Definite Integral ∫_{0}^{∞} dx exp(−x^2−a exp(b x^2)) I've been trying without success to do $$\int_0^\infty dx\; \exp(-x^2) \exp(-a\exp(bx^2)).$$
It's not in my integral tables. Wolfram online integrator won't do it. It doesn't seem to be amenable to a contour integral method, and the method of integrating $e^{-x^2}$ alone doesn't work either. I don't know if this is the kind of question asked here, but any help would be appreciated.
Thanks, Eric
 A: If you expand the $\exp(-a \exp(b x^2))$ as a power series in the variable $a \exp(b x^2)$ you will get a rather nice series, each term of which is a gaussian integral, so is easy to integrate. I don't have mathematica in front of me as I type, but this should give you about as nice a form as you might hope for (and the sum might be doable in closed form).
EDIT
THe other point is that if you make the substitution $\u = \exp(b x^2),$ then your integral becomes the integral from $1$ to $\infty$ of a power of $\log u$ times a power of $u,$ which should be amenable to contour integration...
A: The manipulations below work assuming that $a>0$ and $b<0$.  Using Igor Rivin's idea, one does get a gaussian, but it requires $b<0$.  The resulting sum is
$$ \frac{\sqrt{\pi}}{2} \sum_{k=0}^{\infty} \frac{\left(-a\right)^k}{k!\sqrt{1-kb}}$$
which does not seem to have a closed form.  Sure, one can rewrite $ \frac{1}{\sqrt{1-kb}}$ a series in $k$, but that doesn't help because the resulting term (in $k$) is not hypergeometric, so swapping the order of summation still leads to a dead end.  The above sum might be the best that can be done.
