I am having trouble evaluating the following double integral:

$$\int\limits_0^\pi\int\limits_0^{2\pi}\exp\left[a\sin\theta\cos\psi+b\sin\theta\sin\psi+c\cos\theta\right]\sin\theta d\theta\, d\psi$$

I will be satisfied with an answer involving special functions (e.g. modified Bessel function).

I tried expanding the product terms using the product-to-sum identities, as well applied other trigonometric identities, to no avail. Can anyone help?

I suspect that the solution will involve a modified Bessel function $I_n(x)$ with argument in the form $\sqrt{a^2+b^2+c^2}$. The reason I think so is because the above integral is related to a likelihood function used in evaluation of a non-coherent detection of signals that are corrupted by additive white Gaussian noise (for example, see Proakis "Digital Communications" 4th Edition pp. 304-306). However, these signals are in the two-dimensional domain (they are complex numbers) and the corresponding integral is as follows: $\int_{0}^{2\pi}\exp[a\sin\theta+b\cos\theta]d\theta=2\pi I_0(\sqrt{a^2+b^2})$. The double integral that I am trying to evaluate is the extension of this problem to a three-dimensional space, with the corresponding change from polar to spherical coordinate system.


$$\begin{align}a \sin{\theta} \cos{\psi} + b \sin{\theta} \sin{\psi} &= (a \cos{\psi} + b \sin{\psi}) \sin{\theta}\\ &= \sqrt{a^2+b^2} \cos{(\psi- \psi_0)} \sin{\theta} \end{align}$$


$$\tan{\psi_0} = \frac{b}{a}$$

Now, when we integrate the periodic function over $\psi \in [0,2 \pi)$, the result does not depend on $\psi_0$ and we get

$$\int_0^{2 \pi} d\psi \: e^{\sqrt{a^2+b^2} \sin{\theta} \cos{(\psi- \psi_0)}} = \int_0^{2 \pi} d\psi \: e^{\sqrt{a^2+b^2} \sin{\theta} \cos{\psi}}$$

This turns out to be

$$2 \pi I_0(\sqrt{a^2+b^2} \sin{\theta})$$

Now the integral is

$$\int_0^{\pi} d\theta \sin{\theta} I_0(\sqrt{a^2+b^2} \sin{\theta}) e^{c \cos{\theta}}$$

  • $\begingroup$ Thanks for the answer, however, I am confused by statement "The integral is independent of $\psi$ for the same reason as you find in the 2d polar integral", as I am not sure how one gets rid of $\psi$. Am I missing some trig identity? Again, thanks! $\endgroup$ – M.B.M. Apr 15 '13 at 2:58
  • $\begingroup$ @M.B.M. I will elaborate. $\endgroup$ – Ron Gordon Apr 15 '13 at 3:02
  • $\begingroup$ @M.B.M.: I am afraid I spoke too soon; when actually working the problem through, i found an error I had made. Not sure if what you have is true in general. $\endgroup$ – Ron Gordon Apr 15 '13 at 3:39
  • $\begingroup$ Hmm, this is the problem I was having as well (not being to get rid of the modified Bessel in the integral). However, I didn't get the $\sin\theta$ in front of the $I_0(\cdot)$, just inside... $\endgroup$ – M.B.M. Apr 15 '13 at 3:55
  • $\begingroup$ I just went through G&R, no dice. Perhaps there is a way to circumvent this? As I said in OP, I'd like to extend the non-coherent detection from 2d to 3d. My conjecture is that one should still use square-law (i.e. test statistic $a^2+b^2+c^2$). To prove this I need to show that the integral in my OP evaluates to a function that is monotonically non-decreasing with $a^2+b^2+c^2$ for real $a$, $b$, $c$. If I do this, I'll be well on the way towards answering a question that long riddled me: stats.stackexchange.com/questions/20146/… $\endgroup$ – M.B.M. Apr 15 '13 at 4:43

First, I would like to thank Ron Gordon for the partial answer above, and, very importantly, for pointing out the error in the original formulation question in the comments. Here I show a closed-form expression for the last integral in his answer:

$$\int_0^{\pi} d\theta \sin{\theta} I_0(\sqrt{a^2+b^2} \sin{\theta}) e^{c \cos{\theta}}$$

Let $t=-\cos\theta$. Then $dt=d\theta\sin\theta$, $\theta=\arccos(-t)\Rightarrow\sin\theta=\sqrt{1-t^2}$, and the limits of integration $\{-1,1\}$. Substitution all of this yields the following integral:


Gradshteyn&Ryzhik, 7th edition has the following integral as formula 6.616.5: $\int_{-1}^{1}e^{-ax}I_0(b\sqrt{1-x^2})dx=2(a^2+b^2)^{-1/2}\sinh\sqrt{a^2+b^2}$.

Thus, the integral in my question evaluates to:


  • $\begingroup$ If this is correct numerically, then this is awesome! How is application to signals going? $\endgroup$ – Yuriy S Jul 12 '16 at 23:59
  • $\begingroup$ @You'reInMyEye This was long time ago! :) I think it worked for the case I was evaluating. Certainly computed a lot faster than the integral... $\endgroup$ – M.B.M. Jul 17 '16 at 11:48
  • $\begingroup$ M.B.M. Still very nice. I'm trying to see if this may be useful for quantum mechanics applications. $\endgroup$ – Yuriy S Jul 17 '16 at 11:52

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