how can I show this integral diverges?

Question

I want to show $E(T_a)=\infty$

$$E(T_a)=\int_0^{\infty}{{x|a|}\over\sqrt{2\pi}}x^{-3/2}e^{-a^2/x}dx$$

to show this I need to show this integral diverges.

I know gamma function that

$$\Gamma (\alpha)=\int_0^\infty x^{\alpha -1}e^{-x}dx$$

By benefit from gamma function, how can I show this integral diverges ?

Thank you for helping

Answer 1

As suggested by mickep, making the change of variable $$u:=a^2/x, \qquad |a|x^{-1/2}=u^{1/2}, \qquad dx=-\frac{a^2}{u^2}du $$ gives $$\begin{align} \int_0^{\infty}{{x|a|}\over\sqrt{2\pi}}x^{-3/2}e^{-a^2/x}dx&={{|a|}\over\sqrt{2\pi}}\int_0^{\infty}x^{-1/2}e^{-a^2/x}dx\\\\ &={{1}\over\sqrt{2\pi}}\int_0^{\infty}\frac1{u^{3/2}}e^{-u}\:du\\\\ &\geq{{1}\over\sqrt{2\pi}}\int_0^{\ln 2}\frac1{u^{3/2}}\times\frac12\:du\\\\ &\geq{{1}\over\sqrt{2\pi}}\left[-\frac1{\sqrt{u}}\right]_0^{\ln 2}\\\\ &\geq +\infty \end{align} $$ and your integral is divergent.

Answer 2

Since $e^{-a^2/x} \to 1$ at $\infty,$ we have $e^{-a^2/x} > 1/2$ for large $x.$ It follows that the integrand is on the order of $1/\sqrt x$ for large $x.$ Since $\int_b^\infty (1/\sqrt x)\,dx = \infty$ for all $b>0,$ the integral diverges.