How do I evalutate $\int_{0}^{\infty}\frac{b}{\sqrt{2\pi t^3}} e^{-\sigma t -\frac{b^2}{2t}}dt$? A bit of context first, this is in attempt to verify the distribution of the hitting time $\tau_b=\inf\{t>0,B_t=b\}$ where $B_t$ is one dimensional Brownian Motion started at the origin. Now, this attempt arises from observing that $e^{\sigma B_t-\frac{\sigma^2 t}{2}}$ is a Martingale, then using the Optimal Sampling Theorem for $\tau_b$. The distribution in question is known to be $p(t)=\frac{be^{\frac{-b^2}{2t}}}{\sqrt{2\pi t}}$. The Optional Sampling result tells us that the Laplace Transform is $\tilde{p}(\sigma)=e^{-b\sqrt{2\sigma}}$. I wanted to know whether this can be additionally verified by evaluating explicitly  $\int_{0}^{\infty}\frac{b}{\sqrt{2\pi t^3}} e^{-\sigma t -\frac{b^2}{2t}}dt$
 A: From the integral representation of the modified Bessel function:
$$K_{\nu}\left(z\right)=\tfrac{1}{2}(\tfrac{1}{2}z)^{\nu}\int_{0}^{\infty}\exp%
\left(-t-\frac{z^{2}}{4t}\right)\frac{\mathrm{d}t}{t^{\nu+1}}, $$
we obtain the integral as a $K_{\frac{1}{2}}\left(\left|b\right|\sqrt{2\sigma}\right)$ function. It can be expressed as
$$\frac{e^{-\left|b\right|\sqrt{2\sigma}}}{\left|b\right|}$$ as $K_{1/2}(z)$ as a simple expression.
A: Here is one possible derivation of the integral (I'm sure it can be done simpler) using a standard trick of differentiating wrt a parameter and using Glasser's theorem. 
Define 
$$f(\sigma) = \frac{1}{\sqrt{2\pi}}\int_0^\infty \frac{1}{t^{3/2}}e^{-\sigma t - \frac{b^2}{2t}}{\rm d}t$$
Considering $\frac{df(\sigma)}{d\sigma}$ and performing the substitution $u = t^{1/2}$ we get
$$f'(\sigma) = -\sqrt{\frac{2}{\pi}}\int_0^\infty e^{-\sigma u^2 - \frac{b^2}{2u^2}}{\rm d}u$$
Next completing the square in the exponential, $-\sigma u^2 - \frac{b^2}{2u^2} = -\sigma\left(u - \frac{|b|}{\sqrt{2\sigma}u}\right)^2 - |b|\sqrt{2\sigma}$, we get
$$f'(\sigma) = -e^{-|b|\sqrt{2\sigma}}\sqrt{\frac{2}{\pi}}\int_0^\infty e^{-\sigma\left(u - \frac{b}{\sqrt{2\sigma}u}\right)^2}{\rm d}u$$
which by Glasser's theorem equals
$$f'(\sigma) = -e^{-|b|\sqrt{2\sigma}}\sqrt{\frac{2}{\pi}}\int_0^\infty e^{-\sigma u^2}{\rm d}u = -e^{-|b|\sqrt{2\sigma}}\sqrt{\frac{1}{2\sigma}} $$
Integrating up using $f(\sigma)\to 0$ as $\sigma\to\infty$ gives us
$$f(\sigma) = \int_\sigma^\infty e^{-|b|\sqrt{2\sigma'}}\sqrt{\frac{1}{2\sigma'}}{\rm d}\sigma' = \frac{e^{-|b|\sqrt{2\sigma}}}{|b|}$$
