I am an undergraduate student writing an expository thesis on the complex-analytic proof of the Prime Number Theorem.

I understand that applying the Mellin Transform to the partial sum of the van Mangoldt function Λ(x) makes it complex-analytic and related to zeta, so we can study the result's asymptotic behaviors and then take the Mellin Inversion ride back to the sum of primes. This makes sense to me on a surface level.

On a deeper level though, I know that the Mellin Transform is a tool for diagonalizing dilations (owing to its kernel being an eigenfunction for dilation). So there seems to be some deeper purpose to applying the Mellin Transform to Λ(x) and π(x), considering that the primes are a multiplicative idea. Perhaps we are "decomposing" prime distribution into some dilative frequency?

I think it would make more immediate intuitive sense to me if applying the Mellin Transform to a function representative of the group of positive integers gave prime distribution, considering that I conceptualize the primes as some kind of multiplicative bases, but it is confusing to me that in fact we should be going the opposite way.

I've been wracking my head with this problem for a few weeks and also have read up on everything I could on the Mellin Transform and the PNT, but no search of mine turned up yet an exact answer to this particular confusion.

If anyone could provide any tips/directions/hints or reading recommendations that might reveal some fundamental intuition about why it is valid to apply the Mellin Transform to prime distribution, I'd really appreciate it.

  • 2
    $\begingroup$ While I'm not sure, from what I understand the main intuition is Dirichlet series — it's fairly natural to consider Dirichlet series of multiplicative functions, and thus of primes $\endgroup$ Apr 11 at 13:43
  • 3
    $\begingroup$ With respect to your comment on "decomposing" prime distribution into some dilative frequency, are you aware of the explicit formula for $\psi(x)=\sum\limits_{n=1}^x \Lambda(n)$ derived from the residues of $-\frac{\zeta'(s)}{\zeta(s)}\, \frac{x^s}{s}$ where $$-\frac{\zeta'(s)}{\zeta(s)}=s\, \mathcal{M}_x[\psi (x)](-s)=s\int\limits_0^\infty \psi(x)\, x^{-s-1}\, dx=\sum\limits_{n=1}^\infty \frac{\Lambda(n)}{n^s}\,,\quad\Re(s)>1$$ and in particular the sum over the non-trivial zeta zeros $\sum\limits_\rho \frac{x^\rho}{\rho}$? $\endgroup$ Apr 11 at 16:13
  • $\begingroup$ Yes, as @StevenClark comments, the "explicit formulas" show a bigger picture. In addition to Riemann's and von Mangoldt's, I think Guinand's 1947 really shows what's going on: it's Fourier inversions, and playing off the Euler product versus the Hadamard product. :) $\endgroup$ Apr 11 at 18:17
  • 1
    $\begingroup$ I wrote up a short relevant pdf "Dirac’s Delta Function and Riemann’s Jump Function J(x) for the Primes p", available at tcjpn.wordpress.com/2011/11/21/jx, that provides a perspective based on the inverse Mellin transform (IMT), which you might find interesting. Of course the $Sl_2$ dilation op is $x\partial_x$ of which $x^{-s}$ has the eigenvalue $-s$, so the IMT is the associated transform just as the Fourier transform is for the $Sl_2$ translation op. $\zeta(s)$ can be regarded as the MT of regularly spaced delta functions. $\endgroup$ Apr 12 at 19:56

3 Answers 3


You observed that the Mellin transform is used to diagonalize dilations.

One can indeed motivate its presence in the theory of prime numbers in this form. However, we're not interested in diagonalizing dilations themselves but rather operators constructed as sums of dilations. You can consider the operator $$ f\mapsto \sum_{n=1}^{\infty} f(nx)$$ (which is rarely well-defined near zero but often well-defined away from $0$) which is the composition of the operators $$ f\mapsto \sum_{k=1}^\infty f(p^k x)$$ for prime $p$. Each of these can be obtained as the formal exponential of the operator $$f \mapsto \sum_{k=1}^\infty f(p^k x)/k$$ (which you might prefer to think of as the integral of the flow $$\frac{d}{dt} f(x,t) =\sum_{k=1}^\infty f(p^k x,t)/k$$ from $t=0$ to $t=1$) and so the whole operator is the formal exponential of $$ f\mapsto \sum_p \sum_{k=1}^\infty f(p^k x)/k. $$

It is this last operator which has the property that understanding it would clearly tell us a lot about how to count the primes, and therefore diagonalizing it would be a reasonable choice, motivating the Mellin transform. The reason for using this operator, rather than anything else, to study sums over primes is that exponentiating it gives a nice operator expressed in terms of sums over all numbers that we can hope to understand using additive Fourier analysis or other tools.


This is too thin an answer really, but it's too long for a comment - I expect there'll be much better answers soon enough but hopefully this is at least a start:

If you have any sum, you wonder what you can do to it. (That's maths right?:D) Whenever you have a sum you can do Fourier analysis on, you can try applying Fourier's Inversion Theorem - for the primes, this is all Mellin inversion, i.e. Perron's formula. For divisor functions, it's Voronoi's summation formula. In each case you started with a sum and now have a new sum - the dual sum.

Maybe the new sum is better, maybe it's not. If it is better and you get new results, then later people can look at it more philosophically/deeply and try to figure out why it worked, what really goes on, etc., probably in the hope of generalising it to other situations, but in practice the reason why it was attempted in the first place was maybe just to see what happens.

So you consider the Dirichlet series just to see what you can say about the new sum after applying an inversion theorem - a contour integral for anything you're looking at the Dirichlet series of. Of course, it was Riemann's insight to realise you can move the contour by extending $\zeta (s)$.

By the way, have a look at Chapter 5 of Koukoulopoulos's book - there are many good books of course but (imo) this one doesn't shy away from really explaining the why's.


The prime numbers encode multiplicative structure. Similarly, the von Mangoldt function $\Lambda$ encodes additive structure, as expressed by the identity $\log n = \sum_{d \mid n} \Lambda(d).$

Intuition for the reason we apply the Mellin transform to $\Lambda$ can be obtained by noting that the Mellin transform converts additive structure into multiplicative structure.

Additive structure means $z^n z^m=z^{n+m}$, multiplicative structure means $n^{-s}m^{-s}=(nm)^{-s}$. The Mellin transform of $z^n$ (after the change of variables $z\mapsto e^{-x}$) is just $n^{-s}$ (up to a factor $\Gamma(s)$). In particular, the Mellin transform of a generating function $\sum_n a_n z^n$ gives the (generalized) zeta function $\sum_n a_n n^{-s}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.