How to tackle the integral $\int_{0}^{\infty} \frac{\ln x}{x^{n}-1} d x$?

Question

In my post, I started to investigate the integral $\displaystyle \int_{0}^{\infty} \frac{\ln x}{x^{2}-1} d x$. Fortunately, $$\displaystyle \int_{0}^{\infty} \frac{\ln x}{x^{2}-1} d x =2 \int_{0}^{1} \frac{\ln x}{x^{2}-1} d x.$$

So we only need to evaluate the integral $J$ using series and integration by part. $\displaystyle \begin{aligned} J\displaystyle & = \int_{0}^{1} \frac{\ln x}{1-x^{2}} d x =\sum_{k=0}^{\infty} \int_{0}^{1} x^{2 k} \ln x d x=\sum_{k=0}^{\infty}\left(\left[\frac{x^{2 k+1} \ln x}{2 k+1}\right]_{0}^{1}-\frac{1}{2 k+1} \int_{0}^{1} x^{2 k+1} \cdot \frac{1}{x} d x\right) \\\displaystyle &=-\sum_{k=0}^{\infty}\frac{1}{(2 k+1)^{2}}=-\frac{\pi^{2}}{8} \end{aligned} \tag*{} $

$$\therefore \displaystyle \int_{0}^{\infty} \frac{\ln x}{x^{2}-1} d x =-2J=\frac{\pi^{2}}{4} $$

However, when I began to increase the power $n$, I found, in Wolframalpha, that there is a pattern for the integral$$ I_{n}=\int_{0}^{\infty} \frac{\ln x}{x^{n}-1} d x $$ $$ \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline I_{n} & \text { Diverges } & \frac{\pi^{2}}{4} & \frac{4 \pi^{2}}{27} & \frac{\pi ^2}{8} & \frac{8 \pi^{2}}{25(5-\sqrt{5})} & \frac{\pi^{2}}{9} & \frac{\pi^{2}}{49} \csc ^{2}\left(\frac{\pi}{7}\right) & \frac{\pi^{2}}{64} \csc ^{2}\left(\frac{\pi}{8}\right) \\ \hline \end{array} $$ By the pattern, let’s guess the formula for $I_n$ as $$ I_{n}=\left(\frac{\pi}{n}\right)^{2}\csc ^{2}\left(\frac{\pi}{n}\right). $$

How to prove it? Is it difficult or interesting? Looking forward to your suggestions and proofs.

Answer 1

In order to make use of power series, I split the integration interval of $I_n$ as:

$$ I_n=\underbrace{\int_{0}^{1} \frac{\ln x}{x^{n}-1} d x}_{J}+\underbrace{\int_{1}^{\infty} \frac{\ln x}{x^{n}-1} d x}_{K} $$

$$ \begin{aligned} J &=-\sum_{k=0}^{\infty} \int_{0}^{1} x^{n k} \ln x d x \\ &=-\sum_{k=0}^{\infty} \int_{0}^{1} \ln x d\left(\frac{x^{n k+1}}{n k+1}\right) \\ & \stackrel{IB P}{=}-\sum_{k=0}^{\infty}\left(\left[\frac{x^{n k+1}\ln x}{n k+1}\right]_{0}^{1}-\int_{0}^{1} \frac{x^{n k}}{n k+1} d x\right) \\ &=\sum_{k=0}^{\infty} \frac{1}{(n k+1)^{2}} \\ &=\frac{1}{n^{2}} \sum_{k=0}^{\infty} \frac{1}{\left(\frac{1}{n}+k\right)^{2}} \end{aligned} $$ Similarly, $$ \begin{aligned} K &=\int_{1}^{\infty} \frac{\ln x}{x^{n}\left(1-\frac{1}{x^{n}}\right)} d x \\ &=\sum_{k=0}^{\infty} \int_{1}^{\infty} \ln x \cdot x^{-n-n k} d x\\ &\stackrel{IBP}{=}\sum_{k=0}^{\infty}\left(\left[\frac{x^{-n-n k+1}\ln x}{-n-n k+1}\right]_{1}^{\infty}-\int_{1}^{\infty} \frac{x^{-n-n k}}{-n-n k+1} d x\right)\\& =\sum_{k=0}^{\infty} \frac{1}{[-n (k+1)+1]^{2}}\\ \end{aligned} $$

Reindexing by replacing $k$ by $-k-1$ gives $$ \begin{aligned} K&=\sum_{k=-\infty}^{-1} \frac{1}{(n k+1)^{2}}=\frac{1}{n^{2}} \sum_{k=-\infty}^{-1} \frac{1}{\left(\frac{1}{n}+k\right)^{2}} \end{aligned} $$

Finally, adding $J$ and $K$ gives $$ \begin{aligned} I_n &=\frac{1}{n^{2}} \sum_{k=-\infty}^{-1} \frac{1}{\left(\frac{1}{n}+k\right)^{2}}+\frac{1}{n^{2}} \sum_{k=0}^{\infty} \frac{1}{\left(\frac{1}{n}+k\right)^{2}} =\frac{1}{n^{2}} \sum_{k=-\infty}^{\infty} \frac{1}{\left(\frac{1}{n}+k\right)^{2}} \end{aligned} $$ By the theorem, $$ \sum_{k=-\infty}^{\infty} \frac{1}{z+k}=\pi \cot (\pi z) $$ Differentiating w.r.t. $z$ yields $$ \sum_{k=-\infty}^{\infty} \frac{1}{(z+k)^{2}}=\pi^{2} \csc^{2}(\pi z) $$ Putting $ \displaystyle z=\frac{1}{n}$ yields the conclusion that

$$\boxed{I_n=\left(\frac{\pi}{n}\right)^{2} \csc ^{2}\left(\frac{\pi}{n}\right)}.$$

Wish you enjoy the proof. Your comments and alternate proofs are warmly welcome.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{2\int_{0}^{1}{\ln\pars{x} \over x^{2} - 1}\dd x} = \left. 2\,\partiald{}{\nu}\int_{0}^{1}{1 - x^{\nu} \over 1 - x^{2}}\,\dd x\right\vert_{\nu\ =\ 0} \\[5mm] \stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,\, & \left. \partiald{}{\nu}\int_{0}^{1}{x^{-1/2}\ -\ x^{\nu/2 - 1/2} \,\,\,\over 1 - x}\,\dd x\right\vert_{\nu\ =\ 0} \\[5mm] = & \ \partiald{}{\nu}\bracks{% \int_{0}^{1}{1\ -\ x^{\nu/2 - 1/2} \,\,\,\over 1 - x}\,\dd x - \int_{0}^{1}{1\ -\ x^{-1/2} \,\,\,\over 1 - x}\,\dd x }_{\nu\ =\ 0} \\[5mm] = & \ \left.\partiald{\Psi\pars{\nu/2 + 1/2}}{\nu} \right\vert_{\nu\ =\ 0} = {1 \over 2}\,\Psi\,'\pars{1 \over 2} = \bbx{\pi^{2} \over 4} \approx 2.4674 \end{align}