On an integral representation of special values of the zeta function at odd integers

Abstract

An integral representation of the $p$-series of odd $p$ is shown;$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^{2p+1}}=(-1{)}^p\frac{(2\pi {)}^{2p}}{\left(2p\right)!}{\int }_0^1{B}_{2p}(t\left)\log \right(\sin \pi t\left)\mathrm{d}t\right(p=1,2,\dots ),$$where $B_{2p}\left(t\right)$ is a Bernoulli polynomial of degree $2p$. As a consequence of this we have $$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^{2p+1}}=(-1{)}^p\frac{(2\pi {)}^{2p}}{\left(2p\right)!}2\left[\sum _{k=0}^p\left(\begin{array}{c}2p\\ 2k\end{array}\right)B_{2p-2k}\left(\frac{1}{2}\right)b_{2k}\right],$$where $b_{2k}={\int }_0^{\frac{1}{2}}{t}^{2k}\log \left(\cos \pi t\right)\mathrm{d}t$, $k=0,1,\dots ,p$.