📝 SummarXiv

Abstract

In this work we study integrals of the form $\int_{0}^{\infty}\frac{\tanh(x)}{x}\mathrm{sech}(x)^{L}\exp(-Tx)dx$. For $L \in \mathbb{N}$, $L \leq 4$ and $T \in \mathbb{R}$ we give explicit expressions in terms of derivatives of the Hurwitz zeta function at negative integers. We use these expressions to evaluate these integrals for $T \in \mathbb{N}_{0}$ exactly. For the special case $T = 0$ we give explicit evaluations for any $L \in \mathbb{N}$ based on the functional equations for $\beta(s)$ and $\zeta(s)$. As it turns out the value of $\int_{0}^{\infty}\frac{\tanh(x)}{x}\mathrm{sech}(x)^{2N + 1}dx$ is a linear combination of $\beta(2)$, ..., $\beta(2N + 2)$ and the value of $\int_{0}^{\infty}\frac{\tanh(x)}{x}\mathrm{sech}(x)^{2N}dx$ a linear combination of $\zeta(3)$, ..., $\zeta(2N + 1)$. We give recursive formulae for the coefficients in these linear combinations.