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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?


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up vote 8 down vote accepted

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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:) Yea... a few hours would be good!!! – Roupam Ghosh Jan 6 '11 at 17:13
Got it, Mellin transform of $\sin(y)$ in $(0,\infty)$ is $\Gamma(s) \sin\frac{\pi s}{2}$ using standard formulas!!! What was your method, that you had in mind? – Roupam Ghosh Jan 6 '11 at 17:32
Hi Roupam! Working through the proof of the Mellin transform identity from first principles is equivalent to my original idea. Well done. – user02138 Jan 10 '11 at 14:37

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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By the way, if you should pick up the book don't just stop at the appropriate page for this problem. The book is excellent and chapter 14 contains one of my all time favourite results, Rademacher's convergent series for the partition function. It's well worth a look. – Derek Jennings Jan 6 '11 at 18:39
Yup... Thanks, thats what I needed... Its on page 82 (39.2)... I was going through the book, its a really good book!!! – Roupam Ghosh Jan 7 '11 at 2:56
Which answer should I accept? Both give part of the solution. :D – Roupam Ghosh Jan 7 '11 at 3:00

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