Can this definite integral be solved in terms of Bessel functions as a function of the constant $a$?

Question

This Integral is used to derive isotropic antenna Directivity of a uniform array.

$$\int_{\phi =0}^{2 \cdot \pi} \int_{\theta =0}^{\pi}\cos(a \cdot \sin(\theta)\cdot \cos(\phi))\cdot \sin(\theta)\cdot d\theta \cdot d\phi $$ Where $a$ is a constant.

Answer 1

Integrate over $\phi$ first. I get $$ \int_0^{2\pi} \cos(a \sin(\theta) \cos(\phi)) \; d\phi = 2\pi J_0(a \sin(\theta)) $$ So now your integral is

$$\eqalign{ 2 \pi \int_0^\pi J_0(a \sin(\theta)) \sin(\theta)\; d\theta &= 2\pi \sum_{k=0}^\infty \frac{(-1/4)^k a^{2k}}{k!^2} \int_0^\pi \sin^{2k+1}(\theta)\; d\theta\cr &= 4 \pi \sum_{k=0}^\infty \frac{(-1)^k a^{2k}}{(2k+1)!}\cr &= 4 \pi \frac{\sin(a)}{a} }$$

Answer 2

This is not a complete answer.

Reorder the double integral so that the inner integral is something very close to the Bessel Integral (1)

$$J_n(x)=\frac{1}{\pi} \int_0^{\pi} \cos(n \tau - x \sin \tau) \,d \tau \tag{1}$$

Use the substitution $\tau=\phi+\frac{\pi}{2}$ in (1) with $n=0$

thus $$J_0(x)=\frac{1}{\pi} \int_{-\pi/2}^{\pi/2} \cos(-x \cos \phi) \,d \tau$$

Utilising angular symmetry and $\cos(x)=\cos(-x)$ we have $$J_0(x)=\frac{1}{2\pi} \int_{-\pi}^{\pi} \cos(x \cos \phi) \,d \tau$$

which leads to your outer final integral according to both Mathematica and Robert Israel being

$$2 \pi\int_0^{\pi } \sin (\theta ) J_0(a \sin (\theta )) \, d\theta=\frac{4 \pi \sin ( a)}{a}=4 \pi\, j_0(a)$$

where $j_0(x)$ (the sinc function) is the first spherical Bessel function.