In Titchmarsh's book "The theory of the Riemann Zeta function" pg. 15 where the functional equation of the zeta function is being derived, I couldn't understand this part: $$\frac{s}{\pi} \sum_{n=1}^{\infty} \frac{(2n\pi)^s}{n} \int_{0}^{\infty} \frac{\sin y}{y^{s+1}} dy = \frac{s}{\pi} (2\pi)^s \{-\Gamma(-s)\}\sin\frac{1}{2}s\pi\zeta(1-s)$$
I could not digest Titchmarsh's reasoning. Can anyone explain this please?
Thanks,
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)