Integral $I=\int_0^1\frac{\ln x}{x^n-1}dx$

Question

Hi I am trying to obtain a closed form for$$ I_n=\int_0^1\frac{\ln x}{x^n-1}dx, \quad n\geq 1. $$ This integral is quite nice and generates many other known closed form results such as $$ \int_0^1\frac{\ln x}{x^2-1} dx=\frac{\pi^2}{8}, \quad \int_0^1\frac{\ln x}{x-1} dx=\frac{\pi^2}{6}. $$ In these cases I use residue methods, but am unsure how to generalize as in this case of $I_n$.

Thank you

Answer 1

We have

$$I_n=\int_0^1\frac{\ln x}{x^n-1}dx=-\sum_{k=0}^\infty\int_0^1x^{nk}\ln x dx=\sum_{k=0}^\infty\frac1{(nk+1)^2}$$

and for $n=1$ and $n=2$ the two sums are known.

Answer 2

(Edited for rigour, credit for this goes to RandomVariable). $$I(\alpha)=-\int_0^b\frac{x^{\alpha}dx}{1-x^n}$$ $$I(\alpha)=-\left(\frac{b^{\alpha+n+1}}{ \alpha+n+1}+\frac{b^{\alpha+2n+1}}{ \alpha+2n+1}+\cdots\right)$$ Now, as $I'(\alpha)=\int_0^b\frac{\log(x) x^{\alpha}dx}{x^n-1}$, or rather, $J(\alpha)= \lim_{b \to 1}I'(\alpha)$ is something the value of which we want to find, $$I'(\alpha)=-\log(b)\left(\frac{b^{\alpha+n+1}}{ \alpha+n+1}+\frac{b^{\alpha+2n+1}}{ \alpha+2n+1}+\cdots\right)+\left(\frac{b^{\alpha+n+1}}{( \alpha+n+1)^2}+\frac{b^{\alpha+2n+1}}{( \alpha+2n+1)^2}+\cdots\right)$$ So $$J(\alpha)=\lim_{b \to 1}I'(\alpha)=\left(\frac{1}{( \alpha+n+1)^2}+\frac{1}{( \alpha+2n+1)^2}+\cdots\right).$$ Then set $\alpha=0$ and we have accordance with the other answer: $$J(1)=\left(\frac{1}{(n+1)^2}+\frac{1}{(2n+1)^2}+\cdots\right).$$

Answer 3

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $$ \Psi\pars{s + 1} + \gamma = \int_{0}^{1}{1 - t^{s} \over 1 - t} \quad\imp\quad \Psi'\pars{1 \over n} = -\int_{0}^{1}{t^{1/n - 1}\ln\pars{t} \over 1 - t}\,\dd t $$

With $\ds{t = x^{n}}$: $$ \Psi'\pars{1 \over n}= -\int_{0}^{1}{\pars{x^{n}}^{1/n - 1}\ln\pars{x^{n}} \over 1 - x^{n}}\,nx^{n - 1} \,\dd x=n^{2}\int_{0}^{1}{\ln\pars{x} \over x^{n}- 1}\,\dd x $$

$$\color{#00f}{\large% \int_{0}^{1}{\ln\pars{x} \over x^{n}- 1}\,\dd x = {1 \over n^{2}}\,\Psi'\pars{1 \over n}} $$ Note that $\ds{\Psi'\pars{1} = {\pi^{2} \over 6}}$ and $\ds{\Psi'\pars{\half} = {\pi^{2} \over 2}}$.

Some details are given in my previous answer.

Answer 4

$$ \begin{align} \int_0^1\frac{\log(x)}{x^n-1}\mathrm{d}x &=\int_0^\infty\frac{u}{1-e^{-nu}}e^{-u}\,\mathrm{d}u\\ &=\int_0^\infty u(e^{-u}+e^{-(n+1)u}+e^{-(2n+1)u}+\dots)\,\mathrm{d}u\\ &=1+\frac1{(n+1)^2}+\frac1{(2n+1)^2}+\dots\\ &=\frac1{n^2}\left(\frac1{(0+1/n)^2}+\frac1{(1+1/n)^2}+\frac1{(2+1/n)^2}+\dots\right)\\[4pt] &=\frac1{n^2}\psi^{\,\prime}\left(\frac1n\right) \end{align} $$ The order of integration and summation can be interchanged using Tonelli's Theorem since everything is positive in the last integral.