Mellin transform

1. Aug 30, 2009

knightrider

Is the following statement correct?

The poles of the Mellin transform of a function analytic, say on a complex domain or the whole complex plane, occur in general at (perhaps non-positive) integer values.

It is true for the Gamma function.

2. Aug 30, 2009

rrogers

No. The common occurence is a result of the gamma function being the Mellin transform of the exponential exp(a*x); which in turn is the solution of differential equations with constant coefficients. In so far as your problems are reducible to this then poles at integer values will show up; but when you deal with more complicated problems the poles will generally be elsewhere.
There is also the result that expansion in terms of a Laurent series and then transforming the terms one by one will result in poles at integer values; corresponding to the powers of x.
Fully explaining the fact that some (actually most) transforms don't have poles there is not within my grasp, but I think it can be explained by Muntz's theorems (the current ones). Where some real and some complex functions can be approximated to arbitrary accuracy by series similiar to the power series.
In summary: I think most functions can be approximated by poles at integer values, but the "true" transforms do not have them there.

Ray

3. Aug 30, 2009

knightrider

Thanks for the response, Rrogers.

I agree with you there should be more restrictions on g for its Mellin transform to have simple poles at (negative) integers. I am experimenting with a few conditions.

Would you be able to give an example where the poles fall on non-integers, with g being analytic (preferably on the whole plane)?

4. Aug 31, 2009

rrogers

Simple process:
Let M(f(x))=F(s) then proceeding formally M(f(x^a))=a^(-1) F(s/a)
Assuming F(s) has poles at integers then let a be irrational. We have moved all of the poles off of the integers.
Alternately the poles can be shifted uniformly by multiplying f(x) by $$x^{\beta}$$
M$$(\frac{1}{1+x})=\frac{\pi}{\sin(s\pi)}$$
M$$(\frac{1}{1+x^a})=\frac{\pi}{a*\sin(\frac{s}{a} \pi)}$$
Or more fundamentally
$$M(e^{-\alpha x})=a^{-s}\Gamma(s)$$
$$M(e^{-\alpha x^{u}})=(u^{-1}) a^{-\frac{s}{u}}\Gamma(\frac{s}{u})$$
$$M(x^{\beta}e^{-\alpha x})=a^{-s-\beta}\Gamma(s+\beta)$$

Or even more fundamentally just place two poles out in left field and invert (they have to occur in complex conjugate pairs for "real" functions):
$$F(s)=\frac{1}{s+(a+bi)}+\frac{1}{s+(a-bi)}$$
then
$$f(x)=x^{(a+ib)}+x^{(a-bi)}=x^{a} \cdot 2 \cdot cos(b \cdot ln(x))$$

Unstated conditions abound for the above, and you should always check things I write. I am really really notorious for misstating things the first time around; even the most obvious.

Ray

5. Aug 31, 2009

knightrider

Good examples, Ray.

On the other hand, the examples you offered where the poles are not on the integers all have branch cuts at the origin. I think restricting the function f to be analytic at least in the vicinity of the origin should force the poles of its Mellin transform in the left half complex plane onto the non-positive integers.

Last edited: Aug 31, 2009
6. Aug 31, 2009

rrogers

I don't see how. Some more explanation or an example would help.
I don't see how the third (partial fraction) example is affected since the source terms are single valued.

Ray

7. Sep 1, 2009

knightrider

For your second line, the partial fraction source $$x^{(a+ib)} =e^{(a+ib)\ln(x)}$$ where $$\ln(x)$$ is multi-valued with value difference $$i 2\pi$$.

I will elaborate on the your first question later.

8. Sep 3, 2009

knightrider

Ray, here is the solution to your first question. For a function analytic in the vicinity of the origin, expand the function in power series within a radius r. Break the Mellin integral into two parts, from 0 to r, and from r to infinity. Term by term integration of the first part produces a meromorphic function with simple poles at non-positive integers, whereas the second integral results in an analytic function. And we are done.

9. Sep 3, 2009

rrogers

I believe you are "begging the question". Suppose the real expansion was the order of x^(1/2)
Now you _can_ expand in terms of the taylor series but it's just an approximation.
Like I said the taylor series can be used to expand functions but it's basically an approximation; not the "truth"; whatever that is.

Ray

10. Sep 3, 2009

knightrider

Ray, I don't quite understand what you are saying.

A complex function is analyticity at a point if and only if it has (non-negative integer) power series expansion in an open disk of that point. The series will not contain terms like x^(1/2). The series is not just approximation but converges uniformly in that disk, therefore can be integrated or differentiated term by term. I suppose what you mean by "truth" is exact equality. The power series equals exactly to the original function in that disk.

11. Sep 4, 2009

rrogers

I haven't double checked but I am pretty sure:
The Taylor series (like other Muntz series) are merely dense in Analytic (and more general) functions. Like the rational numbers are dense in the real line; pi is not a rational number.
Disclosure: I am a unrepentant Platonist.
Well the first book I checked disagree's with me. I will check further for a reference.

Ray

12. Sep 4, 2009

rrogers

"The power series equals exactly to the original function in that disk"
Since I haven't found a reference yet, let's consider your statement. If the Taylor series were anything but a close approximation then you wouldn't have to confine it to a "disk".
Furthermore using it as a substitute for the real function is only valid on that disk. Since it diverges at the first complex pole you can't do the necessary semicircle closure for finding the Mellin transform. Also, as I remember the series doesn't converge anywhere on the edge; even where the original function has well defined finite values.

Ray
.

