0

I'm trying to learn how you can use the Mellin transformation to obtain closed expressions of harmonic sums. There are demonstrations om MSE how show this technique. eg
Proving $\sum_{n =1,3,5..}^{\infty }\frac{4k \ \sin^2\left(\frac{n}{k}\right)}{n^2}=\pi$ and Series $\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}$

The problem I have is that the identity $\mathcal{M}\left(\sum_{k\geq 1}\lambda_k g(\mu_k x),s\right)=\sum_{k\geq 1}\frac{\lambda_k}{\mu_k}\mathcal{M}\left(g(x),s\right)$. Is only a "real" identity when $k$ ranges over a finite set. But when dealing with an infinite series you only have that it gives an expansion in the neighborhood of 0 (sometimes +$\infty$, depending on the situation).

What do I need to do to prove that harmonic sum identity is true on a certain interval? Trying to proof that you can switch sum and integral doesn't work, because then the equality would be true everywhere for all x. I have looked at the solutions of the above examples, in Series $\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}$ the solution is exact for $x\in(0,2\pi)$. Why? How can you formally show this? The argument "because cosine has a period $2\pi$ " is unclear to me.

The Mellin transform technique also requires to calculate the inverse Mellin transform. e.g in previous example you need to calculate the integral $\int_{\frac{5}{2}-i\infty}^{\frac{5}{2}+i\infty} Q(s)/x^s \, ds$. If you do this using the residue theorem with as a contour a rectangle with corners $\frac{5}{2}+iT,\frac{5}{2}-iT,-N-iT, -N+iT$ you will have that if $T\rightarrow \infty$, $N\rightarrow \infty$. only the integral over $(\frac{5}{2}+i\infty,\frac{5}{2}-i \infty)$ remains if $(x\in 2\pi)$. Can this be used to proof exactness on a given interval?

I really find the Mellin transform an interesting and powerful technique, I looked at the survey http://algo.inria.fr/flajolet/Publications/mellin-harm.pdf where they talk about the asymptotics. But now I'm stuck trying to understand how you can have exactness on certain intervals. And how formally to prove this.

Community
  • 1
J. Spel
  • 65
  • When this question appeared a few days ago I pointed out that in the second link I effectively omitted the part where you show that the integral on the left side of the rectangular contour vanishes for $x$ in a certain interval. The first link that you cite does show how to do this part of the computation. – Marko Riedel Nov 19 '15 at 20:15
  • I will answer in a moment but I suggest you enclose the harmonic sum identity in dollar signs first, it shows up as text in my browser. – Marko Riedel Nov 19 '15 at 22:27
  • Since I can't edit the comment above, I looked a those computations and I agree, but you always seem to start the computation by saying $ \mathcal{M}\left(\sum_{k\geq 1}\lambda_k g(\mu_k x),s\right)=\sum_{k\geq 1}\frac{\lambda_k}{\mu_k}\mathcal{M}\left(g(x),s\right)$. While $k$ ranges over an infinite set. This equality does not hold everywhere... Why does it hold when the integral vanishes for certain x? I don't seem to understand the link between the computation of the inverse transform and the exactness of the series. – J. Spel Nov 19 '15 at 22:32
  • Well you may delete the comment and post a corrected one. Anyway, the convergence of the series plays a crucial role in the computation, because we intersect the half-plane of convergence of the series with the fundamental strip of the base function to determine the location of the Mellin inversion integral. Once we have that shift to the left for an expansion about zero and to the right for an expansion about infinity. If these two do not intersect we need to add correction terms to the base function to shift the fundamental strip. These are then removed again at the end of the computation. – Marko Riedel Nov 19 '15 at 22:39
  • The two horizontal segments are taken care of by the so-called smallness property of Mellin transforms which is proved at the beginning of the Szpankowski paper which I have cited several times. – Marko Riedel Nov 19 '15 at 22:39
  • With shift to the left you mean construct a box such that in the limit the inside of the box covers the left plane (left of the line of the mellin inversion integral) or do you mean make a box such that you get an integral over the imaginary axis? – J. Spel Nov 19 '15 at 22:45
  • "If these two do not intersect ": you mean if the half-plane of convergence of the series and the fundamental stip don't intersect? – J. Spel Nov 19 '15 at 22:51
  • There are several possibilities for the shift to the left, we may shift to an abscissa where the integral can be evaluated due to special properties (like an odd function on the line) or we may shift to minus infinity (which is what is done when computing closed forms of Fourier series). Yes it is indeed correct, we intersect the fundamental strip of the base function and the half-plane of convergence of the series. I suggest you read the Szpankowski paper which also explains how to obtain functional equations for harmonic sums from their Mellin transforms. – Marko Riedel Nov 19 '15 at 22:56
  • The process of extending contours to infinity is used in complex variables all the time for example when using the $\pi\cot(\pi z)$ multiplier to evaluate certain series. – Marko Riedel Nov 19 '15 at 22:58
  • Oke thanks a lot – J. Spel Nov 19 '15 at 22:58

0 Answers0