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Logarithmic integral

Calculus (Integrals, Series)
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Tolaso J Kos
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Logarithmic integral

#1

Post by Tolaso J Kos » Tue Nov 10, 2015 2:03 pm

Evaluate the integral:

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

#2

Post by Grigorios Kostakos » Tue Nov 10, 2015 2:13 pm

\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
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Re: Logarithmic integral

#3

Post by Tolaso J Kos » Tue Nov 10, 2015 4:05 pm

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.
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