Integral with polylogarithm

Calculus (Integrals, Series)
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Tolaso J Kos
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Integral with polylogarithm

#1

Post by Tolaso J Kos »

Let $\mathbb{N} \ni m \geq 2$ and let $\zeta$ and ${\rm Li}_m$ denote the zeta and polylogarithmic function respectively. Evaluate the integral:

$$\int_{0}^1 \frac{\zeta(m) - {\rm Li}_{m} (x)}{1-x} (\ln x)^{m-1} {\rm d}x$$
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Re: Integral with polylogarithm

#2

Post by admin »

A solution by Seraphim

We note that

$$\zeta(m) - {\rm Li}_m(x)= \sum_{n=1}^{\infty} \frac{1-x^n}{n^m}$$

Therefore

\begin{align*}
S &= \int_{0}^{1}\frac{\zeta(m)-{\rm Li}_m (x)}{1-x} \log^{m-1} x \, {\rm d}x \\
&= \sum_{n=1}^{\infty} \frac{1}{n^m} \int_{0}^{1}\frac{1-x^n}{1-x} \log^{m-1} x \, {\rm d}x\\
&=\sum_{n=1}^{\infty} \frac{1}{n^m} \sum_{k=0}^{n-1} \int_{0}^{1}x^k \log^{m-1} x \, {\rm d}x \\
&= \sum_{n=1}^{\infty} \frac{1}{n^m} \sum_{k=0}^{n-1} \frac{(-1)^{m-1} (m-1)!}{(k+1)^m}\\
&= (-1)^{m-1} (m-1)! \sum_{n=1}^{\infty} \frac{1}{n^m} \sum_{k=1}^{n} \frac{1}{k^m} \\
&= \frac{(-1)^{m-1} (m-1)!}{2} \left ( \zeta^2(m) + \zeta(2m) \right )
\end{align*}

The last equality holds from equations $4$ and $20$ that can be found here.
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