How to show this integral equals $\pi^2$?

Question

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.

Answer 1

Related problems: (I).

We will use the Mellin transform approach

$$ F(s) = \int_{0}^{\infty} x^{s-1}f(x)dx \implies F'(s)= \int_{0}^{\infty} x^{s-1}\ln(x)f(x)dx. $$

So the whole problem boils down to find the Mellin transform of $\frac{1}{x-1} $ which is given by

$$ F(s) = (-1)^{s+1}\Gamma(s)\Gamma(1-s) , $$

where $\Gamma(z)$ is the gamma function. Differentiating $F(s)$ and taking the limit as $s$ goes to $1/2$ gives the desired result

$$ I = \pi^2 $$

Answer 2

Another (low-tech) way to prove your claim is to consider that: $$\begin{eqnarray*} I &=& 4\int_{0}^{+\infty}\frac{\log x}{x^2-1}\,dx =4\left(\int_{0}^{1}\frac{\log x}{x^2-1}\,dx + \int_{1}^{+\infty}\frac{\log x}{x^2-1}\,dx \right)\\&=&8\int_{0}^{1}\frac{-\log x}{1-x^2}\,dx=8\sum_{k=0}^{+\infty}\int_{0}^{1}(-\log x)\,x^{2k}\,dx=8\sum_{k=0}^{+\infty}\frac{1}{(2k+1)^2}\\&=&8\left(\zeta(2)-\frac{1}{4}\zeta(2)\right)=8\cdot\frac{3}{4}\cdot\frac{\pi^2}{6}=\color{red}{\pi^2}.\end{eqnarray*}$$

Answer 3

Make the substitution $\sqrt{x}=t$ which gives $$4\int_{0}^{\infty}{\frac{\log(t)}{t^{2}-1}dt}$$ Notice the symmetry at $t=1$, or if you prefer spliting up the integral and making the change of Variable $t=\frac{1}{u}$ at $(1,\infty]$

This reduces to $$8\int_{0}^{1}{\frac{\log(u)}{u^{2}-1}du}$$

Expanding the denomenator into a geometric series and applying the monotone convergence theorem we get that $$-8\int_{0}^{1}{\frac{\log(u)}{1-u^{2}}du}=-8\sum_{n=0}^{\infty}\int_{0}^{1}{u^{2n}\log(u)du}=8\sum_{n=0}^{\infty}\frac{1}{(2n+1)^{2}}$$

The later sum is well-known to evaluate at $\frac{\pi^{2}}{8}$

Answer 4

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

Answer 5

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{\infty}{\ln\pars{x} \over \root{x}\pars{x - 1}}\,\dd x}$

\begin{align}&\color{#66f}{\Large% \int_{0}^{\infty}{\ln\pars{x} \over \root{x}\pars{x - 1}}\,\dd x} =\int_{0}^{1}{\ln\pars{x} \over \root{x}\pars{x - 1}}\,\dd x +\int_{1}^{0}{\root{x}\ln\pars{1/x} \over \pars{1/x - 1}} \,\pars{-\,{\dd x \over x^{2}}} \\[3mm]&=-2\int_{0}^{1}{\ln\pars{x} \over \root{x}\pars{1 - x}}\,\dd x =-2\lim_{\mu\ \to\ 0}\partiald{}{\mu}\int_{0}^{1}{x^{\mu - 1/2} - x^{-1/2} \over 1 - x}\,\dd x \\[3mm]&=2\lim_{\mu\ \to\ 0}\partiald{}{\mu}\bracks{% \int_{0}^{1}{1 - x^{\mu - 1/2} \over 1 - x}\,\dd x -\int_{0}^{1}{1 - x^{-1/2} \over 1 - x}\,\dd x} \\[3mm]&=2\lim_{\mu\ \to\ 0}\partiald{\Psi\pars{\mu + 1/2}}{\mu} =2\,\Psi'\pars{\half}=2\,{\pi^{2} \over 2} = \color{#66f}{\LARGE \pi^{2}} \approx {\tt 9.8696} \end{align}

Answer 6

Taking into account the fact that you already received very good answers, I give here just as a curiosity (given by a CAS), $$\int \frac{\ln(x)}{\sqrt{x}(x-1)} dx=-2 \text{Li}_2\left(1-\sqrt{x}\right)-2 \text{Li}_2\left(-\sqrt{x}\right)-\log \left(\sqrt{x}+1\right) \log (x)$$ $$\int_0^a \frac{\ln(x)}{\sqrt{x}(x-1)} dx=\sqrt{a} \left(a \Phi \left(a,2,\frac{3}{2}\right)+4\right)-2 \log (a) \tanh ^{-1}\left(\sqrt{a}\right)$$ and, if $a=1$, the result you already found $\frac{\pi ^2}{2}$ and the remaining coming from what Felix Marin shows at the beginning of his answer.