evaluate a Fraunhofer diffraction integral I need to evaluate the following integral: $$\int_{-\infty}^\infty \text{rect}(\frac{x'}{2w}) \exp\left (\frac{im}{2}\sin \left(k_Gx' \right) \right )\exp\left (-\frac{ik}{z}xx' \right )dx'$$ where $\text{rect}$ is the Rectangle function, as defined here. This integral becomes (because of $\text{rect}$ values): $$\int_{-w}^w \exp\left (\frac{im}{2}\sin \left(k_Gx' \right) \right )\exp\left (-\frac{ik}{z}xx' \right )dx'$$ I may use the following identity for my solution: $$\exp\left (\frac{im}{2}\sin \left(k \right)\right)=\sum_{q=-\infty}^\infty J_q(\frac{m}{2})\exp\left(iqkx\right)$$ Where $J_q$ is Bessel Function of the First Kind. I need some help working it out. This is a Fraunhofer diffraction from a slit with the transfer function $t(x)=\text{rect}(\frac{x}{2w}) \exp\left (\frac{im}{2}\sin \left(k_Gx \right) \right )$.

Make you substitution and pull the sum out of the integral to get

$\displaystyle \sum_q J_q(\frac{m}{2})\int_{-w}^w dx^\prime \exp(i(qk_G-\frac{kx}{z})x^\prime)$

Now each integral is a straightforward integral of a complex exponential:

$\displaystyle \int_{-w}^w dx^\prime ... = \frac{2i\sin((qk_G-kx/z)w)}{qk_G-kx/z}$

So you just have a series of terms, each one being a Bessel function multiplied by the coefficient you get from the integrals. Honestly, I'm now at my limits. I suspect that the sum can be evaluated or well-approximated by some piece of other information you have been given.