I am trying to solve below integration;
$$\int_{0}^{\infty} H_{0}^{1}(pR)\sin(pR)\frac{p}{k^2-p^2} dp$$ here $k,R$ are constants. This is related to the question link.
Below shows my approach to get an answer;
$$\int_{0}^{\infty} H_{0}^{1}(pR)\sin(pR)\frac{p}{k^2-p^2} dp=\frac{1}{2}\int_{-\infty}^{\infty} H_{0}^{1}(pR)\sin(pR)\frac{p}{k^2-p^2} dp$$ Since $p\cos(pR)$ is an odd function, I modified above equation to $$\frac{1}{2}\int_{-\infty}^{\infty} H_{0}^{1}(pR)[\sin(pR)+i\cos(pR)]\frac{p}{k^2-p^2} dp=\frac{1}{2}\int_{-\infty}^{\infty} H_{0}^{1}(pR)[ie^{-ipR}]\frac{p}{k^2-p^2} dp$$ $$\frac{i}{2}\int_{-\infty}^{\infty} H_{0}^{1}(pR)e^{-ipR}\frac{p}{k^2-p^2} dp$$ considering the lower half plane of the semi circle and poles $p=k,-k$ leads to $$-\pi H_{0}^{1}(kR)\cos(kR)$$. Here I have considered zero order Hankel function of first kind $H_{0}^{1}$ as an even function (as $J_{0}(-x)=J_{0}(x)$ and $Re[Y_{0}(-x)]=Re[Y_{0}(x)]$). WolformAlpha does not give an answer for this. Am I doing right here?
