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

score 7 · Accepted Answer · answered Apr 16 '14 at 19:55

7

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.

answered Apr 16 '14 at 19:55

This is very nice, Thanks so simply! However, how do you justify switching the order of summation and integration here? I am interested in a mathematical proof which involves uniform convergence as well as solves the integral. Thanks – Jeff Faraci Apr 16 '14 at 19:59
We can use the theorem of integration term by term. (I don't know what we call it in English) – Apr 16 '14 at 20:02
Okay. Thank you for the solution – Jeff Faraci Apr 16 '14 at 20:03
The last series it shouldn't be $\sum_{k=0}^\infty\frac1{(nk+1)}$ ? – Free X Apr 16 '14 at 20:34
@FreeX No then the summer isn't convergent – Jeff Faraci Apr 16 '14 at 20:56
@Integrals How can I calculate the last integral so ? (I tried integration by parts) – Free X Apr 16 '14 at 20:58
Yes you should integrate by parts@FreeX – Apr 16 '14 at 20:59
@FreeX http://math.stackexchange.com/questions/347815/evaluating-int-0-1-xk-ln-xm-dx-for-integer-k-and-m – Random Variable Apr 16 '14 at 20:59
@FreeX http://math.stackexchange.com/questions/319921/use-integration-by-parts-to-prove-that-int-xn-lnx-dx-fracxn1n1 – Jeff Faraci Apr 16 '14 at 21:00
@Integrals Thank you! – Free X Apr 16 '14 at 21:02
Well done, Sami! – amWhy Apr 17 '14 at 11:42
@SamiBenRomdhane: technically, you need more to justify the interchange of summation and integration in an infinite summation. Term by term integration is fine for a finite sum, but you need something more, say Tonelli's Theorem or Fubini's Theorem for an infinite sum. – robjohn May 02 '14 at 16:40
@robjohn Yes of course we need to justify the interchange of $\int$ and $\sum$ for a rigorous proof and as you said the Fubini's theorem is applied here since $$\sum_{k=0}^\infty\int_0^1\left|x^{nk}\ln x\right| dx<\infty$$ – May 02 '14 at 16:52

Meow · Answer 2 · 2014-04-16T22:38:48.807

3

(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: $$J(1)=\left(\frac{1}{(n+1)^2}+\frac{1}{(2n+1)^2}+\cdots\right).$$

edited Apr 16 '14 at 22:38

answered Apr 16 '14 at 20:08

Meow

6,353

1

How is $x^\alpha$ related to $\ln x$ – Jeff Faraci Apr 16 '14 at 20:09
3

It looks like your series for $I(\alpha)$ is divergent. If the integral does not exist how do you justify your equality? – Peder Apr 16 '14 at 20:21
The problem is that the integral diverges. But there must be some way to justify this approach. – Random Variable Apr 16 '14 at 21:23
@Integrals $\frac{d}{d \alpha}I(\alpha)= \int_0^1 \frac{\ln(x) x^\alpha}{x^n-1}dx$. – Meow Apr 16 '14 at 21:26
I don't know how to justify playing with divergent integrals. – Meow Apr 16 '14 at 21:27
@Alyosha Thanks! – Jeff Faraci Apr 16 '14 at 21:33
1

Perhaps integrate from $0$ to $b$. And then after differentiation, let $b$ go to $1$ from the left. – Random Variable Apr 16 '14 at 21:34
@RandomVariable Is there a rigorous way one could justify that that would make the integral converge (I'm really bad with this, despite trying to learn some analysis)? – Meow Apr 16 '14 at 21:38
The integral converges if the upper limit is less than $1$. So I think you have to do what I suggested in my previous comment. – Random Variable Apr 16 '14 at 21:42
@RandomVariable Oh, and only take the limit after differentiating? – Meow Apr 16 '14 at 22:03
Yes, that's what I meant. Now you're not dealing with a divergent integral and a divergent series. Complete rigor would involve justifying bringing the limit inside of the infinite sum at the end. – Random Variable Apr 17 '14 at 00:05

score 3 · Answer 3 · edited Apr 13 '17 at 12:20

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

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.

robjohn · Answer 4 · 2014-05-02T16:36:20.393

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

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

4 Answers4