Question

$\rm \: Prove \: that : \: \mathbb{\int ^{1}_ 0 \frac{ln {}^{2}(1-x)ln(x)dx }{x} }=- \frac{\pi^4}{180} \\\\ Hint : \sum^ \infty_{n-1}H_n.x^n = \frac{ln(1-x)}{1-x}$
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Answer 1

First, assume we are well aware of the following famous result.

$\[ \zeta(2)=\dfrac{\pi^{2}}{6}, \quad \zeta(4)=\dfrac{\pi^{4}}{90} \]$

Next, by a simple calculation we obtain

$\\ \tt\[ H_{n}:=\sum_{k=1}^{n} \frac{1}{k}=\int_{0}^{1} \frac{1-t^{n}}{1-t} d t . \]$

and

$\\ \tt\[ \frac{\log (1-x)}{1-x}=-\sum_{n=1}^{\infty} H_{n} x^{n} \]$

Finally, define the polylogarithm as

$\\ \tt\[ Li _{s}(x):=\sum_{n=1}^{\infty} \frac{x^{n}}{n^{s}}, \]$

so that it satisfies the recurrence relation

$\pmb{\[ \begin{array}{l} \displaystyle\tt Li _{1}(x)=-\log (1-x) \\ \\ \displaystyle \tt Li _{s+1}(x)=\int_{0}^{x} \frac{ Li _{s}(t)}{t} d t \end{array} \]}$

and the identity

$\tt\[ Li _{s}(1)=\zeta(s) . \]$

The the all-in-one straight calculation goes as follows:

$\\ \tt\[ \int_{0}^{1} \frac{\log x \log ^{2}(1-x)}{x} d x=\]$

$\begin{array}{l} \displaystyle \tt=\int_{0}^{1} \frac{\log (1-x) \log ^{2} x}{1-x} d x= -\sum_{n=1}^{\infty} H_{n} \int_{0}^{1} x^{n} \log ^{2} x d x =-2 \\ \\ \displaystyle \tt\sum_{n=1}^{\infty} \frac{H_{n}}{(n+1)^{3}} \sum_{n=1}^{\infty}\left[\frac{1}{(n+1)^{4}}-\frac{H_{n+1}}{(n+1)^{3}}\right] \\ \\ \displaystyle \tt \sum_{n=0}^{\infty}\left[\frac{1}{(n+1)^{4}}-\frac{H_{n+1}}{(n+1)^{3}}\right] \\ \\ \displaystyle \tt =2\zeta(4)-2 \sum_{n=1}^{\infty} \frac{H_{n}}{n^{3}} \end{array}$

$\begin{array}{l}=2 \zeta(4)-2 \sum_{n=1}^{\infty} \frac{1}{n^{3}} \\ \\ \displaystyle \tt\int_{0}^{1} \frac{1-t^{n}}{1-t} d t=2 \zeta(4)-2 \\ \\ \displaystyle \tt \int_{0}^{1} \frac{\zeta(3)-\operatorname{Li}_{3}(t)}{1-t} d t \\ \tt +\left[2(\zeta(3)-\operatorname{Lin}(t)) \log \left(1-\frac{1}{d}\right)\right] +2 \\ \\ \displaystyle \tt \int_{0}^{1} \frac{\operatorname{Li}_{2}(t) \log (1-t)}{t} d t \\ \\ \displaystyle \tt=2 \zeta(4)-2 \int_{0}^{1} \operatorname{Li}_{2}(t) \frac{d \operatorname{Li}_{2}(t)}{d t} \\ \\ \displaystyle \tt =2 \zeta(4)-\left[\operatorname{Lin}_{2}^{2}(t)\right]_{0}^{1}=2 \zeta(4) \\ \displaystyle \tt -\zeta(2)^{2}=\frac{\pi^{4}}{45}-\frac{\pi^{4}}{36}=-\frac{\pi^{4}}{180} \end{array}$

Answer 2

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