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this is a homework problem I am stuck on: Compute the following integral for $\sigma > 1$ $\displaystyle \lim_{T \to \infty} \frac{1}{2T} \int_{-T}^{T}\left|\zeta{(\sigma + it)}\right|^2dt .$

I tried integrating over the contour given by the semicircle of radius $T$ with diameter from $\sigma - it$ to $\sigma + it$ going "left" and applying the residue theorem, but it does not seem to be going anywhere. I suppose part of my confusion comes from not completely understanding the structure of the function $\left|\zeta{(\sigma + it)}\right|^2$. Am I right to say that this function is holomorphic everywhere except at 1, where it has a pole of order 2 and no simple pole (by squaring the laurent expansion of $\zeta$ at 1)?

I'd rather not see a full solution immediately, but any hints in the right direction would be appreciated.

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    I'm sorry, it should be \sigma > 1.2012-02-28

2 Answers 2

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Hint: Use the fact that $\zeta(s)=\sum_{n=1}^{\infty}n^{-s}$ for $\sigma>1$. Then since all the series converge absolutely, we may freely rearrange everything, and bring the integral inside a double series, leading to the result.

Warning! Spoiler/Solution Follows:

For $\sigma>1$, since $\zeta(s)=\sum_{n=1}^{\infty}n^{-s}$, we may write $\left|\zeta\left(\sigma+it\right)\right|^{2}=\zeta\left(\sigma+it\right)\overline{\zeta\left(\sigma+it\right)}=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\left(mn\right)^{-\sigma}n^{-it}m^{it}.$ Now, we bring the integral inside the sum and split into cases depending on whether or not $\frac{m}{n}=1$. When $m\neq n$ then $\int_{-T}^{T}n^{-it}m^{it}dt=\int_{-T}^{T}\left(\frac{m}{n}\right)^{it}dt=\frac{i\left(\frac{m}{n}\right)^{-iT}-i\left(\frac{m}{n}\right)^{iT}}{\log\left(\frac{m}{n}\right)}.$ It is then not hard to show that $\sum_{n=1}^{\infty}\sum_{m\neq n}\left(mn\right)^{-\sigma}\frac{i\left(\frac{m}{n}\right)^{-iT}-i\left(\frac{m}{n}\right)^{iT}}{\log\left(\frac{m}{n}\right)}$ is bounded in absolute value for a fixed $\sigma>1$, and hence factor of $\frac{1}{2T}$ them implies that the off diagonal elements contribute $0$ in the limit. When $n=m$ , the integral is exactly $2T$ , so we conclude that $\lim_{T\rightarrow \infty} \frac{1}{2T}\int_{-T}^{T}\left|\zeta\left(\sigma+it\right)\right|^{2}dt=\sum_{n=m}\left(mn\right)^{-\sigma}=\sum_{n=1}^\infty (n^2)^{-\sigma}=\zeta(2\sigma).$

Edit: Corrected last line.

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    To Eric above: shouldn't that be $\zeta(2\sigma)$? Since when $m = n$, the summand simplifies to $(n^2)^{-\sigma} = n^{-2\sigma}$?2012-03-01
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Someone may have some better hints, but perhaps this will help (I'm looking at at a calculation of $\int_{-T}^T |\zeta(\sigma+it)|^{2k}\ dt$, in case there are discrepancies in my answer). First, note that $|\zeta(s)|^2$ is not holomorphic (since it is real-valued), so trying to use holomorphy is probably not helpful. Instead, let's be direct. Write $\zeta(s)$ as a partial sum (of terms less than $T$) plus error, $S+E$. Now, your integral is $\int_0^T (S+E)(\bar S+\bar E)$. Expand the integrand and bound (in $T$) all the terms other than $\int_0^T |S|^2$ (they should all vanish when you divide by $T$ and take the limit). Then bound that term (just multiply everything out and collect the correct terms together, this step may be difficult). I hope I haven't given you more than you want. I can provide further steps if you need them.

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    Since you have absolute convergence, the step with the error is unnecessary. One can simply expand everything out entirely and switch orders.2012-02-29