}}(x-1)} \ dx - i \pi \ \text{Res} [f(z), e^{2 \pi i}] =0  $$
where
$$ \text{Res}[f(z), e^{2 \pi i}]= \lim_{z \to e^{2 \pi i}} \frac{\ln^{2}z}{\sqrt{z}} = 4 \pi^{2} .$$
Equating the imaginary parts on both sides of the equation, $$ 4 \pi \int_{0}^{\infty} \frac{\ln x}{\sqrt{x}(x-1)} \ dx -  4 \pi^{3} = 0$$
or $$\int_{0}^{\infty} \frac{\ln x}{\sqrt{x}(x-1)} \ dx = \pi^{2}.$$
EDIT:
Because of the $\sqrt{z}$ in the denominator, integrating $ \displaystyle g(z) = \frac{\ln z}{\sqrt{z}(z-1)}$ around the same contour would work as well.
Specifically, $$ \int_{0}^{\infty} \frac{\ln x}{\sqrt{x}(x-1)} \ dx + \text{PV} \int_{\infty}^{0} \frac{\ln x + 2 \pi i }{\sqrt{x e^{2 \pi i}}(x-1)} \ dx - i \pi \ \text{Res} [g(z), e^{2 \pi i}] =0$$ where $$\text{Res}[g(z),2 \pi i] = -2 \pi i .$$
Equating the real parts on both sides of the equation, $$2 \int_{0}^{\infty} \frac{\ln x}{\sqrt{x} (x-1)} \ dx - 2 \pi^{2} = 0 $$
or $$\int_{0}^{\infty} \frac{\ln x}{\sqrt{x}(1-x)} \ dx = \pi^{2}. $$