The inverse Mellin Transform

[Hoplite]

To invert the Mellin transformed function F(s), the equation is,
f(x) = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} x^{-s}F(s)ds
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.

[Tide]

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).

[Hoplite]

Ahhhh, thanks Tide. I understand now that we need both of \int_{0}^{1}x^{s-1}f(x)dx and \int_{1}^{\infty}x^{s-1}f(x)dx 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.)

[Tide]

Glad to be of help - good luck on your exam!