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Recently I derived an expression for a particular probability density function. The expression contains the integral $$ f(t,v,a) = \int_0^t \frac{{\rm e}^{-a^2 z}}{\sqrt{z} (z+v)} \,dz = 2a \int_0^{a\sqrt{t}} \frac{{\rm e}^{-x^2}}{x^2+a^2 v} \,dx \;, $$ where $t>0$, $v>0$ and $a \in \mathbb{R}$, and I would like to rewrite it in terms of named functions (such as error functions and exponential integrals). It seems innocuous but I've tried every integral substitution I can think of without success. The Wolfram Mathematica Online Integrator didn't help, nor did Abramowitz & Stegun's well-known book.

I was about to give up when I stumbled upon the NIST Digital Library of Mathematical Functions, and in particular the page http://dlmf.nist.gov/7.7 where it is said ``Integrals of the type $\int {\rm e}^{-z^2} R(z) \,dz$, where $R(z)$ is an arbitrary rational function, can be written in closed form in terms of the error functions and elementary functions.'' Okay, how do I do this?

Two final comments: Differentiation under the integral sign led me to $$ f(t,v,a) = \frac{\pi}{\sqrt{v}} {\rm e}^{a^2 v} {\rm erfc} \left( a \sqrt{v} \right) - \frac{4}{\sqrt{v}} {\rm e}^{a^2 v} \int_{a\sqrt{v}}^\infty \int_{\frac{\sqrt{t} q}{\sqrt{v}}}^\infty {\rm e}^{-p^2} {\rm e}^{-q^2} \,dp \,dq \;, $$ but this doesn't seem to be helpful. I can evaluate the integral in the special case $t=v$: $$ f(v,v,a) = \frac{\pi}{2 \sqrt{v}} {\rm e}^{a^2 v} \left( 1 - \left( {\rm erf} \left( a \sqrt{v} \right) \right)^2 \right) \;, $$ but this also doesn't seem helpful.

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The resulting function is closely related to the Owen's T-function, and thus to be cumulative distribution function of bivariate normal distribution. –  Sasha Dec 2 '11 at 18:05
    
Aha, so then $f(t,v,a) = \frac{4 \pi}{\sqrt{v}} {\rm e}^{a^2 v} T \left( \sqrt{2 v} a, \frac{\sqrt{t}}{\sqrt{v}} \right)$ where $T(h,\hat{a})$ is Owen's T-function. I guess that's the best one can do. –  djws Dec 2 '11 at 20:46

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