I am studying Scattering theory but I am stuck at this point on evaluating this integral
$G(R)={1\over {4\pi^2 i R }}{\int_0^{\infty} } {q\over{k^2-q^2}}\Biggr(e^{iqR}-e^{-iqR} \Biggl)dq$
Where $ R=|r-r'|$
This integral can be rewritten as
$G(R)={1\over {4{\pi}^2 i R }}{\int_{-\infty}^{\infty} } {q\over{k^2}-{q^2}}{e^{iqR}}dq$
Zettili did this integral by the method of contour integration in his book of 'Quantum Mechanics'.He uses residue theorems and arrived at these results.
$G_+(R)={ -1e^{ikR}\over {4 \pi R}}$ and $G_-(R)={ -1e^{-ikR}\over {4 \pi R}}$
I don't get how he arrived at this result. The test book doesn't provide any detailed explanations about this. But I know to evaluate this integral by pole shifting.
My question is how to evaluate this integral buy just deform the contour in complex plane instead of shifting the poles? I done it by deforming the contour and I get $G (R)=-{1\over {4 \pi} } {cos (kR) \over R}$