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.

  • $\begingroup$ The resulting function is closely related to the Owen's T-function, and thus to be cumulative distribution function of bivariate normal distribution. $\endgroup$ – Sasha Dec 2 '11 at 18:05
  • $\begingroup$ 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. $\endgroup$ – djws Dec 2 '11 at 20:46




$=\dfrac{2\sqrt{t}}{v}\Phi_1\left(\dfrac{1}{2},1,\dfrac{3}{2};-\dfrac{t}{v},-a^2t\right)$ (according to About the confluent versions of Appell Hypergeometric Function and Lauricella Functions)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.