calculate :$\int_0^\infty\ e^{-a^2(y^2+\frac{p^2}{4a^2y^2})}\,dy$

Question

Trying to calculate $$\int_0^\infty\ e^{-a^2(y^2+\frac{p^2}{4a^2y^2})}dy.$$ When I searched it on Google, I found the formula $$\int_{-\infty}^{\infty} \ e^{-a(x+b)^{2}}dx=\sqrt\frac{\pi}{a},$$ but when I apply the aforementioned formula I got $\frac{\sqrt\pi}{2a}{e^{pa}}$, but the answer in my book is $\frac{\sqrt\pi}{2a}{e^{-pa}}$. Can I get help?

Answer 1

$I=\int_0^\infty e^{-a^2(y^2+\frac{p^2}{4a^2y^2})}dy$ $＝\int_0^\infty e^{-a^2[(y-\frac{p}{2ay})^2]-pa}dy$ $=e^{-pa}\int_0^\infty e^{-a^2(y-\frac{p}{2ay})}dy$

let $y-\frac{p}{2ay}=s$,then $y=\frac{1}{4a}\Big(2as+\sqrt{(2as)^2+8ap}\Big)$

then $dy=\frac{1}{4a}\Big(2a+\frac{2as}{\sqrt{(2as)^2+8ap}}\Big)ds$

$I=e^{-pa}\int_{-\infty}^{\infty}e^{-a^2s^2}\frac{1}{4a}\Big(2a+\frac{2as}{\sqrt{(2as)^2+8ap}}\Big)ds=\frac{e^{-pa}}{4}\Big(\int_{-\infty}^{\infty} e^{-a^2s^2}dy +\int_{-\infty}^\infty\frac{2as e^{-a^2s^2}}{\sqrt{(2as)^2+8ap}}dy\Big) $

$=\frac{1}{4}e^{-pa}\int_{-\infty}^\infty e^{-a^2s^2}dy=\frac{\sqrt{\pi}}{2a}e^{-pa}$