Prove that the integral $\int_{0}^{\infty}\dfrac{a}{\sqrt{2\pi t^{3}}}e^{-\frac{a^{2}}{2t}}dt=1$ for $a>0$.

Question

I am at the end of a proof and I need to prove that $$\int_{0}^{\infty}\dfrac{b}{\sqrt{2\pi t^{3}}}e^{-\frac{b^{2}}{2t}}dt=1\ \text{for all}\ b>0.$$

I tried to use integration by part, but it fails as follows:

$$\int\dfrac{b}{\sqrt{2\pi t^{3}}}e^{-\frac{b^{2}}{2t}}dt=\dfrac{b}{\sqrt{2\pi}}\int t^{-\frac{3}{2}}e^{-\frac{b^{2}}{2t}}dt,$$let $u=t^{-3/2}$ so that $du=-\frac{3}{2}t^{-5/2}dt$, and let $e^{-b^{2}/2t}dt=dv$ so that $v=-\frac{2}{b^{2}}e^{-b^{2}/2t}$, but then we would have $$-t^{-3/2}\frac{2}{b^{2}}e^{-b^{2}/2t}-\int \frac{2}{b^{2}}e^{-b^{2}/2t}\frac{3}{2}t^{-5/2}dt$$ so the iterating like this will not get rid of anything..

Answer 1

Use the substitution $x = \frac{1}{\sqrt{2t}} \implies dx = -\frac{1}{2}\frac{1}{\sqrt{2t^3}}dt$:

$$I = 2\int_0^\infty \frac{b}{\sqrt{\pi}}e^{-b^2x^2}\:dx = \int_{-\infty}^\infty \frac{b}{\sqrt{\pi}}e^{-b^2x^2}\:dx = \frac{b}{\sqrt{\pi}}\cdot \frac{\sqrt{\pi}}{b} = 1$$

from the known value of the Gaussian integral.

Answer 2

The integrand is continuous on $(0,\infty).$ Furthermore it tends to $0$ as $t\to 0^+.$ And it is $= O(1/t^{3/2})$ as $t\to \infty.$ Thus the integral converges.