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.

Solution 1:

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*}

Solution 2:

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.