After finding the formula for the integral in my post,
\begin{aligned} \int_{0}^{1} \frac{\ln x}{x^{n}-1} d x &= \frac{1}{n^2} \zeta\left(2, \frac{1}{n}\right) \end{aligned}
I plan to go further with higher power of $\ln x$,
$$
I(p, n)=\int_0^1 \frac{\ln ^p x}{1-x^n} d x,\\
$$where $p\in \mathbb N$ and $n>0$.
Using the power series for $|x|<1$,
$\displaystyle \frac{1}{1-x}=\sum_{k=0}^{\infty} x^k\tag*{} $
we can express the integral as
$\displaystyle I(p, n)=\int_0^1 \frac{\ln ^p x}{1-x^n} d x=\sum_{k=0}^{\infty} \underbrace{ \int_0^1 x^{n k} \ln ^p x d x}_{I_k} \tag*{} $
Noting that
$\displaystyle \begin{aligned}I_k & =\left.\frac{\partial^p}{\partial a^p} \int_0^1 x^a d x\right|_{a=n k} \\& =\left.\frac{(-1)^{p} p!}{(a+1)^{p+1}}\right|_{a=n k} \\& =\frac{(-1)^p p !}{(n k+1)^{p+1}}\end{aligned} \tag*{} $
Plugging back yields
$\displaystyle \begin{aligned}I(p, n) & =\sum_{k=0}^{\infty} \frac{(-1)^p p !}{(n k+1)^{p+1}} \\& =(-1)^p p ! \sum_{k=0}^{\infty} \frac{1}{(n k+1)^{p+1}} \\& =\frac{(-1)^p p !}{n ^{p+1}} \sum_{k=0}^{\infty} \frac{1}{\left(k+\frac{1}{n}\right)^{p+1}} \\& =\boxed{\frac{(-1)^p p !}{n^{p+1}}\zeta \left(p+1, \frac{1}{n}\right)}\end{aligned}\tag*{} $
For examples,
$\displaystyle \begin{aligned}& I(2,2)=\frac{(-1)^2 \times 2 !}{2^3} \zeta\left(3, \frac{1}{2}\right)=\frac{7 \zeta(3)}{4}\approx{2.1035996} \\& I(3,5)=\frac{(-1)^3 \times 3 !}{5^4} \zeta\left(4, \frac{1}{5}\right)=-\frac{6}{625} \zeta\left(4, \frac{1}{5}\right)\approx {-6.005192}\end{aligned}\tag*{} $
My question:
Do we have other simpler alternative methods?
Best Answer
Performing the change of integration variables from $x$ to $t$ via $x=\mathrm{e}^{-t/n}$ and using $(25.11.25)$, we obtain $$ I(p,n) = \frac{{( - 1)^p }}{{n^{p + 1} }}\int_0^{ + \infty } {\frac{{t^p {\rm e}^{ - t/n} }}{{1 - {\rm e}^{ - t} }}{\rm d}t} = \frac{{( - 1)^p p!}}{{n^{p + 1} }}\zeta\! \left( {p + 1,\tfrac{1}{n}} \right), $$ provided $p\ge 0$ is an integer and $n>0$.