Perron's formula (Passing a limit under the integral)

Question

I want to understand why assuming that $\sum_{n \ge 1} \frac{a_n}{n^s}$ converges uniformly for $\mathrm{Re}(s) > \sigma > 0$ with $c > \sigma$ implies that $$ \sum_{n \le x} \, \!\!^* a_n = \frac 1{2\pi i}\int_{c-i\infty}^{c+i\infty} \sum_{n \ge 1} \frac{a_n}{n^s} \frac{x^s}{s} \, ds. $$ I've managed to show that for $c > 0$, $$ \frac 1{2\pi i}\int_{c-i \infty}^{c + i \infty} \frac{y^s}s \, ds = \begin{cases} 0 & \text{ if } 0 < y < 1 \\ 1/2 & \text{ if } y = 1 \\ 1 & \text{ if } y > 1 \\ \end{cases} $$ so we can write $$ \sum_{n \le x} \, \!\!^* a_n = \frac 1{2\pi i} \sum_{n \ge 1} \int_{c-i\infty}^{c+i\infty} a_n \left( \frac xn \right)^s \frac{ds}s \overset{!}{=} \frac 1{2\pi i}\int_{c-i\infty}^{c+i\infty} \sum_{n\ge 1} \frac{a_n}{n^s} \frac{x^s}s \, ds. $$ But that $!$ that I put there means I don't understand why the sum can go under the integral sign. Any ideas about that part? Thanks.

Answer 1

Even the basic identity, about integrating $y^s/s$ on a vertical line, requires qualification to be truly sensible, since the integral is certainly not absolutely convergent. One way to be completely up-front about it is to compute $\int_{c-iT}^{c+iT} {y^s\over s}\,ds$ and keep track of the error (from the ideal answers you give) in terms of $y$ and $T$. A finite-extent integral can certainly be interchanged with the sum over $n$. Then summing the Dirichlet series gives an estimable error, which goes to $0$ as $T$ goes to $+\infty$.