13. Sep 4, 2009

rrogers

At last I found positive references. The case of 1/(1+x^2) expanded around zero. It certainly is well defined but the taylor series starts making bigger and bigger errors after x>2. None the less it does have a Mellin xform.
The discussion of this point is on Page 151 of Buck's "Advanced Calculus" 1978. I can dig up the Bibtex entry if you want.

Ray

14. Sep 5, 2009

knightrider

Let's deal with the issues one at a time.

1. Have you read carefully read my last post and noticed that, I cut the integrating region, i.e., the positive real axis, into two parts, [0 r) and [r,\infty), where r is less than the radius of convergence --- beyond which the power series diverges ----- of the power (Taylor) series around 0? I integrate the power series term by term in the first interval and integrate the function in the original form (not Talyor series!) the rest of the real axis.

Note that the poles appear only when you analytically continue the fundamental region of the s complex plane post Mellin transform to the left of the fundamental region.

I will elaborate on these points if you still object to the treatment.

2. On the issue of denseness of the Taylor series. The correct statement is the set of finite partial sum of a power series is a dense subset of the set of the analytic function. The set of Taylor series, or power series, includes all infinite sums and thus the limit points. So the set is complete and equal to the analytic function set.

3. Each power series around a specified point has a radius of convergence, inside which the series converges to the original analytic function, and beyond which it diverges. The radius of convergence of your function 1/(1+x^2) around 0 is 1. That is why you found the Taylor series "starts making bigger and bigger errors after x>2". In fact it will do so not only for x>2, but once |x|>=1.

15. Sep 5, 2009

rrogers

I guess we disagree about "equality".
When you get a chance, I do think you should look at the Muntz polynomials and think about Muntz's closure theorem.

Ray

16. Sep 5, 2009

knightrider

Hmm, to pin point the dividing line, let me ask you two questions:

1. do you agree with the following statement?
$$f(x) \stackrel{\text{def}}{=}\frac{1}{1-x} = \sum_{n=0}^\infty x^n, \forall |x|<1,$$
where $$x$$ is complex.

2. what is your definition of an analytic function?

17. Sep 5, 2009

rrogers

1) Basically no. Restricted to the reals and [0,1) they have the same value. But I can do things to the left hand, integrate in a tight circle in the complex plane around 1, that I can't do to the series; and get the same value. That and the restriction lead me too say they are different functions.
2) I would stick to the old definition; the Cauchy condition of having a derivative in the complex plain.

18. Sep 5, 2009

knightrider

1. Please note I did not ask anything about the situation |x|>=1, but only |x|<1, the open disk of unit radius centered around 0 of the complex plain. If you integrate round 1, you are going outside the unit disk where question 1 has no comment. You are not answering the question I am asking. Can we strictly stick to the question for now?

2. You mean being continuously differentiable on an open set in the complex plane, right? We will come back to expand on this topic later.

19. Sep 6, 2009

rrogers

1) I agree the two expressions have the same values in the limited domain.
2) Yeap

Ray

20. Sep 6, 2009

knightrider

OK. What is then your objection to my previous statement "The power series equals exactly to the original function in that disk", where the disk is of radius of convergence, specifically in reference to question 1? Do we not agree on the meaning of equality? Note the power series is the same of the Taylor series.

21. Sep 6, 2009

rrogers

I'll skip discussing the formal definition of "equality". In my mind it wanders off into topologies and equivalence relations. Not to mention, including Hyperreals; which I would have to study a lot.
Reset:
As I interpret it, you propose to replace the Mellin part of:

$$F(s)=\frac{1}{s+(a+bi)}+\frac{1}{s+(a-bi)}$$
$$f(x)=x^{(a+ib)}+x^{(a-bi)}=x^{a} \cdot 2 \cdot cos(b \cdot ln(x))$$

With a series of poles on the negative integers; and keep all of the functionality that Mellin transform theory provides.
You would have to demonstrate that to me.

Ray

22. Sep 6, 2009

knightrider

Could we please not wander and waste time, could we please be effective and stay on the topic at hand?

Let me restate my claim and see if you agree with it.

f(x) is a function defined on the positive real axis so that its Mellin transform F(s) exist in a fundamental strip $$(0,\beta)$$, where \beta is a positive real number. f(x) is also analytic at 0 (meaning in a neighbourhood of 0 on the complex plane). The F(s) is analytically continuable to a meromorphic function in the left half complex plane (Re(s)<=0) with simple poles at and only at the nonpositive integers.

Last edited: Sep 7, 2009
23. Sep 7, 2009

knightrider

Ray, I have a question concerning your example above in post #21 of a Mellin transform. There does not exist an s such that the Mellin transform converges as a complex valued function on the complex plane. I wonder if this could be defined as a distribution, such as the Dirac delta function, instead of a function in the normal sense. If it is so, perhaps you could elaborate on how this distribution is exactly defined.

24. Sep 8, 2009

rrogers

"I have a question concerning your example above in post #21 of a Mellin transform. There does not exist an s such that the Mellin transform converges as a complex valued function on the complex plane."
Could you explain why you think that? It seems to fit all of the requirements. Existance of the integral and such.
I am studying the relationship between Distributions and Mellin xforms. Learning a lot about both, but the source materials I have don't seem to be "correct". But I haven't worked my way through the alternate domains and convolutions yet; so I could very well (probably) be wrong.

Ray

25. Sep 8, 2009

knightrider

Try integrating $$\int_0^\infty x^{a-1} dx$$ and tell me what real number a would make the integral convergent.