If You Can Solve This, You Are Truly Brilliant! 2 of X

$\displaystyle \int_0^{\infty} \frac{1}{\sqrt{y}} e^{-(\frac{a}{y} + by ) } dy = \sqrt{\frac{\pi}{b}} e^{-2\sqrt{ab}}$

$\mathbf{Proof \; : }$

$\displaystyle A= \int_0^{\infty} \frac{1}{\sqrt{y}} e^{-(\frac{a}{y} + by ) } dy$

$\displaystyle \overset{{\color{#D61F06}{y \to \sqrt{\frac{a}{b}} y }} }{=} \sqrt{\sqrt{\frac{a}{b}}} \int_0^{\infty} \frac{1}{\sqrt{y}} e^{-\sqrt{ab} ( \frac{1}{y}+y)} dy$

$\displaystyle \overset{ {\color{#D61F06}{y \to \frac{1}{y} }} }{=}\sqrt{\sqrt{\frac{a}{b}}} \int_0^{\infty} \frac{1}{y\sqrt{y}} e^{-\sqrt{ab} ( \frac{1}{y}+y)} dy$

Hence:

$\displaystyle 2A = \sqrt{\sqrt{\frac{a}{b}}} \int_0^{\infty} \bigg( \frac{1}{\sqrt{y}} + \frac{1}{y\sqrt{y}} \bigg) e^{-\sqrt{ab} ( \frac{1}{y}+y)} dy$

$\displaystyle A = \sqrt{\sqrt{\frac{a}{b}}} \int_0^{\infty} \bigg( \frac{1}{2\sqrt{y}} + \frac{1}{2y\sqrt{y}} \bigg) e^{-\sqrt{ab} ( \sqrt{y} - \frac{1}{\sqrt{y}})^2 -2\sqrt{ab} } dy$

$\displaystyle \overset{{\color{#D61F06}{\sqrt{y} - \frac{1}{\sqrt{y}} \to y }}}{=} \sqrt{\sqrt{\frac{a}{b}}} \int_{-\infty}^{\infty} e^{-\sqrt{ab}y^2 - 2\sqrt{ab}} dy$

This can be derived from the Gaussian Integral :

$\displaystyle \boxed{A = \sqrt{\frac{\pi}{b}} e^{-2\sqrt{ab}} }$ .

Set $a=b=1985$ , you get the desired answer.