# Mellin transform of time-shifted function

The Mellin transform of a function $f(x)$ can be written as

$$\mathcal M[f(x);z]=\int_0^\infty f(x)x^{z-1} dx$$

Is there a simple expression for the Mellin transform of the function $f(x-x_0)$? For example

$$\mathcal M[e^{-x^2}] = \frac{1}{2}\Gamma\left(\frac{z}{2}\right)$$

What would $\mathcal M[e^{-(x-x_0)^2}]$ or more generally $\mathcal M[e^{-(x-x_0)^n}]$ be? This may, for example, find application in the Mellin transform of the Gaussian probability density function.

• obviously, the Mellin transform of $e^{-a(x-x_0)}$ is just $e^{a x_0}$ times the Mellin transform of $e^{-a x}$; what's the problem? – Carlo Beenakker Jun 25 '14 at 8:33
• Thank you Carlo, that was maybe a too simple example. I changed the example. – aslan Jun 25 '14 at 8:45
• well, the Mellin transform of $e^{-(x-x_0)^2}$} is a hypergeometric function, it won't get any simpler than that I'm afraid. – Carlo Beenakker Jun 25 '14 at 8:49
• It seems that if $x_0 < 0$, then you are introducing information that is not available to the original transform. Since Mellin is really a transform over the group $\mathbb{R}^\times_{>0}$, rescaling the domain is much more natural. – S. Carnahan Jun 25 '14 at 10:33