13
$\begingroup$

This question inspired me to ask the following.

Prove that

$$I_n = \int_0^\infty \frac{\ln x}{x^n-1}\,dx = \left(\frac{\pi}{n\sin\left(\frac{\pi}{n}\right)}\right)^2,$$

for $\Re(n)>1$.

For some cases there is a nice specific form of $I_n$. For example

$$\begin{align} I_2 & = \frac{\pi^2}{4} \\ I_3 & = \frac{4\pi^2}{27} \\ I_4 & = \frac{\pi^2}{8} \\ I_5 & = \frac{2\left(5+\sqrt 5\right)\pi^2}{125} \\ I_6 & = \frac{\pi^2}{9} \\ I_7 & = \frac{2\pi^2}{49\left(1-\sin\left(\frac{3\pi}{14}\right)\right)} \\ I_8 & = \frac{\left(2+\sqrt 2\right)\pi^2}{32} \end{align}$$

$\endgroup$
  • 5
    $\begingroup$ $\displaystyle\int_0^\infty\dfrac{x^{^{k-1}}}{1-x^{^n}}dx~=~\dfrac\pi n\cot\bigg(k~\dfrac\pi n\bigg)$ $\endgroup$ – Lucian Oct 24 '14 at 20:19
  • 2
    $\begingroup$ Just to put @Lucian's comment clear, for non-experts like me, we define $f(k) = \int_0^\infty \dfrac{x^{k-1}}{1-x^n}dx$, then $f'(k) = \int_0^\infty \dfrac{ \log x x^{k-1}}{1-x^n}dx$, so the integral in question is $-f'(1)$ $\endgroup$ – Petite Etincelle Oct 24 '14 at 20:49
12
$\begingroup$

$$ \begin{align} \int_0^\infty\frac{\log(x)}{x^n-1}\mathrm{d}x &=\int_{-\infty}^\infty\frac{x}{e^{nx}-1}e^x\,\mathrm{d}x\\ &=\int_0^\infty\frac{x}{e^{nx}-1}e^x\,\mathrm{d}x+\int_0^\infty\frac{x}{1-e^{-nx}}e^{-x}\,\mathrm{d}x\\ &=\int_0^\infty x(e^{(1-n)x}+e^{(1-2n)x}+e^{(1-3n)x}+\dots)\,\mathrm{d}x\\ &+\int_0^\infty x(e^{-x}+e^{(-1-n)x}+e^{(-1-2n)x}+\dots)\,\mathrm{d}x\\ &=\frac1{(n-1)^2}+\frac1{(2n-1)^2}+\frac1{(3n-1)^2}+\dots\\ &+1+\frac1{(n+1)^2}+\frac1{(2n+1)^2}+\frac1{(3n+1)^2}+\dots\\ &=\frac1{n^2}\sum_{k\in\mathbb{Z}}\frac1{\left(k+\frac1n\right)^2}\\ &=\frac{\pi^2}{n^2}\csc^2\left(\frac\pi{n}\right) \end{align} $$ where the last step uses the derivative of $(7)$ from this answer

| cite | improve this answer | |
$\endgroup$
8
$\begingroup$

We can use a contour integral in the complex plane, as I showed here for the case $n=3$. Now, however, we use

$$\oint_C dz \frac{\log^2{z}}{z^n-1}$$

where $C$ is the modified keyhole contour about the positive real axis of outer radius $R$ and inner radius $\epsilon$.

Let's evaluate this integral over the contours. As before, there are $8$ pieces to evaluate, as follows:

$$\int_{\epsilon}^{1-\epsilon} dx \frac{\log^2{x}}{x^n-1} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\log^2{\left (1+\epsilon e^{i \phi}\right )}}{(1+\epsilon e^{i \phi})^n-1} \\ + \int_{1+\epsilon}^R dx \frac{\log^2{x}}{x^n-1} + i R \int_0^{2 \pi} d\theta \, e^{i \theta} \frac{\log^2{\left (R e^{i \theta}\right )}}{R^n e^{i n \theta}-1} \\ + \int_R^{1+\epsilon} dx \frac{(\log{x}+i 2 \pi)^2}{x^n-1} + i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{(\log{\left (1+\epsilon e^{i \phi}\right )}+i 2 \pi)^2}{(1+\epsilon e^{i \phi})^n-1} \\ + \int_{1-\epsilon}^{\epsilon} dx \frac{(\log{x}+i 2 \pi)^2}{x^n-1} + i \epsilon \int_{2 \pi}^0 d\phi \, e^{i \phi} \frac{\log^2{\left (\epsilon e^{i \phi}\right )}}{\epsilon^n e^{i n \phi}-1} $$

