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Consider that I am some height $r$ above a cylinder (measured from the centre) that has its axis in the direction of $z$. I wish to integrate a 2D Gaussian (of SD $\sigma$) along some part of its length and have the following double integral: $$ \int_0^{2\pi} \int_0^L \exp\left[-(a^2 + r^2 + z^2 - 2ar\cos\phi)/2\sigma^2\right] \mathrm{d}z\, \mathrm{d}\phi, $$ where $a$ is the radius of the cylinder. I can then rearrange this to form: $$ \int_0^{2\pi} \exp \left[ -(a^2 + r^2 - 2ar\cos\phi) / 2\sigma^2 \right] \mathrm{d}\phi \int_0^L \exp \left[ -z^2 / 2\sigma^2 \right] \mathrm{d}z. $$ The second integral gives $$ \DeclareMathOperator{\erf}{erf} \sqrt{\frac{\pi}{2}} \sigma \erf \left( \frac{L}{\sqrt{2} \sigma} \right), $$ but I can't see any way of doing the first integral. Wolfram|Alpha also seems to agree. Have I missed a trick, or is this indeed not do-able with standard functions? Is it possible that a solution could be found if I parameterised my problem differently?

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Use the fact that

$$\int_0^{2 \pi} d\phi \, e^{x \cos{\phi}} = 2 \pi I_0(x) $$

where $I_0$ is the modified Bessel function of the first kind of zeroth order.

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  • $\begingroup$ Great, thanks a lot! $\endgroup$ – Student's Teapot Oct 23 '15 at 14:27

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