What is $\int_0^{\infty}\!e^{-x^2}e^{-ae^{bx^2}}\,dx$? I've been trying without success to evaluate
$$
{\Large\int_0^{\infty}\!e^{-x^2}e^{\Large \,-ae^{\,bx^2}}\,dx}.
$$
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'm trying to solve a PDE that shows up in a financial mathematics context and have made a lot of progress. If I could do this integral, I would have a closed form. Any halep would be appreciated.
 A: It seems that the only approach to solve this integral is to apply the series decomposition approach.
Note that the following procedure works only when $a=0$ and $(a\neq0$ and $b\leq0)$ . (Think why?)
$\int_0^\infty e^{-x^2}e^{-ae^{bx^2}}~dx$
$=\int_0^\infty e^{-x^2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(ae^{bx^2})^n}{n!}dx$
$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^na^ne^{(bn-1)x^2}}{n!}dx$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^na^n\sqrt{\pi}}{2n!\sqrt{1-bn}}$
A: I seem to be late to the party and this is really just a comment on Harry Peter's answer, but while there seems to be no closed form solution of your integral, there are closed form bounds on it (at least under the hypotheses $a \geq 0$ and $b < 0$). In fact, if we denote the value of your integral by $J(a,b)$, then
$$
\frac{\sqrt\pi}{2} e^{-a / \sqrt{1+b}} \leq J(a,b) \leq \frac{\sqrt\pi}{2} e^{-a}.
$$
This may be good enough for what you have in mind. (Also note the right-hand side bound is particularly nice since it is independent of $b$.)
I'll sketch the idea and leave the final verifications to you. Consider Harry's series solution
$$
J(a,b) = \frac{\sqrt\pi}{2}\sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \frac{a^n}{\sqrt{1-bn}}.
$$
Ignore for the moment that $(-1)^n$ isn't always positive. Then we could bound each term in the sum as
$$
\frac{(-1)^n}{n!} \frac{a^n}{\sqrt{(1-b)^n}}
\leq
\frac{(-1)^n}{n!} \frac{a^n}{\sqrt{1-bn}}
\leq
\frac{(-1)^n}{n!} a^n,
$$
which would give the bounds I wrote. To get around that $(-1)^n$ is sometimes negative we consider the series expansions of each of the bounds, sum them as even $+$ odd terms and try to bound those, that is, we show that
\begin{align}
\frac{\sqrt\pi}{2}
\sum_{n=0}^{\infty} 
\Bigl( \frac{1}{(2n)!}\frac{a^{2n}}{(1+b)^n} 
&- \frac{1}{(2n+1)!}\frac{a^{2n+1}}{\sqrt{(1+b)^{2n+1}}} \Bigr)\\
&\leq 
\frac{\sqrt\pi}{2}\sum_{n=0}^{\infty} \Bigl( 
\frac{1}{(2n)!} \frac{a^{2n}}{\sqrt{1-b2n}}
-\frac{1}{(2n+1)!} \frac{a^{2n+1}}{\sqrt{1-b(2n+1)}}
\Bigr) \\
&\leq 
\frac{\sqrt\pi}{2}
\sum_{n=0}^{\infty} 
\Bigl(\frac{1}{(2n)!} a^{2n} - \frac{1}{(2n+1)!} a^{2n+1} \Bigr)
\end{align}
by proving the inequalities hold for each of the $n$-th terms. We can simplify each term by cancelling the $\pi$ factors, multiplying by $(2n+1)!$ and factoring $a^{2n}$ out. Then it is not hard, but takes a little elementary work, to show the inequalities holds for each term. (Note that we do not show that any of these is always $\geq 0$ -- indeed for a fixed $n$ and big $a$ that won't be true.)
