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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.

QuantumGirl
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    what contour are you using, and, which simple pole is found within it? – James S. Cook Jul 22 '15 at 14:33
  • @James S. Cook poles are $p=k,-k$. I used half circle. Correct me if I am wrong. I am doing a mistake here, but I still could not figure it out. Thanks – QuantumGirl Jul 22 '15 at 14:51
  • I don't understand the notation in the last equation. – joriki Jul 22 '15 at 15:07
  • @joriki I modified the last equation. Hope now it is clear:) – QuantumGirl Jul 22 '15 at 15:15
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    Only one of the exponentials is small on a semicircle in either half-plane. If $R$ and $k$ are real, use $\frac{\sin (pR)}{R} = \operatorname{Im} \frac{\exp (ip\lvert R\rvert)}{\lvert R\rvert}$ and a semicircle in the upper half-plane. Otherwise, split the integrand and use the upper half-plane for one part and the lower half-plane for the other, each with the exponential that is small on the respective semicircle. – Daniel Fischer Jul 22 '15 at 15:21
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    @DanielFischer: I think that's an answer rather than a comment. – joriki Jul 22 '15 at 15:24
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    @joriki Only if I don't find a duplicate. – Daniel Fischer Jul 22 '15 at 15:25
  • @user50724: Are you aware you can write proper limits like this? \lim_{z\to kR} – joriki Jul 22 '15 at 15:26
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    Some questions where the problem of using the upper resp. lower half-plane for each of the exponential terms is touched: 1, 2, 3. Not sure whether any of those qualifies as a duplicate candidate. – Daniel Fischer Jul 22 '15 at 15:52
  • @DanielFischer many thanks for the explanation :) – QuantumGirl Jul 23 '15 at 01:43

1 Answers1

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Assuming that $R$ is a positive real number, $$\begin{eqnarray*}\color{red}{PV}\int_{-\infty}^{+\infty}\frac{z\sin(R z)}{k^2-z^2}\,dz &=&\frac{1}{2}\,\color{red}{PV}\int_{-\infty}^{+\infty}\left(\frac{\sin(Rz)}{k-z}-\frac{\sin(Rz)}{k+z}\right)\,dz\\&=&\frac{1}{2}\,\color{red}{PV}\int_{-\infty}^{+\infty}\left(\frac{\sin z}{R k-z}-\frac{\sin z}{R k+z}\right)\,dz\\&=&\frac{1}{2}\,\text{Im}\left(-2\pi i\,e^{ikR} \right)\\&=&-\pi\,\text{Re}\left(e^{ikR}\right)\\&=&\color{red}{-\pi\cos(R\cdot \text{Re}\, k)\exp(-R\cdot\text{Im}\, k)}.\end{eqnarray*}$$

Jack D'Aurizio
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  • Thanks for this Principal value approach. Would you mind giving a small explanation on arriving from step 2 to step 3? Thanks – QuantumGirl Jul 23 '15 at 01:42
  • @user50724: we just have to compute $$PV\int_{\mathbb{R}}\frac{e^{iz}}{z},dz,$$ translate the variable $z$, then take the imaginary part. – Jack D'Aurizio Jul 23 '15 at 14:26