As $R \to \infty$, the fourth integral vanishes as $\log^2{R}/R^{n-1}$. As $\epsilon \to 0$, the second integral vanishes as it is $O(\epsilon^n)$, while the eighth integral vanishes as $\epsilon \log^2{\epsilon}$. This leaves the first, third, fifth, sixth and seventh integrals, which in the above limits, become

$$PV \int_0^{\infty} dx \frac{\log^2{x} - (\log{x}+i 2 \pi)^2}{x^n-1} + i \frac{4 \pi^3}{n}$$

The residue computation is a little more involved, because we now have $n-1$ poles at which we need to evaluate residues. The contour integral is thus

$$i 2 \pi \sum_{k=1}^{n-1} \frac{-4 \pi^2 k^2/n^2}{n e^{i 2 (n-1) \pi k/n}} = -i \frac{8 \pi^3}{n^3} \sum_{k=1}^{n-1} k^2 \, e^{-i 2 (n-1) \pi k/n} $$

The sum is doable, but the algebra is a bit hideous. The result is

$$\sum_{k=1}^{n-1} k^2 \, e^{-i 2 (n-1) \pi k/n} = \frac12 \left ( \frac{n}{\sin^2{\frac{\pi}{n}}} - n^2\right ) - i \frac12 n^2 \cot{\frac{\pi}{n}} $$

Equating real and imaginary parts of both equations for the contour integral yields

$$\int_0^{\infty} dx \frac{\log{x}}{x^n-1} = \frac{\pi^2}{n^2 \sin^2{\frac{\pi}{n}}} $$

$$PV \int_0^{\infty} dx \frac{1}{x^n-1} = -\frac{\pi}{n} \cot{\frac{\pi}{n}} $$

| cite | improve this answer | |
$\endgroup$
5
$\begingroup$

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \color{#66f}{\large \overbrace{\int_{0}^{\infty}{\ln\pars{x} \over x^{n} - 1}\,\dd x}^{\ds{\color{#c00000}{x^{n}\ \mapsto\ x}}}}\ &=\ -\int_{0}^{\infty}{\ln\pars{x^{1/n}} \over 1 - x}\,{1 \over n}\,x^{1/n - 1}\,\dd x =-\,{1 \over n^{2}}\int_{0}^{\infty}{x^{1/n - 1}\ln\pars{x} \over 1 - x}\,\dd x \\[5mm]&=-\,{1 \over n^{2}}\lim_{\mu\ \to\ 0}\partiald{}{\mu} \int_{0}^{\infty}{x^{\mu + 1/n - 1} - x^{1/n - 1} \over 1 - x}\,\dd x \\[5mm]&=-\,{1 \over n^{2}}\lim_{\mu\ \to\ 0}\partiald{}{\mu} \\&\bracks{ \int_{0}^{1}{x^{\mu + 1/n - 1} - x^{1/n - 1} \over 1 - x}\,\dd x +\int_{1}^{0}{x^{-\mu - 1/n + 1} - x^{-1/n + 1} \over 1 - 1/x}\, \pars{-\,{\dd x \over x^{2}}}} \\[5mm]&=-\,{1 \over n^{2}}\lim_{\mu\ \to\ 0}\partiald{}{\mu} \int_{0}^{1}{x^{\mu + 1/n - 1} - x^{-\mu - 1/n}\over 1 - x}\,\dd x \\[5mm]&={1 \over n^{2}}\lim_{\mu\ \to\ 0}\partiald{}{\mu}\bracks{ \int_{0}^{1}{1 - x^{\mu + 1/n - 1} \over 1 - x}\,\dd x -\int_{0}^{1}{1 - x^{-\mu - 1/n} \over 1 - x}\,\dd x} \\[5mm]&={1 \over n^{2}}\lim_{\mu\ \to\ 0}\partiald{}{\mu}\bracks{ \Psi\pars{\mu + {1 \over n}} - \Psi\pars{-\mu - {1 \over n} + 1}} \end{align} where $\ds{\Psi\pars{z}}$ is the Digamma Function $\color{#000}{\bf 6.3.1}$.

\begin{align}\color{#66f}{\large \int_{0}^{\infty}{\ln\pars{x} \over x^{n} - 1}\,\dd x} &={1 \over n^{2}}\bracks{\Psi'\pars{1 \over n} + \Psi'\pars{-\,{1 \over n} + 1}} \end{align}

With the Euler Reflection Formula $\color{#000}{\bf 6.4.7}$ $$ \Psi'\pars{1 - z} + \Psi'\pars{z} = - \pi\,\totald{\cot\pars{\pi z}}{z} =\pi^{2}\csc^{2}\pars{\pi z} $$

we'll find $$ \color{#66f}{\large \int_{0}^{\infty}{\ln\pars{x} \over x^{n} - 1}\,\dd x ={\pi^{2} \over n^{2}}\,\csc^{2}\pars{\pi \over n}} $$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.