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