how about this log generalization?

[galactus]

Derive a general closed form for:


$$\int_{0}^{\infty}\left(\frac{\log(x)}{x-1}\right)^{n}dx$$

$$=n\sum_{m=1}^{\infty}\frac{1}{m^{n}}\left(\prod_{k=1}^{n-2}(m+k)+\prod_{k=1}^{n-2}(m-k)\right)$$

or whatever form you come up with that gives the correct solution.

e.g. Let $n=6$ and we get:

$$\int_{0}^{\infty}\left(\frac{\log(x)}{x-1}\right)^{6}dx=2\pi^{2}+\frac{14}{3}\pi^{4}+\frac{32}{105}\pi^{6}=12\zeta(2)+420\zeta(4)+288\zeta(6)$$

[Tolaso J Kos]

Here is a solution. Let us first use some lemmas:

Lemma 1: It holds that:

$$\frac{1}{\left ( 1-x \right )^m}= \frac{1}{(m-1)!}\sum_{k=m}^{\infty} \frac{(k-1)!}{(k-m)!} x^{k-m}$$

Proof: Left to the reader.

Lemma 2: It holds that:

$$\int_{0}^{1} x^{k-m}\log^n x \, {\rm d}x = (-1)^n \frac{n!}{(k+1-m)^{n+1}}$$

Proof: Well, we have successively:

\begin{align*}
\int_0^1 x^{k-m} \log^n x \, {\rm d}x &\overset{y=-\log x}{=\! =\! =\! =\! =\!} (-1)^n \int_{0}^{\infty} y^n e^{-(k+1-m)y} \, {\rm d}y\\
&\!\!\!\!\!\!\!\!\!\!\overset{(k+1-m)y=u}{=\! =\! =\! =\! =\! =\! =\! =\!} \frac{(-1)^n}{\left ( k+1-m \right )^{n+1}} \int_{0}^{\infty}x^n e^{-x} \, {\rm d}x\\
&= \frac{(-1)^n}{\left ( k+1-m \right )^{n+1}} \Gamma(n+1)\\
&= \frac{(-1)^n n!}{\left ( k+1-m \right )^{n+1}}
\end{align*}

Now, back to the problem. Let $m, n \in \mathbb{N}$. We have that:

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

Now equating $m=n$ we have that:

$$\int_{0}^{1} \frac{\log^n x}{(1-x)^n} \, {\rm d}x = (-1)^{2n-1} n \sum_{j=1}^{n-1} (-1)^j s (n-1, j) \zeta(n-j+1) , \; n \geq 2$$

while for $n=1$ we have that $\displaystyle \int_0^1 \frac{\log x}{1-x}\, {\rm d}x = -\frac{\pi^2}{6}$.

In general:

$$\int_{0}^{1} \frac{\log^n x}{(1-x)^m} \, {\rm d}x = \left\{\begin{matrix}
(-1)^n n! \zeta(n+1) &, &m=1 \\\\
\displaystyle \frac{(-1)^{n+m-1} n!}{(m-1)!} \sum_{j=1}^{m-1} (-1)^j s (m-1, j) \zeta(n-j+1)& , & m \geq 2
\end{matrix}\right.$$

where $s(m, n)$ are the Stirling numbers of first kind. Definition is equation $8$ at the link.

Edit: Whoops. I solved another problem. The initial states otherwise. Well split the integral apart from $0$ to $1$ and $1$ to $\infty$. I leave the second part for now .