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

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.

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I must say that $$\frac 1{2\pi i}\int_{c-i \infty}^{c + i \infty} \frac{y^s}s \, ds$$ doesn't not converge for $y=1$, but his Cauchy principal value $$\lim_{T \to \infty} \frac 1{2\pi i}\int_{c-i T}^{c + i T} \frac{y^s}s \, ds$$ exist and it's equal to $1/2$. You can find here detail version of proof, but there is small problem: it is on Serbian, but there is a lot formula so you should made it :-) – Cortizol Aug 30 '13 at 19:27

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 $+\infty$.
So you're saying I should just re-work out the proof I did for the integral $\int_{c-i\infty}^{c+i \infty}$ y^s/s \, ds$and keep track of the error when I'm summing? I thought it would be more easy than that, I was too lazy to try it that way. This approach does seem hard though. – Patrick Da Silva May 8 '13 at 2:29 Yes, probably you should just redo the intuitive version more scrupulously. Yes, it requires some labor, but it's not profoundly difficult. That it requires something is not unreasonable, given the not-absolute-convergence. In some scenarios, integrating against$y^s/s(s-1)(s-2)...(s-\ell)$with$\ell\ge1$is sufficient, and does give better convergence. But, in fact, often this more-smoothed "sample" gives substantially-less-interesting results, unfortunately. Thus, all the more with hindsight, the trouble it takes to be sure of the simpler-but-not-absolutely-convergent case is warranted. – paul garrett May 8 '13 at 2:41 Um, you seem to put$y^2$all the time, I meant$y^s$, you know that right? Or is it just a frequent typo of yours :P – Patrick Da Silva May 8 '13 at 2:42 Heh... "2" or "s"... kinda similar? :) – paul garrett May 8 '13 at 2:43 So should I try to bound $$\left| \int_{c-iT}^{c+iT} \sum_{ n \ge 1} \frac{a_n}{n^s} \frac{x^s}s \, ds - \sum_{n \le x} \,\!\!^* a_n \right|$$ by some function of$T$that goes to zero as$T\$ goes to infinity? – Patrick Da Silva May 8 '13 at 2:50