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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.

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  • $\begingroup$ 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? $\endgroup$ – Carlo Beenakker Jun 25 '14 at 8:33
  • $\begingroup$ Thank you Carlo, that was maybe a too simple example. I changed the example. $\endgroup$ – aslan Jun 25 '14 at 8:45
  • $\begingroup$ 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. $\endgroup$ – Carlo Beenakker Jun 25 '14 at 8:49
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    $\begingroup$ 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. $\endgroup$ – S. Carnahan Jun 25 '14 at 10:33

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