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## Logarithmic integral

Calculus (Integrals, Series)
Tolaso J Kos
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### Logarithmic integral

Evaluate the integral:

$$\int_0^1 \frac{\ln (1-x) \ln x}{1-x}\, {\rm d}x$$
Imagination is much more important than knowledge.
Grigorios Kostakos
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### Re: Logarithmic integral

\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*}
Grigorios Kostakos
Tolaso J Kos
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### Re: Logarithmic integral

Hey Grigoris,

thank you for your solution. Here is another one which actually is able to crack with more ease the more general case:
$$\int_0^1 \frac{\ln (1-x) \ln^k x}{1-x}\, {\rm d}x, \;\;\; k \geq 1$$

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*}

The Euler sum $\displaystyle \sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{n^2}$ is evaluated as follows:

\begin{aligned} \sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n^2} &=\sum_{n=1}^{\infty}\frac{1}{n^2}\int_{0}^{1}\frac{1-x^n}{1-x}\,{\rm d}x \\ &= \int_{0}^{1}\frac{1}{1-x}\sum_{n=1}^{\infty}\left ( \frac{1}{n^2}-\frac{x^n}{n^2} \right )\,{\rm d}x\\ &=\int_{0}^{1}\frac{\frac{\pi^2}{6}-{\rm Li_2}(x)}{1-x}\,{\rm d}x \\ &= \cancelto{0}{ -\ln(1-x)\left ( \frac{\pi^2}{6}-{\rm Li_2}(x) \right )\bigg|_0^1}+\int_{0}^{1}\frac{\ln^2 (1-x)}{x}\,{\rm d}x\\ &\overset{u=1-x}{=\! =\! =\! =\!}\int_{0}^{1}\frac{\ln^2 u}{1-u}\,{\rm d}u \\ &= \int_{0}^{1}\ln^2 x\sum_{k=0}^{\infty}x^k\,{\rm d}x\\ &= \sum_{k=0}^{\infty}\int_{0}^{1}x^k \ln^2 x \,{\rm d}x\\ &= \cdots \\ &=2\sum_{k=0}^{\infty}\frac{1}{\left ( k+1 \right )^3}=2\zeta(3) \end{aligned}

In the above calculations we made use of the integral representation of the $n$ - th Harmonic number which is:
$$\mathcal{H}_n = \int_0^1 \frac{1-x^n}{1-x}\, {\rm d}x$$
Imagination is much more important than knowledge.