51
$\begingroup$

How can we prove that:

$$\int_0^1 \frac{\log(x)\log(1-x)\log^2(1+x)}{x}dx=\frac{7\pi^2}{48}\zeta(3)-\frac{25}{16}\zeta(5)$$ where $\zeta(z)$ is the Riemann Zeta Function.

The best I could do was to express it in terms of Euler Sums. Let $I$ denote the integral.

$$I=-\frac{\pi^2}{24}\zeta(3)+2\sum_{r=2}^\infty \frac{(-1)^r (H_r)^2}{r^3}-2\sum_{r=2}^\infty \frac{(-1)^r H_r}{r^4}+2 \sum_{r=2}^\infty \frac{(-1)^r H_r^{(2)}H_r}{r^2}-2\sum_{r=2}^\infty \frac{(-1)^r H_r^{(2)}}{r^3}$$

where $\displaystyle H_r^{(n)}=\sum_{n=1}^r \frac{1}{k^n}$. I am unable to simplify these sums further. Does anyone have any idea on how to solve this integral?

Please see here for more details.


Some time ago I was able to solve the simpler integral:

\begin{align*} \int_0^1 \frac{\log(1-x)\log(x)\log(1+x)}{x}dx &=-\frac{3 \pi^4}{160}+\frac{7\log(2)}{4}\zeta(3)-\frac{\pi^2 \log^2(2)}{12} +\frac{\log^4(2)}{12} \\ &\quad+ 2 \text{Li}_4 \left(\frac{1}{2} \right) \end{align*} where $\text{Li}_n(z)$ is the Polylogarithm.

$\endgroup$
51
+50
$\begingroup$

Let the considered integral be denoted by $I$. Our starting point is to reduce the number of logarithms of different arguments in the integrand. Thus, using the fact that $6ab^2=(a+b)^3-2a^3+(a-b)^3$ we obtain \begin{align*} 6I&=\underbrace{\int_0^1\frac{\log x}{x}\log^3(1-x^2)dx}_{x\leftarrow \sqrt{u}} -2\int_0^1\frac{\log x}{x}\log^3(1-x)dx+ \underbrace{\int_0^1\frac{\log x}{x}\log^3\left(\frac{1-x}{1+x}\right)dx}_{x\leftarrow\frac{1-u}{1+u}} \\ &= \frac{1}{4}\int_0^1\frac{\log u}{u}\log^3(1-u)du -2\int_0^1\frac{\log x}{x}\log^3(1-x)dx+ 2\int_0^1\frac{\log^3 u}{1-u^2}\log\left(\frac{1-u}{1+u}\right)du \\ &=\frac{-7}{4}\int_0^1\frac{\log x}{x}\log^3(1-x)dx+ 2\int_0^1\frac{\log^3 x}{1-x^2}\log\left(\frac{1-x}{1+x}\right)dx \\ \end{align*} Thus, $$ I=-\frac{7}{24}J+\frac{1}{3}K \tag{1} $$ with $$ J=\int_0^1\frac{\log(1- x)}{1-x}\log^3x\,dx,\quad K=\int_0^1\frac{1}{1-x^2}\log\left(\frac{1-x}{1+x}\right)\log^3 x\,dx \tag{2} $$ Now, for $x\in(-1,1)$, we have $$ \frac{\log(1-x)}{1-x}=-\left(\sum_{k=0}^\infty x^k\right)\left(\sum_{k=1}^\infty \frac{x^k}{k}\right) =-\sum_{n=1}^\infty H_nx^n $$ where $H_n=\sum_{k=1}^n1/k$ is the $n$-th Harmonic number. Similarly, $$ \frac{1}{1-x^2}\log\left(\frac{1-x}{1+x}\right) =-\left(\sum_{k=0}^\infty x^{2k}\right)\left(\sum_{k=1}^\infty \frac{-2x^{2k-1}}{2k-1}\right) =-2\sum_{n=1}^\infty \widetilde{H}_nx^{2n-1} $$ where $$\widetilde{H}_n=\sum_{k=1}^n\frac{1}{2k-1} =H_{2n}-\frac{1}{2}H_n$$ Thus, using the fact that $\int_0^1x^n(-\log x)^3\,dx=-6/(n+1)^4$ we conclude that \begin{align*} J&=\sum_{n=1}^\infty H_n\int_0^1x^n(-\log x)^3\,dx=6\sum_{n=1}^\infty \frac{H_n}{(n+1)^4}\\ &=6\sum_{n=0}^\infty \frac{1}{(n+1)^4}\left(H_{n+1}-\frac{1}{n+1}\right)=6A-6\zeta(5)\tag{3} \end{align*} with \begin{equation*} A=\sum_{n=1}^\infty\frac{H_n}{n^4}\tag{4} \end{equation*} Similarly, \begin{align*} K&=\sum_{n=1}^\infty (2H_{2n}-H_n)\int_0^1x^{2n-1}(-\log x)^3\,dx=6\sum_{n=1}^\infty\frac{2H_{2n}-H_n}{(2n)^4}\\ &=6\sum_{n=1}^\infty\frac{(1+(-1)^{2n})H_{2n}}{(2n)^4} -\frac{3}{8}\sum_{n=1}^\infty\frac{H_n}{n^4}\\ &=6\sum_{n=1}^\infty\frac{(1+(-1)^{n})H_{n}}{n^4} -\frac{3}{8}\sum_{n=1}^\infty\frac{H_n}{n^4}\\ &=\frac{45}{8}\sum_{n=1}^\infty\frac{H_{n}}{n^4} +6\sum_{n=1}^\infty\frac{(-1)^{n}H_{n}}{n^4}=\frac{45}{8}A+6B\tag{5} \end{align*} with \begin{equation*} B=\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}\tag{6} \end{equation*} Combining (1) with (3) and (5) we find that \begin{equation*} I=\frac{1}{8}A+2B+\frac{7}{4}\zeta(5)\tag{7} \end{equation*} Now, sums $A$ and $B$ are known (see here ) (in a more general setting), and we have $$ A=3\zeta(5)-\zeta(2)\zeta(3),\qquad B=-\frac{59}{32}\zeta(5)+\frac{1}{2}\zeta(2)\zeta(3) $$ Thus $$ I=\frac{7}{8}\zeta(2)\zeta(3)-\frac{25}{16}\zeta(5). $$ which is the announced result.$\qquad\square$

$\endgroup$

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.