How to compute the integral $\displaystyle\int_0^\infty\frac{1}{\sqrt{2\pi t}}\cdot\exp\bigg({\frac{-a}{2t}-bt}\bigg)dt$ How to get the following for $a,b>0$? $$\int_0^\infty\frac{1}{\sqrt{2\pi t}}\cdot\exp\bigg({\frac{-a}{2t}-bt}\bigg)dt=\frac{1}{\sqrt{2b}}\cdot \exp\big(-\sqrt{2ab}\big) $$
The context for this identity is its use in computing the resolvent of Brownian motion in the stochastic calculus book by Le Gall (screenshot below).  I figured there had to be an elementary way to compute it, i.e. without using the fact Le Gall alludes to about the Laplace transform of the hitting time of Brownian motion. As is indicated in the comments, this is clearly a duplicate so I wouldn't object to its being closed for that reason.  Although maybe those who posted the excellent answers below should've been given the chance to migrate their answers to the post of which this is a duplicate, if they wish. 
 A: Substitute $s^2=t$
$$I = \int_0^\infty \frac{2}{\sqrt{2\pi}}\exp\left(-b\left[s^2+\frac{a}{2bs^2}\right]\right)ds = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}\exp\left(-b\left[s^2+\frac{a}{2bs^2}\right]\right)ds$$
by even symmetry. Then complete the square on the term in the exponential
$$s^2 + \frac{a}{2bs^2} = \left[s-\sqrt{\frac{a}{2b}}\frac{1}{s}\right]^2 + \sqrt{\frac{2a}{b}}$$
which gives us the new integral
$$\frac{1}{\sqrt{2\pi}}\exp\left(-\sqrt{2ab}\right)\int_{-\infty}^\infty \exp\left(-b\left[s-\sqrt{\frac{a}{2b}}\frac{1}{s}\right]^2\right)ds$$
Then use the following theorem (in this form attributed to Glasser's Master Theorem, but known to Cauchy)
$$\int_{-\infty}^\infty f\left(x-\frac{k}{x}\right)dx = \int_{-\infty}^\infty f(x)dx$$
for $k>0$. Thus the integral simplifies to
$$\frac{1}{\sqrt{2\pi}}\exp\left(-\sqrt{2ab}\right)\int_{-\infty}^\infty \exp\left(-bs^2\right)ds = \frac{1}{\sqrt{2b}}\exp\left(-\sqrt{2ab}\right)$$
using the Gaussian integral result.
A: The following change of variables
$$u=\sqrt{t},\qquad t=u^2,\qquad dt =2udu$$
gives
\begin{align*}
I:&= \int_0^\infty\frac{1}{\sqrt{2\pi t}}\cdot\exp\bigg({\frac{-a}{2t}-bt}\bigg)dt\\
&=\frac{1}{\sqrt{2\pi}} \int^\infty_0\frac{1}{u} e^{-\frac{a}{u^2}-b u^2}2udu=\sqrt{\frac{2}{\pi}}\int^\infty_0 e^{-\frac{a}{u^2}-b u^2}\,du=F(a)
\end{align*}
The method used here can be then applied to estimate the integral $I$. Here is more or less what we get when we carry out that aforementioned method:
$$F'(a)=-\sqrt{\frac{2}{\pi}}\int^\infty_0 e^{-\big(\frac{a}{u^2}+ bu^2\big)}\frac{1}{u^2}\,du 
$$
The change of variables
$$v=\frac{\sqrt{a}}{\sqrt{b} u},\qquad u=\frac{\sqrt{a}}{\sqrt{b} v},\qquad du=-\frac{\sqrt{a}}{\sqrt{b}v^2}$$
leads to
$$F'(a)=-\sqrt{\frac{2a}{b\pi}}\int^\infty_0  e^{-\Big(bv^2 +\frac{a}{v^2}\Big)}\frac{b}{a}\,dv=-\sqrt{\frac{b}{a}}F(a)$$
The first order ODE we get satisfies the following initial condition:
$$F(0)=\sqrt{\frac{2}{\pi}}\int^\infty_0e^{-bu^2}\,du=\sqrt{\frac{1}{2b}}$$
Hence
$$
F'(a)+\sqrt{\frac{b}{a}}F(a)=0
$$
which is equivalent to
$$
e^{2\sqrt{ba}}\Big(F'(a)+\sqrt{\frac{b}{a}}F(a)\Big)=\frac{d}{da}\Big(e^{2\sqrt{ba}}F(a)\Big)=0
$$
and so
$$
e^{2\sqrt{ab}}F(a)-F(0)=e^{2\sqrt{ab}}F(a)-\sqrt{\frac{1}{2b}}=0
$$
