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Under which conditions is the function

$$ g(s)=a^{c(s-1)}\Gamma(s),\qquad a>0,c\in \mathbb{R} $$

the Mellin transform of a probability density function $f$? If $c=-1$, then $f$ is the exponential density. But what if $c\neq -1$?

Thanks in advance.

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  • $\begingroup$ The inverse Mellin transform of this expression is particularly easy to do. Work it out, and you should be able to discern some answer to your question. Note that $(s-1)! = (\pi / sin(\pi s) ) (1/(-s)!)$ for determining the singularities for inversion. $\endgroup$ Sep 1, 2015 at 17:17
  • $\begingroup$ @TomCopeland I tried to apply naively the Mellin inversion theorem (without taking into account the conditions on g) and the result, if I am not mistaken, should be $a^{-c}exp^{-a^{-c}x}$. This is a probability density function for every a>0 and real c. Am I right? $\endgroup$
    – axl
    Sep 2, 2015 at 13:35
  • $\begingroup$ Do some numerical checks to bolster your confidence and / or catch mistakes. I use an old version of MathCad to numerically corroborate my symbolic results. It'll help you understand the forward and inverse Mellin transform better. $\endgroup$ Sep 2, 2015 at 14:21
  • $\begingroup$ In the future, you might try asking questions at this level on Math StackExchange first. $\endgroup$ Sep 2, 2015 at 14:23
  • $\begingroup$ @TomCopeland thank you for helping me. I am not so much into complex analysis. I tried asking on Math StackExchange, but got no answer. $\endgroup$
    – axl
    Sep 2, 2015 at 14:32

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