WolframAlpha verifies the following result:
For $a > 0$, $$\int_a^\infty x^{1/2} e^{-x} \, dx = \int_{\sqrt{a}}^\infty e^{-x^2} \, dx + \sqrt{a} e^{-a}.$$
Is there a simple proof of this fact?
I tried to prove this using probability. If $X \sim \chi^2_3$, then $$P(X \ge a) = \frac{1}{\sqrt{2\pi}} \int_a^\infty x^{1/2} e^{-x} \, dx,$$ and if $Z \sim N(0,1)$, then $$P(Z > \sqrt{a/2}) = \frac{1}{\sqrt{2 \pi}} \int_{\sqrt{a/2}}^\infty e^{-u^2/2} \, du = \frac{1}{\sqrt{\pi}} \int_{\sqrt{a}}^\infty e^{-x^2}\,dx.$$
I'm unable to see the connection, and also unsure of how to deal with the other term $\sqrt{a} e^{-a}$.
Hint:
I thought Integration by Parts helps?
$$\int_a^{\infty}\sqrt{x}e^{-x}dx=\underbrace{-\sqrt{x}e^{-x}\Bigg\vert_a^{\infty}}_{\sqrt{a}e^{-a}}+\underbrace{\int_a^{\infty}\frac{e^{-x}}{2\sqrt{x}}dx}_{I_1}$$
What happens when you substitute $\displaystyle \sqrt{x}=t$ in $I_1$ ?