Sir,
While studying on the convolution theorem using Green's function (using weyl identity) and doing some problems related to this, I have faced the following integration and I want to find out the closed form answer: $$I=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{\exp{isk\Big[\sqrt{1-p^2-q^2}\Big]}}{\sqrt{1-p^2-q^2}}\exp{-ik(px+qy)}\,dp\,dq$$ Where, $s \in \mathbb R(+)$, $k$ and $\beta \in \mathbb R(+)$ are constants.
After searching some texts related to Weyl identity and Sommerfeld identity, what I did is as follows:
I took $k_r^2= k^2(p^2+q^2)$; where $kp=k_r \cos \phi$ and $kq=k_r \sin \phi$. Therefore, the above integral becomes, $$I=\frac{1}{k^2}\int_0^{\infty}\int_0^{2\pi}\frac{\exp{is\sqrt{k^2-k_r^2}}}{\sqrt{k^2-k_r^2}}\exp{ik_r(x\cos \phi+y\sin \phi)} \times k_r \, dk_r \,d\phi$$ After doing the $\phi$ integral first it gives, $$I=\frac{2\pi}{k^2}\int_0^{\infty}\frac{\exp{is\sqrt{k^2-k_r^2}}}{\sqrt{k^2-k_r^2}} J_0(k_r\sqrt{x^2+y^2}) k_r \, dk_r$$ Now two cases will arise here 1) when $k^2>k_r^2$ and 2) when $k_r^2\geq k^2$ i.e. $\sqrt{k^2-k_r^2}=i\sqrt{k_r^2-k^2}$ Accordingly, I can write, $$I=I_1+I_2$$, Where, $$I_1=\lim{a\rightarrow k}\int_0^a\frac{\exp{is\sqrt{k^2-k_r^2}}}{\sqrt{k^2-k_r^2}} J_0(k_rR) k_r \, dk_r$$ where, $R=\sqrt{x^2+y^2}$and $$I_2=\int_k^{\infty}\frac{\exp{-s\sqrt{k_r^2-k^2}}}{\sqrt{k_r^2-k^2}} J_0(k_rR) k_r \, dk_r$$ Now, if I consider the further substitution i.e. for $I_1$, such that $k^2-k_r^2=w^2$, then, it becomes, $$I_1=\int_0^k\exp{i(sw)} J_0(\sqrt{k^2-w^2}R) \, dw$$ and similarly, taking substitution, $k_r^2-k^2=w^2$, $$I_2=\int_k^{\infty}\exp{(-sw)} J_0(\sqrt{k^2+w^2}R) \, dw$$ Would you kindly check the procedure I did and then suggest me what to do further or give me any relevant references such that I can get the closed form answer of this integral.
Thanking you..