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, $s<<r$ 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..


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