Recent content by knightrider

Ray, it does not expend much to write down the full functional value, a simple 0, on [0,1] for f(x) in you previous several post. It would have saved a lot of unnecessary back and forth and time with a smidgen of care if not rigour. Anyway, your example still does not negate my claim, unless...

Try integrating \int_0^\infty x^{a-1} dx and tell me what real number a would make the integral convergent. What's your comment on #22?

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...

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...

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...

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...

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?

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...

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...

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...

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.

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...

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...

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.