Integration involving complicated exponential form

Question

I'm trying to simplify the following: $\int_0^ts^{-\frac{3}{2}}e^{-\frac{(a+bs)^2}{2s}}~ds$

Basic substitution always gives a $s^{-\frac{1}{2}}~ds$ counterpart which I don't know how to get rid of. Is there anyway to reduce this integral to some expression based on integration of normal density?

Answer 1

$\int_0^ts^{-\frac{3}{2}}e^{-\frac{(a+bs)^2}{2s}}~ds$

$=\int_0^t\dfrac{e^{-\frac{a^2+2abs+b^2s^2}{2s}}}{s^\frac{3}{2}}ds$

$=e^{ab}\int_0^t\dfrac{e^{-\frac{a^2}{2s}-\frac{b^2s}{2}}}{s^\frac{3}{2}}ds$

$=2e^{ab}\int_0^t\dfrac{e^{-\frac{a^2}{2s}-\frac{b^2s}{2}}}{s}d(\sqrt s)$

$=2e^{ab}\int_0^\sqrt{t}\dfrac{e^{-\frac{a^2}{2s^2}-\frac{b^2s^2}{2}}}{s^2}ds$

$=2e^{ab}\int_\sqrt{t}^0e^{-\frac{a^2}{2s^2}-\frac{b^2s^2}{2}}~d\left(\dfrac{1}{s}\right)$

$=2e^{ab}\int_\frac{1}{\sqrt{t}}^\infty e^{-\frac{a^2s^2}{2}-\frac{b^2}{2s^2}}~ds$

$=\dfrac{\sqrt\pi e^{ab}}{\sqrt2|a|}\left(e^{|a||b|}~\text{erfc}\left(\dfrac{|a|}{\sqrt{2t}}+\dfrac{|b|\sqrt t}{\sqrt2}\right)+e^{-|a||b|}~\text{erfc}\left(\dfrac{|a|}{\sqrt{2t}}-\dfrac{|b|\sqrt t}{\sqrt2}\right)\right)$ (according to http://dlmf.nist.gov/7.7#E7)