I need to integrate below function; $$\int_{-\infty}^{\infty} \frac{\sin(pR)}{R}\frac{p}{k^2-p^2} dp$$ here $k,R$ are constants. Since this is an even function of $p$, I tried applying the residue theorem.
$$\int_{-\infty}^{\infty} \frac{e^{ipR}-e^{-ipR}}{2iR}\frac{p}{k^2-p^2} dp$$ Now taking $z=pR, dz=R dp$; $$\int \frac{e^{iz}-e^{-iz}}{2iR}\frac{z}{k^2R^2-z^2} dz$$
$$\frac{\pi i}{2i}\frac{1}{R}[{(\lim{z \rightarrow kR})} \ \frac{(e^{iz}-e^{-iz})z}{k^2R^2-z^2}(z-kR) +{(\lim{z \rightarrow -kR})} \frac{(e^{iz}-e^{-iz})z}{k^2R^2-z^2}(z+kR)]$$ This gives me zero as the answer. However, since this is an even function integration cannot be zero. Any small help on this highly appreciated.
Assuming that $R$ is a positive real number, $$\begin{eqnarray*}\color{red}{PV}\int_{-\infty}^{+\infty}\frac{z\sin(R z)}{k^2-z^2}\,dz &=&\frac{1}{2}\,\color{red}{PV}\int_{-\infty}^{+\infty}\left(\frac{\sin(Rz)}{k-z}-\frac{\sin(Rz)}{k+z}\right)\,dz\\&=&\frac{1}{2}\,\color{red}{PV}\int_{-\infty}^{+\infty}\left(\frac{\sin z}{R k-z}-\frac{\sin z}{R k+z}\right)\,dz\\&=&\frac{1}{2}\,\text{Im}\left(-2\pi i\,e^{ikR} \right)\\&=&-\pi\,\text{Re}\left(e^{ikR}\right)\\&=&\color{red}{-\pi\cos(R\cdot \text{Re}\, k)\exp(-R\cdot\text{Im}\, k)}.\end{eqnarray*}$$