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Crosspost from math.stackexchange. Have a look at the great answers there, even though they do not quite answer the question completely.

Define $$G(\theta) = \int\limits_0^\infty \int\limits_0^{2\pi} \frac{r\,e^{-r^2/2t}}{\sqrt{1-(\sin\theta\sin r \cos\varphi + \cos \theta \cos r)^2}} \mathrm{d} r \,\mathrm{d} \varphi$$ Clearly, for $\theta=0$, this does not converge. However, I would like to obtain an asymptotic expansion for $\theta\searrow 0$.

How could I approach this problem? The function $$F(r, \theta) = \int_0^{2\pi} \frac{\mathrm{d}\varphi}{\sqrt{1-(\sin\theta\sin r \cos\varphi + \cos \theta \cos r)^2}}$$ looks somewhat like an elliptic integral, but I have no experience in dealing with these and I have no idea how this could help.

Any thoughts on how to approach this problem?

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  • $\begingroup$ did you check the book "Asymptotic expansion of integrals" amazon.co.uk/Asymptotic-Expansions-Integrals-Dover-Mathematics/… $\endgroup$ Jan 7, 2015 at 8:54
  • $\begingroup$ The series for $\theta\rightarrow 0$ is very easy to compute with any software. Maple yields $\frac{2\pi}{\sin r}+\frac{\pi}{2}\frac{1}{\sin^2r|\sin r|}\theta^2+O(\theta^4)$ but the corresponding integrals in $r$ become more and more intractable due the infinite poles arising at the denominator of increasing order. Maybe, it would be more interesting to understand the behavior of the integral in $r$ taking the upper integration limit to a variable and studying what happens when this limit increases. $\endgroup$
    – Jon
    Jan 9, 2015 at 14:29
  • $\begingroup$ Thank you! But what exactly do you mean? The coefficients of your $\theta$ expansion diverge when integrated over $r$ as soon as the upper limit is greater than $\pi$. It seems that one has to somehow evaluate the $r$ integral first... $\endgroup$ Jan 9, 2015 at 16:02
  • $\begingroup$ This is exactly what I am saying. The series in $\theta$ worsens convergence at any order. But if you permit the upper integration limit in $r$ to be finite one can study the integral in $r$ increasing this limit. $\endgroup$
    – Jon
    Jan 9, 2015 at 17:35

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