Sir, I have been trying to find out the time average of the Intensity distribution of a scattering problem but I could not find the desired closed form answer of the following integral, which shows the averaging:
$$I_{avg}=\frac{A^2}{2\pi}\int_0^{2\pi} \operatorname{sinc}^2{(\alpha \sin{\phi})} \operatorname{sinc}^2{(\alpha \cos{\phi})}\,d\phi$$
Where, $A$ is the amplitude and taken as a constant, $\alpha$ is also a constant and $\alpha>0$.
Sir, would you kindly suggest me any method or any relevant reference from where I can get some help.
Possible ideas (?) in terms of approximation.
$$I=\int_0^{2\pi} \text{sinc}(\alpha \sin (\phi ))^2 \text{sinc}(\alpha \cos (\phi ))^2\,d\phi=8\int_0^{\frac \pi4} \text{sinc}(\alpha \sin (\phi ))^2 \text{sinc}(\alpha \cos (\phi ))^2\,d\phi$$
Now, using series around $\phi=0$ $$ \text{sinc}(\alpha \sin (\phi ))^2 \text{sinc}(\alpha \cos (\phi ))^2=\sum_{n=0}^p a_n \phi^{2n}+O(\phi^{2p+2})$$ would make $$I \sim 8 \sum_{n=0}^p a_n \frac{\left(\frac{\pi}{4}\right)^{2 n+1}}{2 n+1}+O(\phi^{2p+3})$$
Using only $$a_0=\frac{\sin ^2(\alpha )}{\alpha ^2}\qquad \text{and} \qquad a_1=-\frac{\sin (\alpha ) \left(\left(\alpha ^2-3\right) \sin (\alpha )+3 \alpha \cos (\alpha )\right)}{3 \alpha ^2}$$ for a few values of $\alpha$ some results $$\left( \begin{array}{ccc} \alpha & \text{estimate} & \text{solution} \\ 0.5 & 5.77842 & 5.77776 \\ 1.0 & 4.47144 & 4.46264 \\ 1.5 & 2.86061 & 2.82881 \\ 2.0 & 1.45419 & 1.39610 \end{array} \right)$$
Adding the next term, the results would be $$\left( \begin{array}{ccc} \alpha & \text{estimate} & \text{solution} \\ 0.5 & 5.77759 & 5.77776 \\ 1.0 & 4.46040 & 4.46264 \\ 1.5 & 2.82134 & 2.82881 \\ 2.0 & 1.38668 & 1.39610 \end{array} \right)$$
Since you use Mathematica, generate as many coefficients as you wish for better and better results.
Edit
Still in terms of approximations, it seems that we could very accurately "fit" the integrand using $$f(\phi)= \text{sinc}(\alpha \sin (\phi ))^2 \text{sinc}(\alpha \cos (\phi ))^2=b_0+b_1 \phi^2+b_2 \phi^4+b_3 \phi^6+b_4 \phi^8$$ the coefficients $b_i$ being exactly computed in order to match the values of $$f(0) \qquad f\left(\frac{\pi }{4}\right)\qquad f'\left(\frac{\pi }{4}\right)=0\qquad f''(0)\qquad f''\left(\frac{\pi }{4}\right)$$ The formulae are not very complex (notice that the first and fourth conditions immediately give $b_0$ and $b_1$).
For example, for $\alpha=2$, the maximum error is $3.63 \times 10^{-5}$ and the integral of the approximation is $1.39600$ while numerical integration gives $1.39610$