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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?

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We get $G_\pm$ from the pole at $\pm k$, enclosed in a semicircular contour on $\operatorname{sgn}\Re q=\pm\operatorname{sgn}k$:$$G_\pm=\frac{1}{4\pi^2iR}2\pi i\left.\left(\frac{-q}{q\pm k}e^{iqR}\right)\right|_{q=\pm k}=\frac{-1}{4\pi R}e^{\pm ikR}.$$

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  • $\begingroup$ How we draw such a semicircular contour, if we try to to draw such a contour it will pass through poles , it's not possible then how will you deform it @J.G. $\endgroup$ – ROBIN RAJ Oct 31 '19 at 7:32
  • $\begingroup$ @ROBINRAJ Draw a diagram: the "infinite diameter" is the imaginary axis, so the real poles aren't on it. $\endgroup$ – J.G. Oct 31 '19 at 7:36
  • $\begingroup$ Can you draw that contour,@J.G. $\endgroup$ – ROBIN RAJ Oct 31 '19 at 12:29

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