Find a sum, terms given as ratios of improper integrals

Question

Given $$\alpha_k =\frac{\displaystyle \int_0^\infty \frac {x^{k-1}} {e^x -1 } dx }{ \displaystyle \int_0^\infty \frac{x^{k-1}}{e^x} dx }; \quad k\in N$$ find $$\frac {\alpha_2}{1} -\frac{\alpha_6}{15} +\frac {\alpha_{10}}{125} -\frac{\alpha_{14}}{875} +\frac {\alpha_{18}}{5625} -\frac {\alpha_{22}}{34375} +\cdots $$

Answer 1

Hint. One may observe that $$ \begin{align} \int _0^\infty \frac {x^{k-1}}{e^{ x }-1 } dx &=\Gamma(k)\zeta(k) \tag1\\\\ \int _0^\infty \frac {x^{k-1}}{e^{ x }} dx &=\Gamma(k) \tag2 \end{align} $$ and that the first terms of the sequence $\displaystyle \left\{(2k+1)\times5^k \right\} $ are given by $$ 1,15,125,875,5625,34375,\ldots $$ Then the series you are looking for is $$ S=\sum_0^{\infty}(-1)^k\frac{\zeta(4k+2) }{2k+1}\left(\frac15\right)^k. \tag3 $$ From the standard power series expansion of the digamma function, $$ \psi(1+z)=-\gamma-\sum_0^{\infty}(-1)^k\zeta(k+1)z^k,\quad |z|<1, $$ one may deduce that $$ \begin{align} 2\sum_0^{\infty}\frac{\zeta(4k+2) }{2k+1}z^{4k+2}&=\log \left(\Gamma(1+z)\Gamma(1-z) \right)-\log \left(\Gamma(1+iz)\Gamma(1-iz) \right) \tag4\\\\ &=\log \left(\frac{\pi z}{\sin (\pi z)}\right)-\log \left(\frac{\pi z}{\sinh (\pi z)}\right)\\\\ &=\log \left(\frac{\sinh (\pi z)}{\sin (\pi z)}\right). \end{align} $$ Hence

$$ \sum_0^{\infty}\frac{\zeta(4k+2) }{2k+1}z^{4k}=\frac{1}{2z^2}\log \left(\frac{\sinh (\pi z)}{\sin (\pi z)}\right), \quad |z|<1,\tag5 $$

from which you may obtain $(3)$.