Convergence of $\int_{0}^{\infty} \frac{\log\left(x\right)}{x^{2} - 1}{\rm d}x$

Question

I'm trying to solve the following complex analysis problem:

• $$ \mbox{Show that}\ \forall\ n > 1\mbox{, the integral}\quad \int_{0}^{\infty}\frac{\log\left(x\right)}{x^{n} - 1}{\rm d}x\quad converges $$ and in that case compute the integral using residue theory.
• The problem here is not about computing the integral but more on proving that it converges when $n = 2$.
• If $n>2$ you can easily check that if $$ \left\{\begin{array}{l} {\displaystyle\operatorname{f}_{n}\left(x\right) = \frac{\log\left(x\right)}{x^{n} - 1}\,,} \\[2mm] {\displaystyle\alpha \in (0,1)\ \mbox{and}\ \beta \in \left(1,n\right)} \\[2mm] {\displaystyle\mbox{then}\ \lim_{x\ \to\ 0^{+}} \,\,\operatorname{f}_{n}\left(x\right)\,x^{\alpha} = 0\quad\mbox{and}\quad \lim_{x\ \to\ \infty}\,\,\operatorname{f}_{n}(x)\,x^{\beta}=0} \end{array}\right. $$
• By applying the $\epsilon,\delta$ definition of limit, $\operatorname{f}_{n}$ can be integrated in $\left(0,\infty\right)$ $\mbox{when}\ n > 2$.
• This method does not work if $n = 2$, or at least that's what I get. How can I prove that the integral is finite in that case ?.
• I've tried to control $\operatorname{f}_{2}$ by showing that for $x > 1,\ $ $x - 1\geq \log\left(x\right)$ but this implies that $\operatorname{f}_{2} \sim 1/x$ which is a problem.

Answer 1

Hint: For $\int_1^{\infty}$ use the fact that $\log x =2\log \sqrt x\le 2\sqrt x$.

For $\int_0^{1}$ use the fact that $\log x =-2\log \frac 1 {\sqrt x}\ge - 2\frac 1 {\sqrt x}$.

Answer 2

You have to verify what happens when $x\to 1$ and when $x\to +\infty$. Note that $x=1$ is a removable singularity, in fact: $$\lim_{x\to 1}\frac{\log x}{x^2-1}=\lim_{t \to 0} \frac{\log(t+1)}{t(t+2)}=\lim_{t\to 0}\frac{t}{t(t+2)}=\frac12$$ As $x\to +\infty$, you have: $$\int_c^{\infty}\frac{\log x}{x^2-1}dx\sim \int_c^{\infty}\frac{\log x}{x^2}dx=\int_c^{\infty}\frac{2\log \sqrt x}{x^2}dx<\int_c^{\infty}\frac{2 \sqrt x}{x^2}dx<+\infty $$

Answer 3

Computation part.

$$I(n)=\int_{0}^{+\infty} \frac{\log(x)}{x^n-1}dx\stackrel{x=e^t}{=}\int_{-\infty}^\infty\frac{te^t}{e^{nt}-1}dt=\int_{0}^\infty\frac{t(e^t+e^{(n-1)t})}{(e^{nt}-1)}dt\stackrel{t=\frac xn}{=}\frac1{n^2}\int_0^\infty\frac{x(e^{\frac{1-n}nx}+e^{-\frac1n x})}{1-e^{-x}}dx=\frac1{n^2}\left(\zeta(2,\frac{n-1}n)+\zeta(2,\frac1n)\right)$$ where $\zeta(s,a)$ denotes the Hurwitz's zeta function and they can be computed by contour integration. In particular $\zeta(2,a)=\sum_{k=0}^\infty\frac1{(k+a)^2}.$ Hence, $$I(n)=\sum_{k=0}^\infty\frac{1}{(kn+n-1)^2}+\frac{1}{(kn+1)^2}$$ which is also equal to $\frac1{n^2}\csc^2(\tfrac{\pi}n)$ as noticed by Claude Leibovici. See also my latest question which is accidentaly related to this question.

For $n=2$, we have $I(2)=\frac{\pi^2}4.$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\iverson}[1]{\left[\left[\,{#1}\,\right]\right]} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\left.\int_{0}^{\infty}{\ln\pars{x} \over x^{n} - 1}\dd x\right\vert_{\, n\ >\ 1}} = \left.\lim_{\nu\ \to\ 0}\,\,\partiald{}{\nu}\int_{0}^{\infty}{1 - x^{\nu} \over 1 - x^{n}}\,\dd x\right\vert_{\, -1\ <\ \Re\pars{\nu}\ <\ n\ -\ 1\ >\ 0} \\[5mm] & \sr{x^{n}\ \mapsto\ x}{=} {1 \over n}\lim_{\nu\ \to\ 0}\,\,\partiald{}{\nu} \int_{0}^{\infty}{\,\, x^{\color{red}{1/n}\ -\ 1}\ -\ x^{\color{red}{\nu/n\ +\ 1/n}\ -\ 1}\,\,\, \over 1 - x}\,\dd x \\[5mm] & \mbox{Note that}\ \left.\pars{1 - x}^{-1} \,\right\vert_{\, -1\ <\ x\ <\ 1}\,\,\, = \sum_{k = 0}^{\infty}\color{red}{\Gamma\pars{1 + k}\cos\pars{\pi k}}\,{\pars{-x}^{k} \over k!} \\[5mm] & \mbox{Therefore,}\ \\[2mm] & \color{#44f}{\int_{0}^{\infty} {\ln\pars{x} \over x^{n} - 1}\dd x} \\ = & \ {1 \over n}\lim_{\nu\ \to\ 0}\,\,\partiald{}{\nu}\ \overbrace{\bracks{\rule{0pt}{5mm}\Gamma\pars{k}\Gamma\pars{1 - k} \cos\pars{-\pi k}\,\,\,}_{\,\nu/n\ +\ 1/n}^{\,1/n}} ^{\ds{Ramanujan's\ Master\ Theorem}} \\[5mm] = & \ \left.{\pi \over n}\lim_{\nu\ \to\ 0}\,\,\partiald{}{\nu}\cot\pars{\pi k}\ \right\vert_{\,\nu/n\ +\ 1/n}^{\,1/n}\qquad\qquad \left\vert\substack{\ds{Euler}\\[1mm]\ds{Reflection}\\[1mm]\ds{Formula}}\right. \\[5mm] = & \ {\pi \over n}\lim_{\nu\ \to\ 0}\,\,\partiald{}{\nu}\braces{\cot\pars{\pi \over n} - \cot\pars{\pi\bracks{{\nu \over n} + {1 \over n}}}} \\[5mm] = & \ {\pi \over n}\lim_{\nu\ \to\ 0}\ \braces{{\pi \over n}\csc^{2}\pars{\pi\bracks{{\nu \over n} + {1 \over n}}}} = \bbx{\color{#44f}{{\pi^{2} \over n^{2}}\,\csc^{2}\pars{\pi \over n}}} \\ & \end{align}