TITLE: Perron, Fourier
QUESTION [17 upvotes]: Perron´s formula is in some sense just Fourier inversion, but I have never seen proven it that way in a textbook. I take this must be because the conditions for the Fourier inversion formula to hold may be difficult to verify in this case. Or are they? Is it feasible to prove Perron´s formula using mainly just the fact that the Fourier transform is self-dual?

REPLY [4 votes]: I think these answers may be elaborated upon together. Assuming $\sum |a_n| n^{-c}$ converges, the formula to which BR refers explains the sense in which the "traditional" form of Perron's formula is a Fourier integral. By a change of variables, you have 
$$\frac{1}{2\pi}\int_{-T}^{T}\left(\sum_{1}^{\infty}\frac{a_n}{n^{c+iy}}\right)\frac{e^{ixy}dy}{c+iy}=e^{-cx}\sum_{n< e^x}a_n+O\left(\frac{1}{T}\sum_{1}^{\infty}\frac{|a_n|}{n^c|x-\log n|}\right)$$
(see Titchmarsch for a generalization that leads to PNT, p61-63). As $T\rightarrow\infty$, the (sharp) estimate on the r.h.s. shows that the convergence is not uniform, even though the interchange of limits on the l.h.s. is justified because the Dirichlet series converges absolutely (hence uniformly). 
This is obviously where the Schwartz function $\phi$ to which Ben refers justifies appeal to Fourier duality, instead of having to justify interchanging limits beyond the scope of dominated convergence. It also gives you more parameters to tweak for other purposes, if required. Yet, without introducing a further limit, the resulting formula is not Perron's so that doesn't resolve your question (but perhaps Perron's formula is obsolete?). 
As I understand it, this was the precisely the controversy with Riemann's statement of the Fourier expansion of the prime counting function $J(x)$ - he just appealed to Fourier duality and left it there, but that was ultimately verified by Von-Mangoldt, so perhaps there is a proof. Certainly Fourier-Stieltjes applies when the coefficients are positive, so maybe that can be deployed by someone who knows more about measure than me.