Laplace transform of inverse gaussian distribution Can someone write in details how i can derive the Laplace transform of the Inverse Gaussian distribution? I think i am missing something during the calculation of the interval which gives the Laplace transform $L(s)=E(e^{-sX})$, so can anyone help by providing the exact calculations?
 A: This is not an easy problem. Lets get rid of some of the parameters so that we will not fill the screen with them. $\lambda=2$ and $\mu=1$. We need to find:
$$\frac{1}{\sqrt \pi}\int _0 ^{\infty}t^{-3/2} e^ {-\frac{(t-1)^2}{t}-st} \,dt$$
Expand the exponent:
$$\frac{1}{\sqrt \pi}\int _0 ^{\infty}t^{-3/2} e^ {-t+2-\frac1t-st} \,dt$$
$e^{-t}$ corresponds to shift in laplace domain, we will take this into account at the last step. We take $e^2$ out:
$$\frac{1}{\sqrt \pi}e^2\int _0 ^{\infty}t^{-3/2} e^ {-\frac1t}e^{-st} \,dt$$
It turns out the laplace transform of $t^{-3/2} e^ {-\frac1t}$ is $\sqrt{\pi } e^{-2 \sqrt{s}}$. EDIT: For completeness, you can check this solution to see how you can obtain this result yourself: enter link description here Or you can look here for a very detailed inverse transformation by Ron Gordon: enter link description here
So we get 
$$e^{2-2 \sqrt{s+1}}$$
In more general:
$$e^{\frac{\lambda }{\mu }-\frac{\sqrt{\frac{\lambda }{\mu ^2}+2 s}}{\sqrt{\frac{1}{\lambda }}}}$$
