Compute $ \int_{0}^{1}\frac{\ln(x) \ln^2 (1-x)}{x} dx $ Compute
$$ \int_{0}^{1}\frac{\ln(x) \ln^2 (1-x)}{x} dx $$
I'm looking for some nice proofs at this problem. One idea would be to use Taylor expansion and then integrating term by term. What else can we do? Thanks.
 A: The Taylor expansion approach gives you $-2 \sum_{k=1}^\infty H_k/(k+1)^3$ where $H_k = \sum_{n=1}^k 1/n$.  Wolfram Alpha says this is $-\pi^4/180$, but I don't know how it gets that.
A: @Chri's sister: see here http://www.artofproblemsolving.com/Forum/viewtopic.php?f=296&t=353720&p=1921474&hilit=Borwein#p1921474 
A: This answer is from my old calculation.
First, assume we are well aware of the following famous result.
$$\zeta(2) =\frac{\pi^{2}}{6}, \quad \zeta(4) =\frac{\pi^{4}}{90}$$
Next, by a simple calculation we obtain
$$ H_{n} := \sum_{k=1}^{n} \frac{1}{k} =\int_{0}^{1}\frac{1-t^{n}}{1-t}\, dt. $$
and 
$$ \frac{\log (1-x)}{1-x}\ =\ -\sum_{n=1}^{\infty}H_{n}x^{n}. $$
Finally, define the polylogarithm as
$$ \mathrm{Li}_{s}(x) :=\sum_{n=1}^{\infty} \frac{x^n}{n^s}, $$
so that it satisfies the recurrence relation
$$ \mathrm{Li}_{1}(x) =-\log (1-x) , \quad \mathrm{Li}_{s+1}(x) =\int_{0}^{x}\frac{\mathrm{Li}_{s}(t)}{t}\, dt $$
and the identity
$$ \mathrm{Li}_{s}(1) =\zeta(s). $$
The the all-in-one straight calculation goes as follows:
\begin{align*}
\int_{0}^{1}\frac{\log x\log^{2}(1-x)}{x}\, dx
& = \int_{0}^{1}\frac{\log (1-x)\log^{2}x}{1-x}\, dx
  = -\sum_{n=1}^{\infty}H_{n}\int_{0}^{1}x^{n}\log^{2}x\, dx\\
& = -2\sum_{n=1}^{\infty}\frac{H_{n}}{(n+1)^{3}}\\
& = 2\sum_{n=1}^{\infty}\left[\frac{1}{(n+1)^{4}}-\frac{H_{n+1}}{(n+1)^{3}}\right]
  = 2\sum_{n=0}^{\infty}\left[\frac{1}{(n+1)^{4}}-\frac{H_{n+1}}{(n+1)^{3}}\right]\\ 
& = 2\zeta(4)-2\sum_{n=1}^{\infty}\frac{H_{n}}{n^{3}}\\ 
& = 2\zeta(4)-2\sum_{n=1}^{\infty}\frac{1}{n^{3}}\int_{0}^{1}\frac{1-t^{n}}{1-t}\, dt
  = 2\zeta(4)-2\int_{0}^{1}\frac{\zeta(3)-\mathrm{Li}_{3}(t)}{1-t}\, dt\\ 
& = 2\zeta(4)+\left[2 (\zeta(3)-\mathrm{Li}_{3}(t))\log (1-t)\right]_{0}^{1}+2\int_{0}^{1}\frac{\mathrm{Li}_{2}(t)\log (1-t)}{t}\, dt\\ 
& = 2\zeta(4)-2\int_{0}^{1}\mathrm{Li}_{2}(t)\frac{d\mathrm{Li}_{2}(t)}{dt}\, dt\\ 
& = 2\zeta(4)-\left[\mathrm{Li}_{2}^{2}(t)\right]_{0}^{1}
  = 2\zeta(4)-\zeta(2)^{2} 
  = \frac{\pi^{4}}{45}-\frac{\pi^{4}}{36}
  = -\frac{\pi^{4}}{180}\\ 
& = -\frac{1}{2}\zeta(4).
\end{align*}
A: In this answer I will make use of a Maclaurin series expansion for the term $\ln^2 (1 - x)$, which I show here to be
$$\ln^2 (1 - x) = 2 \sum_{n = 2}^\infty \frac{H_{n - 1} x^n}{n},$$
and the well-known Euler sum of
$$\sum_{n = 1}^\infty \frac{H_n}{n^3} = \frac{1}{2} \zeta^2 (2),$$
several proofs for which can be found here. 
From the above Maclaurin series expansion for $\ln^2 (1 - x)$ the integral can be written as
\begin{align*}
\int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = 2 \sum_{n = 2}^\infty \int_0^1 x^{n - 1} \ln x \, dx.
\end{align*}
The integral that appears to the right can be readily found by parts. The result is
$$\int_0^1 x^{n - 1} \ln x \, dx = -\frac{1}{n^2}.$$
Thus
$$\int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = -2 \sum_{n = 2}^\infty \frac{H_{n - 1}}{n^3}.$$
From properties of harmonic numbers, since
$$H_n = H_{n - 1} + \frac{1}{n},$$
the integral becomes
$$\int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = 2 \sum_{n = 2}^\infty \frac{1}{n^4} - 2 \sum_{n = 2}^\infty \frac{H_n}{n^3} = 2 \sum_{n = 1}^\infty \frac{1}{n^4} - 2 \sum_{n = 1}^\infty \frac{H_n}{n^3}.$$
As 
$$\sum_{n = 1}^\infty \frac{1}{n^4} = \zeta (4) \quad \text{and} \quad \sum_{n = 1}^\infty \frac{H_n}{n^3} = \frac{1}{2} \zeta^2 (2),$$
we have
$$\int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = 2 \zeta (4) - \zeta^2 (2) = - \frac{\pi^4}{180} = -\frac{1}{2} \zeta (4),$$
as expected.
