Integral, definite integral How can we prove
$$
\int_0^1 \frac{\ln x \cdot \ln(1+x)}{1+x}dx=-\frac{\zeta(3)}{8}?
$$
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
 A: 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:
$u=log(u), \;\ dv=\frac{log(1-1/u)}{u}, \;\ du=\frac{1}{u}du, \;\ v=Li_{2}(1/u)$
(as a note, $\int\frac{log(1-1/u)}{u}du=Li_{2}(1/u)$ is a rather famous integral related to the dilog).
$$\int\frac{log(u)log(1-1/u)}{u}du=log(u)Li_{2}(1/u)-\int\frac{Li_{2}(1/u)}{u}du$$
Also, note this last integral is simply $$-Li_{3}(1/u)$$
Now, back sub $u=x+1$, and put it altogether using the integration limits 0 to 1.
Hence, we arrive at:
$$ \left|1/3log^{3}(x+1)+log(x+1)Li_{2}\left(\frac{1}{x+1}\right)+Li_{3}\left(\frac{1}{1+x}\right)\right|_{0}^{1}$$
$$=1/3log^{3}(2)+log(2)Li_{2}(1/2)+Li_{3}(1/2)-Li_{3}(1).........(1)$$
Note the identities:  
$$Li_{2}(1/2)=\frac{\pi^{2}}{12}-1/2log^{2}(2)$$
$$Li_{3}(1/2)=7/8\zeta(3)+1/6log^{3}(2)-\frac{\pi^{2}}{12}log(2)$$
sum up (1):
$$1/3log^{3}(2)+log(2)\left(\frac{\pi^{2}}{12}-1/2log^{2}(2)\right)+\left(7/8\zeta(3)+1/6log^{3}(2)-\frac{\pi^{2}}{12}log(2)\right)-\zeta(3)$$
$$=\frac{-\zeta(3)}{8}$$
A: 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.
A: 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.
A: 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:
$$\sum_{n=1}^{\infty}\frac{H_{n}}{2n^{2}}-\sum_{n=1}^{\infty}\frac{1}{2n^{3}}+\sum_{n=1}^{\infty}\frac{H_{n}}{2n^{2}}-\sum_{n=1}^{\infty}\frac{1}{n^{3}}-1/2\zeta(3)=0$$
$$\sum_{n=1}^{\infty}\frac{H_{n}}{n^{2}}-2\sum_{n=1}^{\infty}\frac{1}{n^{3}}=0$$
$$\sum_{n=1}^{\infty}\frac{H_{n}}{n^{2}}=2\zeta(3)$$
A: Note
\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 & = \frac38\int_0^1 \frac{\ln^2t}{1-t}dt
 - \frac12\int_0^1 \frac{\ln^2t}{1-t^2}dt\\ &=\frac38\int_0^1 \frac{\ln^2t}{1-t}dt
 - \frac12\int_0^1 {\ln^2t}\left( \frac1{1-t}-\frac t{1-t^2}\right)dt
\end{align}
Let $t\to t^2$ in the last integral
\begin{align}
I & =- \frac1{16} \int_0^1 \frac{\ln^2t}{1-t}dt
\overset{IBP} = -\frac1{8} \int_0^1 \frac{\ln t\ln(1-t)}{t}dt\\
& =  -\frac1{8} \int_0^1 \ln t \>d\left( \int_0^t\frac{\ln (1-u)}u du\right)
 =  -\frac1{8} \int_0^1 \frac{-\int_0^t\frac{\ln (1-u)}u du}{t} dt \\
&=-\frac1{8} \int_0^1 \frac{Li_2(t)}{t} dt =    -\frac1{8}Li_3(1) = -\frac{\zeta(3)}{8}
\end{align}
A: 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.
