Integral with polylogarithm
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Integral with polylogarithm
Let $\mathbb{N} \ni m \geq 2$ and let $\zeta$ and ${\rm Li}_m$ denote the zeta and polylogarithmic function respectively. Evaluate the integral:
$$\int_{0}^1 \frac{\zeta(m) - {\rm Li}_{m} (x)}{1-x} (\ln x)^{m-1} {\rm d}x$$
$$\int_{0}^1 \frac{\zeta(m) - {\rm Li}_{m} (x)}{1-x} (\ln x)^{m-1} {\rm d}x$$
Imagination is much more important than knowledge.
Re: Integral with polylogarithm
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.
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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