Home » Uncategorized » Inverse zeta(3) limit

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}

Read more

Leave a comment

Who is Tolaso?

Find out more at his Encyclopedia Page.

Donate to Tolaso Network