How to do the integral $\int e^{-\sqrt{ax^2+bx+c}}dx$?

Question

I need to compute the integral

$$ \int e^{-\sqrt{ax^2+bx+c}}dx $$

but I don't know how to do it. I have (unsuccesfully) tried with Mathematica and searched it in the Gradshtein.

Thanks for your help!

Answer 1

By the substitution $t = \sqrt{a} x + b/(2 \sqrt{a})$, this becomes $$ \frac{1}{\sqrt{a}} \int e^{-\sqrt{t^2 + d}}\; dt $$ where $d = c - b^2/(4a)$. Then with $s = \sqrt{t^2 + d}$ it becomes $$ \frac{1}{\sqrt{a}} \int \frac{e^{-s} s}{\sqrt{s^2-d}}\; ds$$

Still no closed form, but for $\sqrt{|d|} < s$ we can expand the integrand in a series in powers of $d$:

$$ \dfrac{e^{-s} s}{\sqrt{s^2-d}} = \sum_{k=0}^\infty \frac{(2k)!}{k!^2} \left(\frac{d}{4s^2} \right)^k e^{-s}$$

And then we can integrate term-by-term using the incomplete Gamma function. Your antiderivative becomes

$$\frac{-1}{\sqrt{a}} \sum_{k=0}^\infty \frac{(2k)!}{k!^2} \left(\frac{d}{4}\right)^k \Gamma(1-2k,s) $$

EDIT: Note that this converges absolutely for $\sqrt{|d|} < s$, since $$ \Gamma(1-2k,s) = \int_s^\infty \frac{e^{-t}}{t^{2k}}\; dt < \frac{1}{s^{2k}} \int_s^\infty e^{-t}\; dt = \frac{e^{-s}}{s^{2k}}$$ and the series for $s/\sqrt{s^2-d}$ converges in this region as $s/\sqrt{s^2-d}$ is analytic in $d$ there.