25
$\begingroup$

How can we prove that:

$$\int_0^1\frac{\log^2(x)\log(1-x)\log(1+x)}{x}dx=\frac{\pi^2}{8}\zeta(3)-\frac{27}{16}\zeta(5) $$

$\endgroup$
1
  • 1
    $\begingroup$ It's curious that ${\tt \mbox{Mathematica 10.0.0.0}}$, in a MacBook Pro, yields $-\ln\left(\, 4\,\right)$ which is clearly ${\large\it\mbox{wrong !!!}}$. $$\verb* Clear[x]; Integrate[Log[x]^2 Log[1 - x] Log[1 + x] /x, {x, 0, 1}]* $$ W&A evaluates it correctly. $\endgroup$ Nov 29, 2014 at 23:10

5 Answers 5

30
$\begingroup$

Performing integration by parts by taking $u=\ln(1-x)\ln(1+x)$, then $$\begin{align} \int_0^1\frac{\ln^2x\ln(1-x)\ln(1+x)}{x}\ dx&=\frac{1}{3}\bigg[\int_0^1\frac{\ln(1+x)\ln^3x}{1-x}\ dx-\int_0^1\frac{\ln(1-x)\ln^3x}{1+x}\ dx\bigg]\\ &=\frac{1}{3}\bigg[I-J\bigg]\\ \end{align}$$ Evaluation of $I$ : $$\begin{align} I&=\int_0^1\frac{\ln^3{x}\big[\ln(1-x^2)-\ln(1-x)\big]}{1-x}\,dx\\ &=\int_0^1\frac{(1+x)\ln^3{x}\ln(1-x^2)}{(1+x)(1-x)}\,dx-\int_0^1\frac{\ln^3{x}\ln(1-x)}{1-x}\,dx\\ &=\int_0^1\frac{\ln^3{x}\ln(1-x^2)}{1-x^2}\,dx+\int_0^1\frac{x\ln^3{x}\ln(1-x^2)}{1-x^2}\,dx-\int_0^1\frac{\ln^3{x}\ln(1-x)}{1-x}\,dx\\ &=\frac{1}{16}\int_0^1\frac{x^{-\frac{1}{2}}\ln^3{x}\ln(1-x)}{1-x}\,dx-\frac{15}{16}\int_0^1\frac{\ln^3{x}\ln(1-x)}{1-x}\,dx\\ &=\frac{1}{16}\int_0^1\frac{x^{-\frac{1}{2}}\ln^3{x}\ln(1-x)}{1-x}\,dx-\frac{15}{16}\int_0^1\frac{\ln^3{x}\ln(1-x)}{1-x}\,dx\\ &=\frac{1}{16}\lim_{\mu \to \frac{1}{2}\,,\,\nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\int_0^1 x^{\mu-1}(1-x)^{\nu-1}\,dx-\frac{15}{16}\lim_{\mu \to 1\,,\,\nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\int_0^1 x^{\mu-1}(1-x)^{\nu-1}\,dx\\ &=\frac{1}{16}\lim_{\mu \to \frac{1}{2}\,,\, \nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\text{B}(\mu,\nu)-\frac{15}{16}\lim_{\mu \to 1\,,\,\nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\text{B}(\mu,\nu)\\ &=12\zeta(5)-\frac{3\pi^2}{8}\zeta(3)-\frac{\pi^4}{8}\ln{2} \end{align}$$ Evaluation of $J$ : \begin{align} J&=\int_0^1\frac{\ln^3{x}\big[\ln(1-x^2)-\ln(1+x)\big]}{1+x}\,dx\\ &=\int_0^1\frac{(1-x)\ln^3{x}\ln(1-x^2)}{(1-x)(1+x)}\,dx-\int_0^1\frac{\ln^3{x}\ln(1+x)}{1+x}\,dx\\ &=\int_0^1\frac{\ln^3{x}\ln(1-x^2)}{1-x^2}\,dx-\int_0^1\frac{x\ln^3{x}\ln(1-x^2)}{1-x^2}\,dx-K\\ &=-\frac{1}{16}\int_0^1\frac{x^{-\frac{1}{2}}\ln^3{x}\ln(1-x)}{1-x}\,dx+\frac{1}{16}\int_0^1\frac{\ln^3{x}\ln(1-x)}{1-x}\,dx-K\\ &=-\frac{1}{16}\lim_{\mu \to \frac{1}{2}\,,\,\nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\int_0^1 x^{\mu-1}(1-x)^{\nu-1}\,dx+\frac{1}{16}\lim_{\mu \to 1\,,\,\nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\int_0^1 x^{\mu-1}(1-x)^{\nu-1}\,dx-K\\ &=-\frac{1}{16}\lim_{\mu \to \frac{1}{2}\,,\,\nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\text{B}(\mu,\nu)+\frac{1}{16}\lim_{\mu \to 1\,,\,\nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\text{B}(\mu,\nu)-K\\ &=\frac{45}{2}\zeta(5)-\frac{5\pi^2}{4}\zeta(3)-\frac{\pi^4}{8}\ln 2-K \end{align} where \begin{align} K&=\int_0^1\frac{\ln^3{x}\ln(1+x)}{1+x}\,dx\\ &=\int_0^1\sum_{k=1}^\infty (-1)^{k-1}H_{k}\,x^k\ln^3x\,dx\tag{1}\\ &=\sum_{k=1}^\infty (-1)^{k-1}H_{k}\int_0^1x^k\ln^3x\,dx\\ &=-6\sum_{k=1}^\infty (-1)^{k-1}\frac{H_{k}}{(k+1)^4}\tag{2}\\ &=-6\sum_{k=1}^\infty (-1)^{k-1}\left[\frac{H_{k+1}}{(k+1)^4}-\frac{1}{(k+1)^5}\right]\tag{3}\\ &=6\sum_{k=1}^\infty (-1)^{k-1}\left[\frac{H_{k}}{k^4}-\frac{1}{k^5}\right]\\ &=9\eta(5)-3\zeta(5)-6\eta(2)\zeta(3)\tag{4}\\ &=\frac{87}{16}\zeta(5)-\frac{\pi^2}{2}\zeta(3)\tag{5} \end{align}

