Integral $ \int_0^1 \frac{\ln x \ln(1+x)}{1+x}dx=-\frac{\zeta(3)}{8} $

Question

How can we prove

$$ \int_0^1 \frac{\ln x \ln(1+x)}{1+x}dx=-\frac{\zeta(3)}{8}? $$

This has been one of the integrals that came out of an integral from another post on here, but no solution to it.

I am not sure how to use a taylor series expansion for the $\ln(1+x)\cdot(x+1)^{-1}$ term, thus I can not simple reduce this integral to the form $$ \int_0^1 x^n \ln x dx $$ I think if I can get the integral in this form, I will be able to recover the zeta function series which is given by $$ \zeta(3)=\sum_{n=0}^\infty \frac{1}{(n+1)^3}. $$ Thanks

Answer 1

I played around with this using parts because it looks like an integral that involves polylogs. Many of these can be done with parts or multiple use of parts.

$$\int\frac{\log(x)\log(1+x)}{1+x}dx$$

Let $$u=x+1$$

$$\int\frac{\log(u-1)\log(u)}{u}du=\int\frac{\log(u)}{u}\left(\log(u)+\log(1-1/u)\right)du$$

$$=\frac{\log^{3}(u)}{3}+\int\frac{\log(u)\log(1-1/u)}{u}du$$

Now, use parts on this last integral:

$u=\log(u), \;\ dv=\frac{\log(1-1/u)}{u}, \;\ du=\frac{1}{u}du, \;\ v=\operatorname{Li}_{2}(1/u)$

(as a note, $\int\frac{\log(1-1/u)}{u}du=\operatorname{Li}_{2}(1/u)$ is a rather famous integral related to the dilogarithm).

$$\int\frac{\log(u)\log(1-1/u)}{u}du=\log(u)\operatorname{Li}_{2}(1/u)-\int\frac{\operatorname{Li}_{2}(1/u)}{u}du$$

Also, note this last integral is simply $$-\operatorname{Li}_{3}(1/u)$$

Now, back sub $u=x+1$, and put it altogether using the integration limits $0$ to $1$.

Hence, we arrive at:

$$ \left.1/3\log^{3}(x+1)+\log(x+1)\operatorname{Li}_{2}\left(\frac{1}{x+1}\right)+\operatorname{Li}_{3}\left(\frac{1}{1+x}\right)\right|_{0}^{1}$$

$$=1/3\log^{3}(2)+\log(2)\operatorname{Li}_{2}(1/2)+\operatorname{Li}_{3}(1/2)-\operatorname{Li}_{3}(1).........(1)$$

Note the identities:

$$\operatorname{Li}_{2}(1/2)=\frac{\pi^{2}}{12}-1/2\log^{2}(2)$$

$$\operatorname{Li}_{3}(1/2)=7/8\zeta(3)+1/6\log^{3}(2)-\frac{\pi^{2}}{12}\log(2)$$

sum up (1):

$$1/3\log^{3}(2)+\log(2)\left(\frac{\pi^{2}}{12}-1/2\log^{2}(2)\right)+\left(7/8\zeta(3)+1/6\log^{3}(2)-\frac{\pi^{2}}{12}\log(2)\right)-\zeta(3)$$

$$=\frac{-\zeta(3)}{8}$$

Answer 2

I think this ties together the aforementioned ideas quite nicely:

Step 1: Integrate by parts. Let $u=\log{x}$ and $dv=\frac{\log(1+x)}{1+x}$. We obtain $v=\frac{1}{2} [\log(1+x)]^2$. Being somewhat careful with the limits, we see that the integral itself is equal to $$ -\frac{1}{2} \int_0^1 \frac{[\log(1+x)]^2}{x}\,dx $$

Step 2: Expand $\log(1+x)$ and $\log(1+x)/x$ into their Taylor series and combine. $$ -\frac{1}{2} \int_0^1\left(\sum_{j=1}^{\infty} (-1)^{j+1} \frac{x^j}{j}\right)\left(\sum_{i=0}^\infty (-1)^i \frac{x^i}{i+1}\right)\,dx = -\frac{1}{2} \sum_{j=1}^\infty \sum_{i=0}^\infty \frac{(-1)^{i+j+1}}{j(i+1)(i+j+1)} $$

