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Using a well known theorem for Dirichlet series can be justified that (for $T>0$) $$\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^T\frac{\zeta(\frac{3}{2}+it)}{\zeta(\frac{3}{2}-it)}dt=\sum_{n=1}^\infty\frac{\mu(n)}{n^3},$$ where $\zeta(s)$ is the Riemann Zeta function and $\mu(n)$ is the Möbius function.

Question. Is it possible do directly the calculations in this example? I am asking if you can show us how deduce the identity directly, first calculating the integral and secondly the limit, to get the identity. Is it possible?

Many thanks.

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    $\begingroup$ You may compute the integral through the residue theorem, then the given identity follows from the fact that $\frac{1}{\zeta(3)}$ can be represented as a Dirichlet series ($\mu$ is the inverse of $1$ with respect to Dirichlet's convolution). $\endgroup$ – Jack D'Aurizio Sep 18 '16 at 21:26
  • $\begingroup$ Very thanks much for your early response @JackD'Aurizio $\endgroup$ – user243301 Sep 18 '16 at 21:28
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    $\begingroup$ In general, one applies the inverse Laplace / Mellin transform to go in the opposite direction, i.e. in order to write a series as an integral. $\endgroup$ – Jack D'Aurizio Sep 18 '16 at 21:28
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$$\frac{\zeta(3/2+it)}{\zeta(3/2-it)} = \sum_{n=1}^\infty n^{-3/2} \sum_{d | n} \mu(d) (d^2/n)^{it}$$ Note it converges absolutely when $t$ is real.

If $x \ne 1$ : $$\lim_{T \to \infty} \frac{1}{2T}\int_{-T}^T x^{it}dt = \lim_{T \to \infty} \frac{1}{2T} \frac{x^{iT}-x^{-iT}}{i\ln x} = 0$$ (with $x=1,x^{it} =1$ you have $\lim_{T \to \infty} \frac{1}{2T}\int_{-T}^T 1 dt = 1$)

Hence

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

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  • $\begingroup$ I've copied the solution and I am going to study it now. Many thanks for your answer, in a word: your answer is very nice and with some calculations and details that one can explore (well, were more than a word.) $\endgroup$ – user243301 Sep 19 '16 at 6:40

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