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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

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    $\begingroup$ There is little reason to expect a closed form expression, even given pleasant endpoints. What is this for? $\endgroup$
    – Will Jagy
    Commented Sep 23, 2011 at 22:47
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    $\begingroup$ Could you provide a bit of context? Is this a numerical integration? Symbolic integration? And I hope you are aware that not all integrals evaluate to elementary expressions. And that MO is for research-level mathematics, so that your question might fall outside what this site covers. If this question is closed, please see the FAQ mathoverflow.net/faq for a list of other Q&A sites you could try. $\endgroup$
    – David Roberts ♦
    Commented Sep 23, 2011 at 22:48
  • $\begingroup$ It can be evaluated numerically quite readily as a function of $a$ and $b$: i583.photobucket.com/albums/ss275/jaspercrowne/… but I guess that's not what you're after? $\endgroup$
    – j.c.
    Commented Sep 23, 2011 at 22:53
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    $\begingroup$ This might be a fine question for MO, but to make it so, please do provide some background. Many integrals show up in many areas of mathematics, and I for one am always interested in hearing about more of them. But it's also easy to write down integrals that do not evaluate to closed expressions. If I understand why this particular integral is important, I'm much more likely to believe that it has a nice evaluation, and I'm much more likely to invest time into answering your question. $\endgroup$
    – Theo Johnson-Freyd
    Commented Sep 24, 2011 at 0:15
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    $\begingroup$ To that end, I have voted that this question be closed temporarily as "too localized". At best, you will revise it, and my complaints will be moot, and you will get a useful answer, and the question will not be closed. Next best, if it is closed before you have time to make revisions, then after revising it you should "flag for moderator attention" and ask that the question be reopened. The point of closing it temporarily is to put some pressure on you to improve the question (I don't think we know each other, so it's hard for me to exert social pressure). $\endgroup$
    – Theo Johnson-Freyd
    Commented Sep 24, 2011 at 0:17

2 Answers 2

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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...

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  • $\begingroup$ It has $e^{-au}$ too. Maple gives me some mess with the WhittakerM function in it. $\endgroup$
    – Brendan McKay
    Commented Sep 25, 2011 at 10:41
  • $\begingroup$ Yes, that's what I meant (I initially was thinking of Laplace transform of a power times power of a log). A mess with a Whittaker function is better than nothing (which is what Mathematica gives me). $\endgroup$
    – Igor Rivin
    Commented Sep 25, 2011 at 14:47
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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.

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  • $\begingroup$ I've tried this :-). The sum isn't really better than the integral for my purposes. And, a>0, b>0 (which I didn't mention in the original post). $\endgroup$
    – Eric Ulm
    Commented Sep 24, 2011 at 12:31
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    $\begingroup$ What ARE your purposes? It would help if you told us... $\endgroup$
    – Igor Rivin
    Commented Sep 24, 2011 at 14:52
  • $\begingroup$ I'm trying to solve a PDE that showed up in a life insurance financial math context. The PDE can be solved numerically straightforwardly, but I've been able to get closed form solutions for some simple but unrealistic mortality laws. I've been working on a more realistic mortality law and have made substantial progress. I could just call the integral above a "special function" in which case I'd be done, but it would be much nicer to do the integral above. I'm sure I could publish it without a solution to that integral, but it would be much nicer with it. $\endgroup$
    – Eric Ulm
    Commented Sep 24, 2011 at 16:24
  • $\begingroup$ Well, the question is: what does the answer need to be good for: getting large $a$ or $b$ asymptotics? Evaluating it for a particular value of $a$ and/or $b$? Closed form is not always the best for those purposes... $\endgroup$
    – Igor Rivin
    Commented Sep 24, 2011 at 20:21
  • $\begingroup$ I guess my quick answer is that I like a closed form for its own sake, i.e. it's satisfying somehow. Overall, getting a solution for a particular value of a and b is most important. I'd be happy if I knew the functional form w.r.t "a" and an integral over "b" (i.e. get the "a" outside the integral). "a" (which depends on the individual's age) would change much more often than "b" (which depends on the parameters of the mortality law, stock market process and interest rates). $\endgroup$
    – Eric Ulm
    Commented Sep 24, 2011 at 20:45

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