Step 3: There are a few ways to go here, but I like $k=i+j+1$ followed by a partial fraction decomposition. Then, $$ -\frac{1}{2} \sum_{k=2}^\infty \frac{(-1)^k}{k} \sum_{j=1}^{k-1} \frac{1}{j(k-j)} = -\sum_{k=2}^\infty \frac{(-1)^k}{k^2} H_{k-1} $$

Step 4: ??? It is not clear to me why this quantity is the desired one, but prior responses seem to indicate as such. Anybody else with thoughts?

[edit] I had an $H_k$ that should have been an $H_{k-1}$. Fixed now.

[edit 2] A more direct approach from the generating function (http://en.wikipedia.org/wiki/Harmonic_number#Generating_functions) of the harmonic sequence: Since $-\sum_{k=1}^\infty H_k (-x)^k = \frac{\log(1+x)}{1+x}$, we have $$ -\int_0^1 \log(x) \sum_{k=1}^\infty (-1)^k H_k x^k\,dx = \sum_{k=1}^\infty \frac{(-1)^k}{(k+1)^2} H_k $$ Definitely simpler, but requires a priori knowledge of the generating function.

Answer 3

A handy thing to note for evaluating $$\sum_{n=1}^{\infty}\frac{H_{n}}{(n+1)^{2}}$$ is to use $$\sum_{n=1}^{\infty}\frac{H_{n}}{(n+1)^{2}}=\sum_{n=1}^{\infty}\frac{H_{n}}{n^{2}}-\zeta(3)$$............[1]

The first sum on the right can be shown in various ways and evaluates to $2\zeta(3)$. If you look around, I am sure it has already been done on the site.

Contours is a fun way to evaluate many Euler sums. A method published by Flajolet and Salvy in their paper "Euler sums and contour integral representations". Use the 'kernel' $\frac{1}{2}\pi\cot(\pi z)(\psi(-z))$ and note the residues for the pole at 0, the positive integers, n, and the negative integers, -n.

The pole at the negative integers is simple and the residue is

$$Res(-n)=\sum_{n=1}^{\infty}\frac{H_{n}}{2n^{2}}-\sum_{n=1}^{\infty}\frac{1}{2n^{3}}$$

The residue at the positive integers is order 2 and is:

$$Res(n)=\sum_{n=1}^{\infty}\frac{H_{n}}{2n^{2}}-\sum_{n=1}^{\infty}\frac{1}{n^{3}}$$

The residue at the pole at 0 is $$\frac{-1}{2}\zeta(3)$$

summing these and setting to 0 gives:

$$\sum_{n=1}^{\infty}\frac{H_{n}}{2n^{2}}-\sum_{n=1}^{\infty}\frac{1}{2n^{3}}+\sum_{n=1}^{\infty}\frac{H_{n}}{2n^{2}}-\sum_{n=1}^{\infty}\frac{1}{n^{3}}-1/2\zeta(3)=0$$

