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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.

Yechiel Asraff
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2 Answers2

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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

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

Robert Israel
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  • Great! Thank you! – Yechiel Asraff Nov 09 '20 at 15:49
  • Hi Robert, can you elaborate how was the last integral calculated? $$ 2\pi \sum_{k=0}^\infty \frac{(-1/4)^k a^{2k}}{k!^2} \int_0^\pi \sin^{2k+1}(\theta); d\theta= 4 \pi \sum_{k=0}^\infty \frac{(-1)^k a^{2k}}{(2k+1)!} $$ I saw the following identity https://math.stackexchange.com/questions/3755787/how-to-evaluate-int-0-pi-sin-n-eta-d-eta?noredirect=1&lq=1 which involves $ \sqrt{\pi } $ which I don't see here. – Yechiel Asraff Apr 01 '22 at 15:36
  • @YechielAsraff You can simplify the $\Gamma$ function from that answer in the case $n=2k+1$. Or do it directly using integration by parts and induction: $$ \int_0^\pi \sin^{2k+1}(\theta) ; d\theta = \dfrac{2k}{2k+1} \int_0^\pi \sin^{2k-1}(\theta); d\theta$$ for $k \ge 1$. – Robert Israel Apr 03 '22 at 05:08
  • Yes, by induction I got $$\int_{0}^{\pi}sin^{2k+1}(\theta )d\theta =\frac{2^{k+1}k!}{(2k+1)\cdot(2k-1)\cdot\cdot\cdot 1}$$ and also found that, $$ (2k+1)\cdot(2k-1)\cdot\cdot\cdot 1=\frac{(2k+1)!}{k!\cdot 2^{k}}$$ – Yechiel Asraff Apr 04 '22 at 13:38
  • So finally I got the following results: $$\int_{0}^{\pi}sin^{2k+1}(\theta )d\theta =\frac{2 \cdot 4^{k} \cdot k!^{2}}{(2k+1)!}=\frac{2^{2k+1} \cdot k!^{2}}{(2k+1)!}$$ – Yechiel Asraff May 22 '22 at 19:12
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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.

James Arathoon
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