Calculate $\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^T\frac{\zeta(\frac{3}{2}+it)}{\zeta(\frac{3}{2}-it)}dt$ as $\sum_{n=1}^\infty\frac{\mu(n)}{n^3}$ 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.

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