# Integral of a Dirichlet Series

I'm stuck at a problem of an exercise list... I'd like some help to solve it :)

The problem: Suppose that the Dirichlet Series

$$A(s)=\lim_{N \to \infty}\sum_{n=1}^Na(n)n^{-s}$$

has abscissa of absolute convergence $\sigma_a<\infty$. Show that:

$$\lim_{T \to \infty}\frac{1}{2T}\int_{-T}^T|A(\sigma + it)|^2dt=\sum_{n=1}^{\infty}|a(n)|^2n^{-2\sigma}$$

for all $\sigma > \sigma_a$.

I have no idea how to even start this, so any hint or start is welcome! :)

Thanks!

Firstly, for $\sigma > \sigma_a$:

\begin{align} |A(\sigma+it)|^2 &= A(\sigma+it)\overline{A(\sigma+it)} = \left(\sum_{n\geqslant 1}a(n)n^{\sigma+it}\right) \left(\sum_{n\geqslant 1}\overline{a(n)}n^{\sigma-it}\right) \\ &= \sum_{m\geqslant 1}\sum_{n\geqslant 1} a(n)\overline{a(m)}n^{\sigma}m^{\sigma}\left(\frac{n}{m}\right)^{it} \\ &= \sum_{n\geqslant 1} |a(n)|^2n^{-2\sigma} + \sum_{\substack{m,n\geqslant 1 \\ m \neq n}} a(n)\overline{a(m)}n^{\sigma}m^{\sigma}\left(\frac{n}{m}\right)^{it}.\end{align}

Now give a look at this series. We have

$$\sum_{m\geqslant 1}\sum_{n\geqslant 1} \left|a(n)\overline{a(m)}n^{\sigma}m^{\sigma}\left(\frac{n}{m}\right)^{it}\right|\leqslant \left(\sum_{m\geqslant 1} |a(m)|m^\sigma\right) \left(\sum_{n\geqslant 1}|a(n)|n^\sigma \right)$$

thus, for fixed $\sigma > \sigma_a$, this series converges uniformly for $t\in\mathbb{R}$. That means that we can integrate term by term, so

$$\frac{1}{2T} \int_{-T}^T |A(\sigma+it)|^2 dt = \sum_{n\geqslant 1} |a(n)|^2n^{-2\sigma} + \sum_{\substack{m,n\geqslant 1 \\ m \neq n}} a(n)\overline{a(m)}n^{\sigma}m^{\sigma}\left(\frac{1}{2T}\int_{-T}^T \left(\frac{n}{m}\right)^{it}dt\right).$$

The observant reader can quickly see that $\frac{1}{2T}\int_{-T}^T \left(\frac{n}{m}\right)^{it}dt = \frac{\sin(T\log(n/m))}{T\log(n/m)}$, which tends to $0$ as $T\to\infty$, then, again, the double series converges uniformly for $T\in\mathbb{R}$, so we can take the limit with respect to $T$ term by term, concluding that the second sum on the right of the above equality tends to $0$ as $T\to\infty$. Hence

$$\lim_{T\to\infty} \frac{1}{2T} \int_{-T}^T |A(\sigma+it)|^2 dt = \sum_{n\geqslant 1} |a(n)|^2n^{-2\sigma}.$$

As required.