Evaluate Integral with $e^{ut}\ \Gamma (u)^{2}$

I am trying to integrate this integral:

$$f(x)=\frac{1}{2\pi j}\int_{c-j\infty}^{c+j\infty}x^{-s}\sigma ^{ms-m}\left [ \frac{\Gamma \left ( \frac{s}{\beta} \right )}{\Gamma \left ( \frac{1}{\beta} \right )} \right ]^{m}ds$$ where $\sigma>0$, $\beta>0$, $m>0$ and both $\beta$ and $m$ are positive integers

For $m=2$, I need help continuing after simplifying to this form below: $$f(x)=\frac{1}{2\pi \sigma^{2} j \Gamma \left ( \frac{1}{\beta} \right )^{2}}\int_{c-j\infty}^{c+j\infty}\left ( \frac{\sigma^{2}}{x} \right )^{s}\Gamma \left ( \frac{s}{\beta} \right )^{2}ds$$

Which I managed to simplify to following form: $$f(x)=\frac{\beta}{2\pi \sigma^{2} j \Gamma \left ( \frac{1}{\beta} \right )^{2}}\int_{c-j\infty}^{c+j\infty}e^{ut}\ \Gamma \left (u \right )^{2}du$$

How to further simplify the following integral using Cauchy's residue theorem for multiple poles due to $\Gamma \left (u \right )^{2}$? $$\int_{c-j\infty}^{c+j\infty}e^{ut}\ \Gamma \left (u \right )^{2}du$$

Thanks

Solution 1:

EDIT

The previous versions of this answer contained an error. I have now corrected the error and have verified its veracity.

Let's look at your last integral, which may be evaluated using a simple Bromwich contour and the residue theorem. Really, for $t \gt 0$,

$$\int_{c-i \infty}^{c+i \infty} du \, e^{u t} \Gamma(u)^2 = i 2 \pi \sum_{n=0}^{\infty} \operatorname*{Res}_{z=-n} e^{z t} \Gamma(z)^2 $$

Now we know that $\Gamma$ has the form, near $z=-n$ of

$$\Gamma(z) = \frac{a_{-1}}{z+n} + a_0 + O(z+n)$$

where $a_{-1} = (-1)^n/n!$ and

$$a_0 = \frac{(-1)^n}{n!} \left ( \sum_{k=1}^n \frac1{k} - \gamma \right ) = \frac{(-1)^n}{n!} \left ( H_n - \gamma \right ) $$

This result may be derived from the relation

$$\Gamma(-n+\epsilon) \Gamma(n+1-\epsilon) = (-1)^{n} \frac{\pi}{\sin{\pi \epsilon}} $$

which means that

$$\Gamma(-n+\epsilon) = \frac{(-1)^n}{n!} \frac1{\epsilon} + \frac{(-1)^n}{n!} \left ( H_n - \gamma \right ) + O(\epsilon)$$

Thus,

$$\operatorname*{Res}_{z=-n} \left [ e^{t z} \Gamma(z)^2 \right ] = a_{-1}^2 t + 2 a_{-1} a_0 = 2 \frac{H_n-\gamma+t/2}{n!^2}$$

Now, the integral we seek is

$$\int_{c-i \infty}^{c+i \infty} du \, e^{u t} \Gamma(u)^2 = i 2 \pi \sum_{n=0}^{\infty} \frac{2 (H_n-\gamma) + t}{n!^2} e^{-n t} $$

Now,

$$\sum_{n=0}^{\infty} \frac{z^n}{n!^2} = I_0 \left ( 2 \sqrt{z} \right ) $$ $$\sum_{n=0}^{\infty} H_n \frac{z^n}{n!^2} =K_0 \left ( 2 \sqrt{z} \right ) + \left ( \frac12 \log{z} + \gamma \right ) I_0 \left ( 2 \sqrt{z} \right )$$

where $I_0$ and $K_0$ are modified Bessel functions of the first and second kind, respectively, of zeroth order. These sums result from the respective expansions of $I_0$ and $K_0$ for small argument.

Subbing $z=e^{-t}$, we finally arrive at the value of the integral:

$$\int_{c-i \infty}^{c+i \infty} du \, e^{u t} \Gamma(u)^2 = i 2 \pi \cdot 2 K_0 \left ( 2 e^{-t/2} \right ) $$

This is a two-sided transform, or a Mellin transform. (This was the original formulation of the problem.) The below derivation illustrates why.

ADDENDUM

I assumed that the first equation above is valid, but did not demonstrate how this is so. To see this, we must show that

$$R \int_{\pi/2}^{3 \pi/2} d\theta \, \left | \Gamma \left (R e^{i \theta} \right )^2 \right | e^{R t \cos{\theta}} $$

vanishes as $R \to \infty$.

Use Stirling's approximation:

$$\left |\Gamma \left (R e^{i \theta} \right )^2 \right | \sim \frac{2 \pi}{R} e^{2 R \log{R} \cos{\theta} -2 R \theta \sin{\theta}} e^{-2 R \cos{\theta}} $$

The term $R \log{R} \cos{\theta}$ dominates the other terms in the exponential, so that the behavior of this term is

$$\left |\Gamma \left (R e^{i \theta} \right )^2 \right | \sim \frac{2 \pi}{R} e^{2 R \log{R} \cos{\theta}} $$

Thus,

$$R \int_{\pi/2}^{3 \pi/2} d\theta \, \left | \Gamma \left (R e^{i \theta} \right ) \right |^2 e^{R t \cos{\theta}} \sim \frac{\pi^2}{R \log{R}} \quad (R \to \infty)$$

which indeed vanishes in the desired limit.

Note that the above criterion is independent of $t$. Thus, the transform is two-sided and not causal as was previously stated.