I need help in solving the following definite integral. I could not find any example like this $$\int_{0}^{2\pi}\int_{0}^{d}\exp\!\Big(\frac{-r^2 +2\alpha\; r\cos\theta}{4\;\sigma^2}\Big)r\; dr\; d\theta$$

  • $\begingroup$ hint Do a change of variable from Polar to Cartesian coordinates, then do a translation in the $x$ direction, then (optionally) change back to Polar coordinates. $\endgroup$ – Willie Wong Sep 1 '11 at 1:56
  • $\begingroup$ I have changed the above problem to the definite integral using cdf, here is the link. math.stackexchange.com/questions/61103/…. Help needed... $\endgroup$ – shaikh Sep 1 '11 at 3:02

Integration with respect to $\theta$ can be carried out explicitly in terms of modified Bessel function of the first kind.

$$ \int_0^{2 \pi} \exp\left( \frac{ \alpha r}{2 \sigma^2} \cos \theta \right) \, \mathrm{d} \theta = 2 \pi I_0 \left( \frac{ \alpha r}{2 \sigma^2} \right) $$

The remaining integral

$$ \mathcal{I}_d = 2 \pi \int_0^d r \cdot \mathrm{e}^{-\frac{r^2}{4 \sigma^2}} \cdot I_0 \left( \frac{ \alpha r}{2 \sigma^2} \right) \mathrm{d} r = 4 \pi \sigma^2 \exp\left( \frac{\alpha^2}{4 \sigma^2} \right) \left(1 - Q_1\left( \frac{\alpha}{\sqrt{2} \sigma}, \frac{d}{\sqrt{2} \sigma} \right)\right) $$ where $Q_1(a,b)$ is the Marcum Q-function.

Notice that $$ \lim_{d \to \infty} \mathcal{I}_d = 4 \pi \sigma^2 \exp\left( \frac{\alpha^2}{4 \sigma^2} \right) $$

  • $\begingroup$ Thanks a lot for your answer.Can you please provide some detail of the second integral. Moreover, the Marcum function is using the infinite sum of the Modified Bessel Function,how to cope with that. Please help... $\endgroup$ – shaikh Sep 1 '11 at 10:09
  • $\begingroup$ Marcum Q-function is supported in Mathematica if you need to work with it numerically. Otherwise use the mathematical properties it has. $\endgroup$ – Sasha Sep 1 '11 at 12:44

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