Putting these together, we will get the desired result.


Explanation :

$(1)$ Use generating function $\displaystyle\sum_{k=1}^\infty (-1)^{k-1}H_{k}\,x^k=\frac{\ln(1+x)}{1+x}$

$(2)$ Use formula $\displaystyle\int_0^1 x^k \ln^n x\ dx=\frac{(-1)^n n!}{(k+1)^{n+1}}\quad,\ n\in\mathbb{Z}_{n\ge0}$

$(3)$ Use property $\displaystyle H_{k}=H_{k+1}-\frac{1}{k+1}$

$(4)$ Use formula $\displaystyle \sum_{k=1}^\infty (-1)^{k-1}\frac{H_{k}}{k^{2n}}=\frac{(2n+1)\eta(2n+1)}{2}-\sum_{k=0}^{n-1}\eta(2k)\zeta(2n+1-2k)\,,\,n\in\mathbb{Z}_{n\ge1}$

$(5)$ Use property of Dirichlet eta function $\displaystyle \eta(s)=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^s}=\left(1-2^{1-s}\right)\zeta(s)$

$\endgroup$
5
  • 1
    $\begingroup$ @FelixMarin I added some explanations in my answer. Thanks $\endgroup$
    – Venus
    Nov 30, 2014 at 11:48
  • $\begingroup$ +1 I think you could have avoided a lot of trouble by directly expanding the integrand as a double sum without integrating by parts. $\endgroup$
    – M.N.C.E.
    Dec 1, 2014 at 6:09
  • $\begingroup$ @M.N.C.E. Thanks. Could you post your answer using that method? I want to see a different approach than mine? Thanks. $\endgroup$
    – Venus
    Dec 1, 2014 at 10:04
  • 2
    $\begingroup$ I can't post a solution now as my laptop is damaged. By applying the aforementioned approach, one can show in a few steps that the integral is \begin{align} \small{2\sum^\infty_{j=1}\frac{(-1)^jH_j}{j^4}- 2\sum^\infty_{j=1}\frac{(-1)^j\left(\zeta(2)-H_j^{(2)}\right)}{j^3}- 2\sum^\infty_{j=1}\frac{(-1)^j\left(\zeta(3)-H_j^{(3)}\right)}{j^2}}\end{align} These sums can be trivially evaluated. I actually prefer this to differentiating the Beta function four times and expanding everything to compute the limit. $\endgroup$
    – M.N.C.E.
    Dec 1, 2014 at 14:38
  • 1
    $\begingroup$ Very nice demonstration.thank $\endgroup$
    – user178256
    Dec 2, 2014 at 21:58
4
$\begingroup$

A generalization of the present integral is given in the book (Almost) Impossible Integrals, Sums, and Series (see page 6), $$\int_0^1\frac{\log(1-x)\log^{2n}(x)\log(1+x)}{x} \textrm{d}x =\frac{1}{2}(2n)!\left(1-\frac{1}{2^{2n+1}}\right)\sum_{k=1}^{2n} \zeta(k+1)\zeta(2n-k+2)$$

$$-(2n)!\sum_{k=1}^{n}\left(1-\frac{1}{2^{2k-1}}\right)\zeta(2k)\zeta(2n-2k+3) + \frac{1}{2^{2n+3}} (2n+3-2^{2n+3})(2n)!\zeta(2n+3),$$ where in his solution the author exploits a classical series representation of $\log(1-x)\log(1+x)$.

