TITLE: Calculation of the integral $\int_0^\infty d x\, x^{2n+1} \tanh(\pi x)\, \log (1 - e^{-2\pi x})$ for non-negative integer $n$
QUESTION [7 upvotes]: In some physics problem I ran into the integral with a non-negative integer $n$ of the form:
\begin{align}
I(n) \equiv \int_0^\infty d x\, x^{2n+1} \tanh(\pi x)\, \log (1 - e^{-2\pi x}) \ ,
\end{align}
which I cannot perform even with Mathematica for general $n$, but for small $n$ the values are given analytically (e.g. in Eq. (3.10) and (3.23) of https://arxiv.org/pdf/1708.00305.pdf):
\begin{align}
\begin{aligned}
I(0) &= - \frac{1}{8}\,\log 2 + \frac{9}{16\pi^2}\,\zeta(3) \ ,\\
I(1) &= - \frac{1}{64}\,\log 2 - \frac{3}{32\pi^2}\,\zeta(3) + \frac{225}{128\pi^4}\,\zeta(5) \ ,
\end{aligned}
\end{align}
It is likely that the integral generally takes the form
\begin{align}
I(n) = c_0\,\log 2 + \sum_{i=1}^{n+1}\,c_i\,\zeta(2i+1)\ .
\end{align}
Can we fix the values of the coefficients $c_i~(i=0,1,\cdots,n+1)$ for integer $n\ge 0$?
Thank you for your help in advance.

REPLY [4 votes]: Yes, every $I(n)$ can be expressed in $\zeta$.
Note that
$$I(n) = \frac{-1}{(2\pi)^{2n+2}}\int_0^1 \left(\frac{1}{u}-\frac{2}{u+1}\right)\log^{2n+1} u \log (1-u) du$$
Such kind of integral is well-known on this site, for example, see links under this question. We have
$$\int_0^1 \frac{\log^{2n+1} u \log (1-u)}{u} du = (2n+1)! \zeta(2n+3)$$
and in term of multiple-zeta function $$\int_0^1 \frac{\log^{2n+1} u \log (1-u)}{1+u} du = -(2n+1)! \zeta(\overline{2n+2},\overline{-1})$$
The follow formula exists:
$$\zeta(\overline{2n+2},\overline{1}) = \sum _{k=0}^{n+1} \zeta (2 k) \bar{\zeta}(-2 k+2 n+3)-\frac{\zeta (2 n+3)}{2}+\bar{\zeta}(1) \bar{\zeta}(2 n+2)-n \bar{\zeta}(2 n+3)-\bar{\zeta}(2 n+3)$$
where $\bar{\zeta}(s) = \left(1-2^{1-s}\right) \zeta (s)$