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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,

2 Answers 2

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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)

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    Hi Rou$p$am! Working through the $p$roof of the Mellin transform identity from first $p$rinci$p$les is equivalent to my original idea. Well done.2011-01-10
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I do not intend to post it, but for those who are interested the proof of the tough part of this question, namely the classic result that

$ \Gamma(s) \sin \left( \frac{\pi s}{2} \right) = \int_0^\infty y^{s-1} \sin y \textrm{ d}y $

where $-1 < Re(s) < 1 $ can be found in "Topics in Analytic Number Theory," by Hans Rademacher (in chapter 6).

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    Which answer should I accept? Both give part of the solution. :D2011-01-07