Inverse Mellin transform of the Gamma function

[Rfael]

Homework Statement

which is the inverse Melin transform of the square of the gamma function

Relevant Equations

$$ \Gamma(s)^2 $$

need a hint to solve this inverse mellin transform involving the square of the Gamma function

 

[Paul Colby]

Ah, it’s Mellin transform, which I haven’t heard of either. But a quick read of the Wiki page suggests contour integration to me. Where are the poles of $\Gamma(s)^2$ in the complex $s$-plane?

 

[julian]

They are mentioned briefly in Arfken, where it is noted that substituting $t=\ln x$ and $i \omega = s -c$ (possibly a typo — perhaps it should be $i \omega = s -ic$?) into the Fourier transform and its inverse leads to

\begin{align*}
G(s) = \int_0^\infty x^{s-1} F(x) dx \qquad \text{the Mellin transformation}
\end{align*}

and

\begin{align*}
F(x) = \frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty} x^{-s} G(s) ds
\end{align*}

 

[julian]

My initial thought was to approach this using complex contour integration due to the structure of the inverse transform. However, there appears to be a possibly more straightforward way to derive the inverse Mellin transform.

Consider

\begin{align*}
\Gamma (s)^2 = \int_0^\infty \left( \int_0^\infty u^{s-1} e^{-u} du \right) t^{s-1} e^{-t} dt
\end{align*}

By applying a change of variables and interchanging the order of integration, this can be rewritten in the form

\begin{align*}
\Gamma (s)^2 = \int_0^\infty x^{s-1} F(x) dx
\end{align*}

where $F(x)$ is an integral over $t$ that is closely related to the integral representation of a special function.

 

[julian]

Alternatively, following the suggestion made by @Paul Colby. Note,

\begin{align*}
\Gamma(s)=\frac{1}{s}\Gamma(s+1)=\frac{1}{s(s+1)}\Gamma(s+2)=\frac{1}{s(s+1)(s+2) \cdots (s+n)}\Gamma(s+n+1)
\end{align*}

What does that tell us about the poles of $\Gamma (s)^2$?

Here are useful formula for working out the residues at the poles:

\begin{align*}
\Gamma (s) = \frac{\Gamma (s+n+1)}{s(s+1) \cdots (s+n)} \quad \text{and} \quad \psi (s) = \frac{\Gamma' (s)}{\Gamma (s)} = -\gamma + \sum_{k=0}^\infty (\frac{1}{k+1} - \frac{1}{k+s})
\end{align*}

where $\psi (s)$ is the digamma function and $\gamma$ is Euler–Mascheroni constant.

Consider the rectangular contour $\mathcal{C}$ with vertices $c \pm iR$ ($c > 0$) and $−(N + 1/2) \pm iR$, where $N$ is a positive integer. Then the poles of $\Gamma(s)^2$ inside this contour are at $0, −1, −2, . . . , −N$. Hence, Cauchy’s residue theorem implies that

\begin{align*}
\frac{1}{2 \pi i} \oint_\mathcal{C} x^{-s} \Gamma (s)^2 ds = \sum_{n=0}^N Res (x ^{−s} \Gamma (s)^2 ,s=−n)
\end{align*}

Let $R$ and $N$ tend to infinity. You need to show that the Stirling's asymptotic formula for the Gamma function squared:

\begin{align*}
\Gamma (s)^2 = \lim_{|s| \rightarrow \infty} \frac{2 \pi}{s} s^{2s} e^{-2s}
\end{align*}

implies the integral on $\mathcal{C}$ minus the line joining $c − iR$ and $c + iR$ tends to zero.

You then obtain a series expansion for $F(x)$ closely related to that of a special function.