$$\sum_{n=1}^{\infty}\frac{H_{n}}{n^{2}}-2\sum_{n=1}^{\infty}\frac{1}{n^{3}}=0$$

$$\sum_{n=1}^{\infty}\frac{H_{n}}{n^{2}}=2\zeta(3)$$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\iverson}[1]{\left[\left[\,{#1}\,\right]\right]} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align*} & \color{#44f}{\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x} \over 1 + x}\dd x} \sr{x\ =\ 1/t\ -\ 1}{=} \\[5mm] = & \ \overbrace{\int_{1/2}^{1}{\ln^{2}\pars{t} \over t}\dd t} ^{\ds{\ln^{3}\pars{2}/3}}\ -\ \int_{1/2}^{1}{\ln\pars{t}\ln\pars{1 - t} \over t}\dd t \\[5mm] = & \ {\ln^{3}\pars{2} \over 3} + \int_{1/2}^{1}\on{Li}_{2}'\pars{t}\ln\pars{t}\,\dd t \\[5mm] \sr{\rm IBP}{=} & {\ln^{3}\pars{2} \over 3} + \bracks{\on{Li}_{2}\pars{1 \over 2}\ln\pars{2} -\int_{1/2}^{1}\overbrace{\on{Li}_{2}\pars{t} \over t}^{\ds{\on{Li}_{3}'\pars{t}}}\,\dd t} \\[5mm] = & \ {\ln^{3}\pars{2} \over 3} + \on{Li}_{2}\pars{1 \over 2}\ln\pars{2} - \on{Li}_{3}\pars{1} + \on{Li}_{3}\pars{1 \over 2} \end{align*} The above $\ds{\,\,\,\left.\rule{0pt}{5mm}\on{Li}_{s}\pars{1 \over 2}\right\vert_{s\ =\ 2, 3}\,\,\,}$ values are given in this link: $$ \mbox{Namely,}\quad \left\{\begin{array}{rcl} \ds{\on{Li}_{2}\pars{1 \over 2}} & \ds{=} & \ds{{\pi^{2} \over 12} - {\ln^{2}\pars{2} \over 2}} \\[2mm] \ds{\on{Li}_{3}\pars{1 \over 2}} & \ds{=} & \ds{{\ln^{3}\pars{2} \over 6} - {\pi^{2}\ln\pars{2} \over 12} + {7\zeta\pars{3} \over 8}} \end{array}\right. $$ and $\ds{\on{Li}_{3}\pars{1} = \zeta\pars{3}}$.

Therefore, $$ \color{#44f}{\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x} \over 1 + x}\dd x} = \bbx{\color{#44f}{-\,{\zeta\pars{3} \over 8}}} \approx -0.1503 $$

Answer 5

\begin{align} I&=\int_0^1 \frac{\ln x \ln(1+x)}{1+x}dx \overset{ibp} =-\frac12 \int_0^1 \frac{ \ln^2(1+x)}{x}dx \\ & = \frac14 \int_0^1 \frac{dx}{x} \left(2{\ln^2(1-x) }-{\ln^2(1-x^2)}-{\ln^2\frac{1-x}{1+x} } \right) \end{align} Substitute ${t=1-x}$, ${t=1-x^2}$ and $ t=\frac{1-x}{1+x}$ in the three integrals, respectively

\begin{align} I & = -\frac18\int_0^1 \frac{\ln^2t}{1-t}dt + \frac12\int_0^1 \frac{t\ln^2t}{1-t^2}\overset{t^2\to t}{dt}\\ & =- \frac1{16} \int_0^1 \frac{\ln^2t}{1-t}dt = -\frac1{8} \text{Li}_3(1) = -\frac{1}{8} \zeta(3) \end{align}

Answer 6

The integral can have the form

$$ I = -\sum_{k=1}^{\infty}\frac{(-1)^k\,H_{k}}{k^2}-\frac{3}{4}\zeta(3), $$

$H_k$ are the harmonic numbers. Try to work out above sum. See a related technique.

Answer 7

Your idea of writing $$\frac{\log (x) \log (x+1)}{x+1}=\sum _{n=1}^{\infty } a_n x^n \log (x)$$ by a Taylor expansion looks good to me almost when you take into account that, for value of $n$ greater or equal to $0$, $$ \int_0^1 x^n \ln x dx=-\frac{1}{(n+1)^2} $$ So $$ \int_0^1 \frac{\ln x \cdot \ln(1+x)}{1+x}dx=-\sum _{n=1}^{\infty } \frac{a_n}{(n+1)^2} $$ But, at this point, I am stuck with the $a_n$ and then with the summation. I made some numerical evaluations and observed that the convergence is not very fast.

I shall wait for answers to learn more.

Thanks for the interesting problem.

Edit

The $a_n$ are $$a_n=(-1)^{n+1} H_n$$ and $$\sum_{n=1}^\infty (-1)^n\,\frac{ H_n}{(n+1)^2}=-\frac{\zeta (3)}{8}$$