# Why is Mellin-inverse of Gamma periodic?

## Specific Case

The periodicity is obvious from computation: $$\cal{M}^{-1}\{\Gamma\}(x) := \frac{1}{2\pi i}\int_{c}\Gamma(s)x^{-s}d s=e^{-x}$$ However, is there a way to see directly from the integral that it should be periodic in $x$ with period $2\pi i$?

One thing you can read off of the integral is that it satisfies a differential equation, although I don't see why this would give periodicity.

One way to tackle this could be by using a $\beta$-integral to prove multiplicativity of the function

## Generalization

The above function naturally lives on $\mathbb{C}/(2\pi i\mathbb{Z})$.

What about $\cal{M}^{-1}\{\prod_{i=1}^n\Gamma(\lambda_i s+\mu_i)\}$ for some $\lambda_i>0,\mu_i\in\mathbb{C}$. In certain cases this function seems to naturally live on a quotient of a Siegel space or on a symmetric space. Is there some general result about this?

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Regarding your second-to-last sentence: it's a function of one variable, so what do you mean? – David Hansen May 31 '11 at 21:58
It's unfortunately a bit vague. I think I want the Mellin-inverse to be a specialization (or restriction) of such a higher-dimensional function (like how for a symmetric space you form a Whittaker function, which involves this inverse Mellin transform) – Ralph Furmaniak May 31 '11 at 23:17
You could go through your favourite proof that $e^{2 \pi i} = 1$ and substitute everything back into the integral directly and see if it works, but obviously that would be truly horrible! If there is a good answer NOT involving exp, this would be interesting (and would, presumably, provide a new proof for exp). Maybe you could choose some properties of exp, the most obvious being the addition formula, and see if you can prove them directly from the integral formula; that might give some idea. – Zen Harper Jun 1 '11 at 7:17
I assume that, by differential equation'', you are referring to the property $f(x)=-f'(x)$ right? This does actually tell you inverse is $e^{-x}$. Indeed, assuming that $s\Gamma(s)=\Gamma(s+1)$ and that differentiating under the integral sign is valid, you have $$-\frac{df}{dx}=\int_{c} \Gamma(s+1)x^{-s-1}ds=\int_{c+1} \Gamma(s)x^{-s}ds.$$ Now, assuming that $\Gamma(s)$ is analytic in the vertical strip $c\leq \sigma\leq c+1$, you can move the path of integration so the integral is equal to $f(x)$. Thus $f(x)=-f'(x)$. – Kevin Smith Mar 21 '12 at 17:42
The fact is that this property implies that $f(x)=e^{-x}$. You can prove this once you observe that, firstly, $f$ must be analytic and then, secondly, that it's Taylor coefficients must be $(-1)^n/n!$. The periodicity then follows from the Taylor series. – Kevin Smith Mar 21 '12 at 17:42