Asymptotics of tail function of product of 2 iid gamma variables

Question

Suppose we have an integral of this form: $\overline F(x)=\frac{\beta^{2\alpha-1}}{\Gamma^2(\alpha)}\int_{0}^{\infty}x^{\alpha-1}e^{-\beta(\frac{x}{y}+y)}dy$, where $\beta>0, \alpha>0$ and $x>0$.
I want to show that it asymptotically converges to a function of this type:
$\overline F(x) = \dfrac{\sqrt{\pi}\beta^{2(\alpha-1)+\frac{1}{2}}}{\Gamma^2(\alpha)}x^{\alpha-\frac{3}{4}}e^{-2\beta x^{\frac{1}{2}}}$
The only method I came up with was the saddle point method, but unfortunately it didn't work. Maybe anyone has some ideas?

Answer 1

Hint $$\int_{0}^{\infty} e^{-\beta(\frac{x}{y}+y)}\,dy=2 \sqrt{x}\, K_1\left(2 \beta \sqrt{x} \right)$$ $$\overline F(x)=\frac{2 \beta }{\Gamma (\alpha )^2}\,x^{\alpha -\frac{1}{2}}\, K_1\left(2 \beta \sqrt{x} \right)$$

Now, uwe the asymptotics of the modified Bessel function of the second kind; it then should be simple.

Answer 2

Laplace's method applies to $\int \exp [\beta f(y)] dy $ where $f(y)$ has a global maximum inside the integration interval.

In our case $$f(y)=-\frac{x}{y} -y $$ which has a global maximum at $y_0 = \sqrt{x}$ , with $f(y_0)=-2 \sqrt{x}$ and $f''(y_0)= -2 \,x^{-1/2}$

Hence

$$ \begin{align} \overline F(x)&=\frac{\beta^{2\alpha-1}}{\Gamma^2(\alpha)}\int_{0}^{\infty}x^{\alpha-1}e^{-\beta(\frac{x}{y}+y)}dy\\ &=\frac{\beta^{2\alpha-1}}{\Gamma^2(\alpha)}x^{\alpha-1}\int_{0}^{\infty}e^{\beta f(y)}dy\\ &\approx \frac{\beta^{2\alpha-1}}{\Gamma^2(\alpha)}x^{\alpha-1} \sqrt{\frac{2 \pi}{ \beta | f''(y_0)|}} \exp(\beta f(y_0))\\ &= \frac{\beta^{2\alpha-3/2}}{\Gamma^2(\alpha)}x^{\alpha-3/4} \sqrt{\pi} \exp(-2\beta x^{1/2}) \end{align} $$

Caveat: this is asymptotically correct for $\beta \to +\infty$ and fixed $x$.