# Bessel function multiplied with an exponential function with a complicated argument

Sir, I am pursuing a study of the near field diffraction theory. While, studying this theory, the following integral appears:

$$I=\int_{0}^{2\pi}\,d\phi\int_{0}^{\infty}\exp i[k r\sin \phi]\times \frac{\exp -ia_0[s^2+r^2+\beta^2+2r\beta \sin \phi]^{1/2}}{[s^2+r^2+\beta^2+2r\beta \sin \phi]^{1/2}}r\,dr$$

Where, $$i=(-1)^{1/2}$$; $$k$$, $$\beta$$, $$s$$ and $$a_0$$ are constants. Further, $$s< and $$s$$ is comparable to $$\beta$$ and $$r dr d\phi$$ represents the area of the circular aperture. The first part i.e. $$\exp i[k r\sin \phi]$$ can be represented as a Bessel function but the remaining part also contains $$r \sin \phi$$ in its argument, so, I could not do the mixed integral.

While searching, I found from the table of integrals by Gradshteyn and Ryzhik,

$$\int_{-\infty}^{\infty}\,dr\frac{\exp ia_0[s^2+r^2+\beta^2]^{1/2}}{[s^2+r^2+\beta^2]^{1/2}}=2i \pi H_0^1[a_0(\sqrt{s^2+\beta^2})]$$

But, I can not use the above integral due to the presence of the factor $$2r\beta \sin \phi$$.

Further, the integral $$I$$ can be written as: $$I=\frac{1}{2s}\frac{d}{ds}\int_{0}^{2\pi}\,d\phi\int_{0}^{\infty}\exp i[k r\sin \phi]\times \exp -ia_0[s^2+r^2+\beta^2+2r\beta \sin \phi]^{1/2}$$

But how the square root in the argument of the exponential term can be removed.

Sir, would you kindly suggest me what to do further to get a closed form answer such that the above integral can be converted to a standard form, prescribed in the table of integrals by Gradshteyn and Ryzhik or any other relevant references, from where I can get some help.Thanking you..