- #1

tworitdash

- 89

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- TL;DR Summary
- I want to compute the integral I have posted here ( https://www.physicsforums.com/threads/spectral-domain-double-integral-with-singularities.989141/ ) This is a sub-problem of the actual problem posted there.

I want to find the analytical solution to the integral given below.

[tex] \int_{-\infty}^{\infty} \frac{ sinc^2(\frac{k_yb}{2})}{\sqrt{k^2 - k_x^2 - k_y^2}}dk_y [/tex]

In other words,

[tex] \int_{-\infty}^{\infty} \frac{ \sin^2(\frac{k_yb}{2})}{(\frac{k_yb}{2})^2\sqrt{k^2 - k_x^2 - k_y^2}}dk_y [/tex]

Can this be solved analytically?

I have seen that the integral [itex] \int_{-\infty}^{\infty} \frac{ J_0(\frac{k_yb}{2})}{\sqrt{k^2 - k_x^2 - k_y^2}})dk_y [/itex] as [itex] \pi J_0(\frac{b}{4} \sqrt{k^2 - k_x^2}) H_0^{(2)}(\frac{b}{4} \sqrt{k^2 - k_x^2)} [/itex] where [itex] J_0 [/itex] is the Bessel function of the first kind of order 0 and [itex] H_0^{(2)} [/itex] is the Hankel function of second kind. Is something similar there for the other integral as well?

[tex] \int_{-\infty}^{\infty} \frac{ sinc^2(\frac{k_yb}{2})}{\sqrt{k^2 - k_x^2 - k_y^2}}dk_y [/tex]

In other words,

[tex] \int_{-\infty}^{\infty} \frac{ \sin^2(\frac{k_yb}{2})}{(\frac{k_yb}{2})^2\sqrt{k^2 - k_x^2 - k_y^2}}dk_y [/tex]

Can this be solved analytically?

I have seen that the integral [itex] \int_{-\infty}^{\infty} \frac{ J_0(\frac{k_yb}{2})}{\sqrt{k^2 - k_x^2 - k_y^2}})dk_y [/itex] as [itex] \pi J_0(\frac{b}{4} \sqrt{k^2 - k_x^2}) H_0^{(2)}(\frac{b}{4} \sqrt{k^2 - k_x^2)} [/itex] where [itex] J_0 [/itex] is the Bessel function of the first kind of order 0 and [itex] H_0^{(2)} [/itex] is the Hankel function of second kind. Is something similar there for the other integral as well?