If I'm reading your question correctly, you'd like to prove the stated equality? If so, perhaps this might orient you a little bit. Write the left hand side as \begin{eqnarray} \frac{s}{\pi} \left( \int_{0}^{\infty} \frac{\sin y}{y^{s+1}} dy \right) (2\pi)^{s} \left( \sum_{n = 1}^{\infty} n^{s-1} \right) \end{eqnarray} To prove that this equals the right side you'll need a definition and an identity. The zeta function is defined as $\zeta(s) = \sum_{n = 1}^{\infty} n^{-s}$ with $\mathbf{Re}(s) > 1$, so the sum above is clearly equal to $\zeta(1-s)$ with $\mathbf{Re}(s) < 0$. The hard part is now showing the following gamma function integral representation \begin{eqnarray} \Gamma(s) = \frac{1}{\sin \frac{\pi s}{2}} \int_{0}^{\infty} \frac{\sin y}{y^{1-s}} dy, \end{eqnarray} where the integral converges if and only if $-1 < \mathbf{Re}(s) < 1$. Once you've got this in hand, then your equality is true on $-1 < \mathbf{Re}(s) < 0$. Now just analytically continue by individually continuing both the gamma function and the zeta functions to their largest respective domains.
(Before I post a proof of the integral representation, I'll give you a few hours to try to work on it yourself. Hint: Use a change of variables on a more common integral representation. More to come)