Inverse zeta(3) limit

Evaluate the limit

$\ell= \lim_{T \rightarrow +\infty} \frac{1}{2T} \int \limits_{-T}^{T} \frac{\zeta(\frac{3}{2}+it)}{\zeta(\frac{3}{2}-it)} \, {\rm d}t$

Solution

We are proving that the limit is $\frac{1}{\zeta(3)}$ . Indeed , one has:

$\frac{\zeta(3/2+it)}{\zeta(3/2-it)} = \sum_{n=1}^\infty n^{-3/2} \sum_{d | n} \mu(d) (d^2/n)^{it}$

If $x \neq 1$ then

$\lim_{T \to +\infty} \frac{1}{2T}\int \limits_{-T}^T x^{it}dt = \lim_{T \to +\infty} \frac{1}{2T} \frac{x^{iT}-x^{-iT}}{i\ln x} = 0$

whereas if $x=1$ then the above limit is $1$ . Thus:

$\begin{aligned} \lim_{T \to +\infty} \frac{1}{2T}\int \limits_{-T}^T \frac{\zeta(3/2+it)}{\zeta(3/2-it)} dt &= \sum_{n=1}^\infty n^{-3/2} \sum_{d | n} \mu(d) \lim_{T \to +\infty} \frac{1}{2T}\int_{-T}^T (d^2/n)^{it}dt \\ &= \sum_{n=1}^\infty n^{-3/2} \sum_{d | n} \mu(d) 1_{n = d^2} \\ &= \frac{1}{\zeta(3)} \end{aligned}$

Inverse zeta(3) limit

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