Evaluate the integral:
$$\int_{0}^{1}\frac{\ln (1-x) \ln x}{1-x}\, {\rm d}x$$
Solution
We are using the following Lemma:
Then for our integral we have successively:
$$ \begin{align*}
\int_{0}^{1}\frac{\ln (1-x)\ln x}{1-x}\, {\rm d}x &=- \int_{0}^{1}\sum_{n=1}^{\infty}\mathcal{H}_n x^n \ln x \, {\rm d}x \\
&= - \sum_{n=1}^{\infty}\mathcal{H}_n \int_{0}^{1}x^n \ln x \, {\rm d}x\\
&= \sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{\left ( n+1 \right )^2}
\end{align*}$$
and we actually have to evaluate this Euler sum. Note that:
$$\mathcal{H}_n = \mathcal{H}_{n+1}- \frac{1}{n+1}$$
Hence the above Euler sum is transformed to:
$$\begin{align*}
\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{\left ( n+1 \right )^2} &=\sum_{n=1}^{\infty}\frac{\mathcal{H}_{n+1}- \frac{1}{n+1}}{\left ( 1+n \right )^2} \\
&= \sum_{n=1}^{\infty}\frac{\mathcal{H}_{n+1}}{\left ( n+1 \right )^2}- \sum_{n=1}^{\infty}\frac{1}{\left ( n+1 \right )^3}\\
&= \sum_{n=2}^{\infty}\frac{\mathcal{H}_n}{n^2} - \sum_{n=1}^{\infty}\frac{1}{\left ( n+1 \right )^3} \\
&= \sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n^2} - 1 - \zeta (3)+1\\
&= \zeta(3)
\end{align*}$$
ending the calculations. We have seen the first Euler sum in the post here.
$$\int_{0}^{1}\frac{\ln (1-x) \ln x}{1-x}\, {\rm d}x$$
Solution
We are using the following Lemma:
Lemma: It holds that: $$\frac{\ln (1-x)}{1-x}= -\sum_{n=1}^{\infty}\mathcal{H}_n x^n$$ where $\mathcal{H}_n$ is the $n$-th harmonic number.
Proof: $$\begin{aligned}\frac{\ln (1-x)}{1-x}&=- \left ( \sum_{n=1}^{\infty}\frac{x^n}{n} \right )\left ( \sum_{n=0}^{\infty}x^n \right )\\
&=- \left ( x+ \frac{x^2}{2}+ \frac{x^3}{3}+\cdots \right )\left ( 1+x+x^2+x^3 +\cdots \right )\\
&=- \left [ x + \left ( 1+ \frac{1}{2} \right )x^2 + \left ( 1+ \frac{1}{2}+ \frac{1}{3} \right )x^3 +\cdots \right ]\\
&= - \sum_{n=1}^{\infty}\mathcal{H}_n x^n
\end{aligned}$$
Then for our integral we have successively:
$$ \begin{align*}
\int_{0}^{1}\frac{\ln (1-x)\ln x}{1-x}\, {\rm d}x &=- \int_{0}^{1}\sum_{n=1}^{\infty}\mathcal{H}_n x^n \ln x \, {\rm d}x \\
&= - \sum_{n=1}^{\infty}\mathcal{H}_n \int_{0}^{1}x^n \ln x \, {\rm d}x\\
&= \sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{\left ( n+1 \right )^2}
\end{align*}$$
and we actually have to evaluate this Euler sum. Note that:
$$\mathcal{H}_n = \mathcal{H}_{n+1}- \frac{1}{n+1}$$
Hence the above Euler sum is transformed to:
$$\begin{align*}
\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{\left ( n+1 \right )^2} &=\sum_{n=1}^{\infty}\frac{\mathcal{H}_{n+1}- \frac{1}{n+1}}{\left ( 1+n \right )^2} \\
&= \sum_{n=1}^{\infty}\frac{\mathcal{H}_{n+1}}{\left ( n+1 \right )^2}- \sum_{n=1}^{\infty}\frac{1}{\left ( n+1 \right )^3}\\
&= \sum_{n=2}^{\infty}\frac{\mathcal{H}_n}{n^2} - \sum_{n=1}^{\infty}\frac{1}{\left ( n+1 \right )^3} \\
&= \sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n^2} - 1 - \zeta (3)+1\\
&= \zeta(3)
\end{align*}$$
ending the calculations. We have seen the first Euler sum in the post here.
Another solution (more elementary) given in mathimatikoi.org:
ReplyDelete\begin{align*} \int_0^1{\frac{\log(1-x)\log{x}}{1-x}\,dx}&\mathop{=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c} {t\,=\,1-x}\\ {-dt\,=\,dx} \\ \end{subarray}} \,\int_0^1{\frac{\log{t}\,\log(1-t)}{t}\,dt}\\ &=-\int_0^1{\frac{\log{t}}{t}\mathop{\sum}\limits_{n=1}^{\infty}\frac{t^n}{n}\,dt}\\ &=-\mathop{\sum}\limits_{n=1}^{\infty}\frac{1}{n}\int_0^1t^{n-1}\log{t}\,dt\\ &=-\mathop{\sum}\limits_{n=1}^{\infty}\frac{1}{n}\Big(-\frac{1}{n^2}\Big)\\ &=\mathop{\sum}\limits_{n=1}^{\infty}\frac{1}{n^3}\\
&=\zeta(3)
\end{align*}