Question

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?

Answer 1

Generally, Perron's formula is the calculation of the inverse Mellin (or Laplace or Fourier) transform of a particular function. When the function's representation as a Mellin transform is known, this is simple. Otherwise, some work in necessary. To a certain extent, I guess it also depends on what you mean by "Perron's formula".

One version of Perron's forumla calculates the inverse Mellin transform of $I_T(s)/s$, where $I_T$ is the indicator function of the strip $|{\rm Im}(s)|< T$, a statement being (from Patterson's book on the zeta function), for $c>0$, $${1\over 2\pi i}\int_{c-iT}^{c+iT}{x^s\over s}ds=\cases{O\big(x^c/T\log(x)\big)& $0< x< 1$\cr 1/2+O(T^{-1}) & $x=1$\cr 1+O\big(x^c/T\log(x)\big)& $x>1$}$$ Since it is a statement about a particular function, you can't really get a general proof. On the other hand, I bet you could extract a certain amount of information as you do in the Paley-Wiener theorem (and similar results).

Wikipedia's version of Perron's formula, which is also an application of the above formula, is amenable to "proof via inversion". Let $g(s)=\sum_{n\ge 1} a_n/ n^s$ be a Dirichlet series converging absolutely for ${\rm Re}(s)>\sigma$. Re-write this as $g(s)=s\int_0^\infty A(x)x^{-s-1}\ dx$, where $A(x)=\sum_{n\le x}a_n$ (with some complication when $x$ is an integer).

If we set $B(x)=A(1/x)$, after changing variables $x\rightarrow x^{-1}$, this becomes $g(s)=s\int_0^\infty B(x)x^{s-1}\ dx=s{\cal M}B(s)$, where $\cal M$ denotes "Mellin transform" (otherwise $g(s)={\cal M}A(-s)$). Divide by $s$ and apply Mellin inversion to both sides (this requires some bound on the decay of $g(s)$ in the region of convergence, which isn't difficult), with $c\gg\sigma$, $${1\over 2\pi i}\int_{c-i\infty}^{c+i\infty}g(s){x^{-s}\over s}\ ds=B(x)$$ Send $x\rightarrow x^{-1}$ to get $${1\over 2\pi i}\int_{c-i\infty}^{c+i\infty}g(s){x^{s}\over s}\ ds=A(x)$$

Answer 2

Harald,

My personal stance on this is that I like to try and avoid using Perron's formula in the "traditional" form. Instead, I like to see the Prime Number Theorem (say) as a statement about $\sum \Lambda(n) \phi(n)$, where $\phi$ is a $C^{\infty}_0$ cutoff function approximating the interval $[1,X]$. To relate this to $\zeta'/\zeta$, you need the Mellin inversion formula for $\phi$ on the vertical line $\Re s = \sigma$, and this really is precisely the same thing as the Fourier inversion formula for the function $e^{\sigma u}\phi(e^u)$. Since everything is a compactly supported smooth function, and in particular a Schwartz function, the analytic issues involved with inverting the Fourier transform are as mild as they can be.

My point of view on this is elaborated upon in in chapter 1 of this course http://www.dpmms.cam.ac.uk/~bjg23/ANT.html.

Answer 3

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.