$\endgroup$
1
$\begingroup$

Different approach:

Let $I$ denotes our integral $\displaystyle\int_0^1\frac{\ln^2x\ln(1-x)\ln(1+x)}{x}\ dx$

By using the identity $4ab=(a+b)^2-(a-b)^2$ and setting $a=\ln(1-x)$ and $b=\ln(1+x)$ we have

$$4I=\int_0^1\frac{\ln^2x\ln^2(1-x^2)}{x}\ dx-\int_0^1\frac{\ln^2x\ln^2\left(\frac{1-x}{1+x}\right)}{x}\ dx$$


The first integral:

$$\int_0^1\frac{\ln^2x\ln^2(1-x^2)}{x}\ dx=\frac18\int_0^1\frac{\ln^2x\ln^2(1-x)}{x}\ dx=\frac14\sum_{n=1}^\infty\frac{H_n}{n+1}\int_0^1 x^n\ln^2x\ dx\\=\frac12\sum_{n=1}^\infty\frac{H_n}{(n+1)^4}=\frac12\sum_{n=1}^\infty\frac{H_n}{n^4}-\frac12\zeta(5)=\boxed{\zeta(5)-\frac12\zeta(2)\zeta(3)}$$


The second integral:

We proved here that $$\ln^2\left(\frac{1-x}{1+x}\right)=-2\sum_{n=1}^\infty\frac{H_n-2H_{2n}}{n}x^{2n}$$

Then we can write $$\int_0^1\frac{\ln^2x\ln^2\left(\frac{1-x}{1+x}\right)}{x}\ dx=-2\sum_{n=1}^\infty\frac{H_n-2H_{2n}}{n}\int_0^1x^{2n-1}\ln^2x\ dx=16\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}-\frac12\sum_{n=1}^\infty\frac{H_n}{n^4}\\ =8\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}+\frac{15}2\sum_{n=1}^\infty\frac{H_n}{n^4}=\boxed{\frac{31}{4}\zeta(5)-\frac72\zeta(2)\zeta(3)}$$

and the answer follows by substituting these boxed results.

$\endgroup$
1
$\begingroup$

Let $I$ denotes our integral $\displaystyle\int_0^1\frac{\ln^2x\ln(1-x)\ln(1+x)}{x}\ dx$

By using the generating function: $$\displaystyle\ln(1+x)\ln(1-x)=\sum_{n=1}^\infty \frac{H_n-H_{2n}}{n}x^{2n}-\frac12\sum_{n=1}^\infty\frac{x^{2n}}{n^2}$$ we can write

\begin{align} I&=\sum_{n=1}^\infty \frac{H_n-H_{2n}}{n}\int_0^1x^{2n-1}\ln^2x\ dx-\frac12\sum_{n=1}^\infty\frac1{n^2}\int_0^1 x^{2n-1}\ln^2x\ dx\\ &=\frac14\sum_{n=1}^\infty \frac{H_n-H_{2n}}{n^4}-\frac1{8}\sum_{n=1}^\infty\frac1{n^5}\\ &=-2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}-\frac74\sum_{n=1}^\infty\frac{H_n}{n^4}-\frac18\zeta(5) \end{align}

By plugging $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}=\frac12\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)$ we get

$$I=\frac34\zeta(2)\zeta(3)-\frac{27}{16}\zeta(5)$$

$\endgroup$
2
  • $\begingroup$ Shater +1 goog method $\endgroup$
    – user178256
    Sep 14, 2019 at 19:53
  • $\begingroup$ thank you glad you like it :) $\endgroup$ Sep 14, 2019 at 19:57
1
$\begingroup$

\begin{align}J&=\int_0^1\frac{\log^2(x)\log(1-x)\log(1+x)}{x}dx\\ U&=\int_0^1\frac{\log(1+x)\log^3 x}{1-x}dx,V=\int_0^1\frac{\log(1-x)\log^3 x}{1+x}dx\\ J&\overset{\text{IBP}}=\frac{1}{3}\Big[\ln^3 x\ln(1-x)\ln(1+x)\Big]_0^1-\frac{1}{3}\int_0^1 \left(\frac{1}{1+x}-\frac{1}{1-x}\right)\ln^3 x\,dx\\ &=\frac{1}{3}\Big(U-V\Big)\\ \end{align} From Integrating $\int_0^1 \frac{\ln(1+x)\ln^3 x}{1+x}\,dx$ with restricted techniques

One obtains: \begin{align*}U&=-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{4}\zeta(2)\zeta(3)+12\zeta(5)\\ V&=\frac{273}{16}\zeta(5)-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)\\ J&=\boxed{\frac{3}{4}\zeta(2)\zeta(3)-\frac{27}{16}\zeta(5)} \end{align*}

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .