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I found out an identity: $$\int_0^\infty\frac{\Phi(x)}{x^2-k^2}J_n(xr)x\operatorname{d}x=\frac{i\pi}{2}\Phi(k)H_n^{(1)}(kr) \text{ -> } Im(k)>0$$ where $Y_n$ is second kind bessel function
Using this identity I can reduce the above integral to following $$-i\frac{L}{8\pi}\int_{-\infty}^\infty J_0(R_0\sqrt{(\frac{\omega_0}{c})^2-k_z^2})H_0^{(1)}(r\sqrt{(\frac{\omega_0}{c})^2-k_z^2})Sinc(\frac{Lk_z}{2})e^{jk_zz}\operatorname{d}k_z \text{ } r>R_0$$

Therefore the problem now is much simpler.
Please help me solve this inverse fourie transform

I found out an identity: $$\int_0^\infty\frac{\Phi(x)}{x^2-k^2}J_n(xr)x\operatorname{d}x=\frac{i\pi}{2}\Phi(k)H_n^{(1)}(kr) \text{ -> } Im(k)>0$$ where $H_n$ is Hankel function of the first kind function
Using this identity I can reduce the above integral to following $$-i\frac{L}{8\pi}\int_{-\infty}^\infty J_0(R_0\sqrt{(\frac{\omega_0}{c})^2-k_z^2})H_0^{(1)}(r\sqrt{(\frac{\omega_0}{c})^2-k_z^2})Sinc(\frac{Lk_z}{2})e^{jk_zz}\operatorname{d}k_z \text{ } r>R_0$$

Therefore the problem now is much simpler.
Please help me solve this inverse fourie transform

I found out an identity: $$\int_0^\infty\frac{\Phi(x)}{x^2-k^2}J_n(xr)x\operatorname{d}x=-\frac{\pi}{2}\Phi(k)Y_n(kr)$$ where $Y_n$ is second kind bessel function
Using this identity I can reduce the above integral to following $$\frac{L}{8\pi}\int_{-\infty}^\infty J_0(R_0\sqrt{(\frac{\omega_0}{c})^2-k_z^2})Y_0(r\sqrt{(\frac{\omega_0}{c})^2-k_z^2})Sinc(\frac{Lk_z}{2})e^{jk_zz}\operatorname{d}k_z$$ In othere words it is inverse fourier transform of $$\frac{L}{4} J_0(R_0\sqrt{(\frac{\omega_0}{c})^2-k_z^2})Y_0(r\sqrt{(\frac{\omega_0}{c})^2-k_z^2})Sinc(\frac{Lk_z}{2})$$ Therefore the problem now is much simpler.
Please help me solve this inverse fourie transform

I found out an identity: $$\int_0^\infty\frac{\Phi(x)}{x^2-k^2}J_n(xr)x\operatorname{d}x=\frac{i\pi}{2}\Phi(k)H_n^{(1)}(kr) \text{ -> } Im(k)>0$$ where $Y_n$ is second kind bessel function
Using this identity I can reduce the above integral to following $$-i\frac{L}{8\pi}\int_{-\infty}^\infty J_0(R_0\sqrt{(\frac{\omega_0}{c})^2-k_z^2})H_0^{(1)}(r\sqrt{(\frac{\omega_0}{c})^2-k_z^2})Sinc(\frac{Lk_z}{2})e^{jk_zz}\operatorname{d}k_z \text{ } r>R_0$$

Therefore the problem now is much simpler.
Please help me solve this inverse fourie transform

I found out an identity: $$\int_0^\infty\frac{\Phi(x)}{x^2-k^2}J_n(xr)x\operatorname{d}x=-\frac{\pi}{2}\Phi(k)Y_n(kr)$$ where $Y_n$ is second kind bessel function
Using this identity I can reduce the above integral to following $$\frac{L}{8\pi}\int_{-\infty}^\infty J_0(R_0\sqrt{(\frac{\omega_0}{c})^2-k_z^2})Y_0(r\sqrt{(\frac{\omega_0}{c})^2-k_z^2})Sinc(\frac{Lk_z}{2})e^{jk_zz}\operatorname{d}k_z$$ In othere words it is inverse fourier transform of $$\frac{L}{4} J_0(R_0\sqrt{(\frac{\omega_0}{c})^2-k_z^2})Y_0(r\sqrt{(\frac{\omega_0}{c})^2-k_z^2})Sinc(\frac{Lk_z}{2})$$ Therefore the problem now is much simpler.

I found out an identity: $$\int_0^\infty\frac{\Phi(x)}{x^2-k^2}J_n(xr)x\operatorname{d}x=-\frac{\pi}{2}\Phi(k)Y_n(kr)$$ where $Y_n$ is second kind bessel function
Using this identity I can reduce the above integral to following $$\frac{L}{8\pi}\int_{-\infty}^\infty J_0(R_0\sqrt{(\frac{\omega_0}{c})^2-k_z^2})Y_0(r\sqrt{(\frac{\omega_0}{c})^2-k_z^2})Sinc(\frac{Lk_z}{2})e^{jk_zz}\operatorname{d}k_z$$ In othere words it is inverse fourier transform of $$\frac{L}{4} J_0(R_0\sqrt{(\frac{\omega_0}{c})^2-k_z^2})Y_0(r\sqrt{(\frac{\omega_0}{c})^2-k_z^2})Sinc(\frac{Lk_z}{2})$$ Therefore the problem now is much simpler.
Please help me solve this inverse fourie transform