Can anyone help me with this very nasty integral: $$\int \frac{e^{-\frac{x^2}{2}}~x\left(-1+2b^2+2x^2\right)}{\sqrt{1-e^{x^2}} \sqrt{b^2+x^2}} dx,$$

where $b\in\mathbb{R}$.

I've tried pretty much everything I have in bag of tools but nothing worked. If somebody could help me I would be very grateful.

  • $\begingroup$ How did this beast come up? $\endgroup$ – Ataraxia Jun 12 '13 at 15:23
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    $\begingroup$ @ZettaSuro Please take a look at this Physics.SE post. $\endgroup$ – PML Jun 12 '13 at 15:25
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    $\begingroup$ the obvious substitution $y=x^2$ should simplify the matter a little bit $\endgroup$ – sds Jun 12 '13 at 15:27
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    $\begingroup$ What is a typical value of $b$? Is $b>2$? One thing you can try to get an approximate solution is to write this as $\int e^{-x^2/2}f(x){dx}$ and then Taylor expand $f(x)$; the poynomial will begin to diverge for $|x|>1$ but gets killed off by the Gaussian. This trick works pretty nicely for $b>2$. $f(x)$ has an even expansion and the Gaussian multiplied by an even power of $x$ can be integrated and expressed in terms of the Gaussian multiplied by a polynomial, and an error function. Up to 4th order works pretty nicely.. $\endgroup$ – Graham Hesketh Jun 12 '13 at 17:24
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    $\begingroup$ In fact the trick works even better if at first you write $ e^{-1/2x^2} (1-e^{x^2})^{-1/2}=e^{-x^2} (e^{-x^2}-1)^{-1/2}$ then write the integral as $\int e^{-x^2}f(x){dx}$. Then the fit is extremely good for $b>1.5$ (ish), even if you just keep the quadratic term in the Taylor expansion. It's probably worth bearing in mind that adding higher orders won't in general give better results where the series is diverging. $\endgroup$ – Graham Hesketh Jun 12 '13 at 20:44

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