The Contour Rule for Inverse Mellin Transforms

In summary, the Mellin transform is used to invert the Mellin transformed function F(s) and is defined by the equation f(x) = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} x^{-s}F(s)ds. The value of c is determined such that F(s) is analytic and typically defines a "vertical strip" of analyticity between a < Re(s) < b. This is similar to the inverse Laplace transform, but with the added requirement that both \int_{0}^{1}x^{s-1}f(x)dx and \int_{1}^{\infty}x^{s-1}f
  • #1
Hoplite
51
0
To invert the Mellin transformed function F(s), the equation is,
[tex]f(x) = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} x^{-s}F(s)ds[/tex]
What is the rule the value of c? I know that in inverse Laplace Transforms, c is any real number large enough that all the residues of F(s) lie to the left of the contour on the complex plane, but this is not necessarially the case with Mellin Transforms.
 
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  • #2
c is such that F(s) is analytic and typically defines a "vertical strip" of analyticity. The Mellin transform is essentially a two sided Laplace transform. In the latter case, the (Laplace) transform exists if the function being transformed is "of exponential order" meaning that Re(s) > b (some number so contour is to the right of all the poles) has to be sufficiently large to provide convergence in the inverse transform. How large Re(s) has to be depends on the particular function under consideration.

One way to look at it is if we "reversed" the Laplace transform (i.e. consider only negative t) then the situation would be reversed and the inverse would exist if Re(s) < a (again, some number so the contour is the left of all the poles) has to be sufficiently large for convergence.

The Mellin transform essentially merges these two idea and has a "strip of analyticity" a < Re(s) < b.

I don't know what your application is but you may want to look at Bleistein and Handlesman (Asymptotic Expansions of Integrals) for a brief summary of the properties (I think it's now available as a Dover edition).
 
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  • #3
Ahhhh, thanks Tide. I understand now that we need both of [tex]\int_{0}^{1}x^{s-1}f(x)dx[/tex] and [tex]\int_{1}^{\infty}x^{s-1}f(x)dx[/tex] to be analytic within a vertical strip, and we run the contour through there.

(As it happens, I don't have a specific application in mind, but I have an exam coming up involving transforms.)
 
  • #4
Glad to be of help - good luck on your exam!
 
  • #5
hi.I m not familier with this transform.and I have exercise,I can not solve them could you help me?

M[(exp(x)+1)-¹]=?


M[(exp(x)-1)-¹]=?
 
  • #6
hi.I m not familier with this transform.and I have exercise,I can not solve them could you help me?

M[(exp(x)+1)-¹]=?


M[(exp(x)-1)-¹]=?
 
  • #7
hi,any one can solve my question?

M[(exp(x)-1)^(-1)]=?
 
  • #8
did you resolve M[(exp(x)+1)-¹]?
I m curious about the solution.
Thanks
 
  • #9
hi,yes,I can solve one of them,any one couldn't solve one of them?!
 
  • #10
Can you post your solution to post any of them.
I tried résolve one but i don't found.
Thank to post your solution.
 
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What is the inverse Mellin transform?

The inverse Mellin transform is a mathematical operation that takes a function in the complex domain and converts it into a function in the real domain.

Why is the inverse Mellin transform important?

The inverse Mellin transform is important because it allows us to study and analyze functions that are defined in the complex domain, which often have more useful properties and applications in science and engineering.

How is the inverse Mellin transform calculated?

The inverse Mellin transform is calculated by integrating the function in the complex domain over a contour that encloses all of its poles, and then applying a series of algebraic manipulations to the resulting integral.

What are some common applications of the inverse Mellin transform?

The inverse Mellin transform has many applications in fields such as signal processing, number theory, and physics. It is often used to solve differential equations, evaluate definite integrals, and study the behavior of complex functions.

Are there any limitations to the inverse Mellin transform?

Yes, the inverse Mellin transform may not exist for all functions in the complex domain. It also requires a certain level of mathematical proficiency and understanding to properly apply and interpret its results.

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