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My question relates to this question, where the integral

$\displaystyle\int_{-\infty}^{\infty}\dfrac{e^{iax}}{x^2-b^2}dx=-\dfrac{\pi}{b}\sin(ab)$, where $a,b\gt 0$

is solved. Now, in many physics texts the so-called Feynman prescription is used, i.e. it is stated that instead uf using little semi-circles in the contour to circumvent the poles one may simply shift the pole a bit into the complex plane. Specifically:

$\displaystyle\int_{-\infty}^{\infty}\dfrac{e^{iax}}{x^2-b^2}dx=\int_{-\infty}^{\infty}\dfrac{e^{iax}}{(x-b)(x+b)}dx=\lim_{\epsilon\rightarrow 0}\int_{-\infty}^{\infty}\dfrac{e^{iax}}{(x-b-i\epsilon)(x+b-i\epsilon)}dx$.

Naïvely, one may now solve this integral easily with a contour that contains the real axis and a semi-circle through the upper half-plane. The latter vanishes, so the integral should directly become

$\displaystyle\lim_{\epsilon\rightarrow 0}\int_{-\infty}^{\infty}\dfrac{e^{iax}}{(x-b-i\epsilon)(x+b-i\epsilon)}dx=\lim_{\epsilon\rightarrow 0}2\pi i(Res_{-b+i\epsilon}f(x) + Res_{b+i\epsilon}f(x))$.

The residues are

$\displaystyle Res_{-b+i\epsilon}=-\frac{e^{-iab}e^{-\epsilon}}{2b}\rightarrow-\frac{e^{-iab}}{2b} \qquad (\epsilon \rightarrow 0)$

and

$\displaystyle Res_{-b+i\epsilon}=\frac{e^{iab}e^{-\epsilon}}{2b}\rightarrow \frac{e^{iab}}{2b} \qquad (\epsilon \rightarrow 0)$.

So, in the end we should get

$\displaystyle\lim_{\epsilon\rightarrow 0}\int_{-\infty}^{\infty}\dfrac{e^{iax}}{(x-b-i\epsilon)(x+b-i\epsilon)}dx=\frac{2\pi i}{2b}(e^{iab}-e^{-iab}) = -\frac{2\pi}{b}\sin ab$.

According to the link on the top this result is wrong. However, I cannot see, where the error is in my "derivation". What puzzles me even more: If we were to shift the poles into the lower half-plane the integral would yield zero since it would not contain the poles anymore.

Thank you very much!

Community
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Dominik
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1 Answers1

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It's true that you can shift the poles instead of deforming the contour, but you only get the same result if the direction in which you shift the poles corresponds to the direction in which you deform the contour. See Wikipedia for different ways of interpreting this integral by choosing different semicircles around the poles, corresponding to different shifts in the denominator.

If you just write the integral like that and don't specify what you mean, it's not well-defined, and you can get any of the three results you quoted by interpreting it differently. Usually, without further specification, one would consider the principal value, which effectively takes the average of the upper and lower semicircles at each pole, thus gets half a contribution from each pole and thus yields the middle one of the three results you quoted.

joriki
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  • Thank you very much for the quick answer, this helps a lot! Although it is kind of intuitive, do you know where I can find a rigorous proof of your first statement? – Dominik Jul 06 '16 at 08:49
  • @Dominik: If you mean the statement that shifting the poles and deforming the contour leads to the same result: That's a consequence of the residue theorem. As long as the poles don't cross the contour, the integral is the sum of the "same" residues (times $2\pi\mathrm i$), so you just have to show that the change in the residue from shifting the pole by $\epsilon$ goes to $0$ with $\epsilon\to0$. – joriki Jul 06 '16 at 08:54
  • It is often the physics of the problem that dictate the interpretation of the problem and the corresponding path forward. – Mark Viola May 30 '25 at 15:58