Integral, definite integral

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.

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):

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

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