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I am in trouble in the calculation of the following double integral: $$\int_0^a d\rho\int_0^{2\pi}d\phi\exp(-ik\rho(\sin(\theta_0)\cos(\phi_0-\phi)+\sin(\theta_1)\cos(\phi_1-\phi)))\rho$$ Many thanks

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Would you care to tell us what sort of problem gave rise to this integral? – Harald Hanche-Olsen Jan 28 '13 at 15:53
@HaraldHanche-Olsen: I know such integrals from experience; they usually have to do with a FT of a function of finite support over a disk. These arise in, for example, diffraction of a plane wave by a transparent disk in an otherwise opaque plane (in this case, the plane wave being oblique to the disk). – Ron Gordon Jan 28 '13 at 15:59
@Harald Hanche Olsen: rigordonna is right. It's an integral related to a Fraunhofer diffraction problem. – Riccardo.Alestra Jan 28 '13 at 16:52
Right. I knew there was something familiar there, but had forgotten. Thanks for the reminder. – Harald Hanche-Olsen Jan 28 '13 at 16:54
up vote 2 down vote accepted

As for the angular integral, you may rewrite the angular terms in the form

$$\int_0^{2 \pi} d \phi \exp{(-i k \rho A \cos{(\phi - \phi')})} $$

where $A$ and $\phi'$ may be expressed in terms of the other angles listed in the original integral. This integral is simply a Bessel function:

$$\int_0^{2 \pi} d \phi \: \exp{(-i k \rho A \cos{(\phi - \phi')})} = 2 \pi J_0{( k \rho A)}$$

The radial integral is then

$$2 \pi \int_0^a d \rho \: \rho J_0{( k \rho A)} = 2 \pi a^2 \frac{J_1{(k A a)}}{k A a}$$


To be explicit, one may find $A$ by determining the coefficients of $\cos{\phi}$ and $\sin{\phi}$; let's say the coefficients are $C$ and $S$, respectively. Then define $\phi' = \arctan{\frac{S}{C}}$ such that

$$C \cos{\phi} + S \sin{\phi} = \sqrt{C^2+S^2} \cos(\phi - \phi')$$

In manipulating the expression in the original integral, one finds that

$$C = \sin{\theta_0} \cos{\phi_0} + \sin{\theta_1} \cos {\phi_1}$$ $$S = \sin{\theta_0} \sin{\phi_0} + \sin{\theta_1} \sin{\phi_1}$$

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