I am studying the following integral: $$\int_{-\infty}^\infty \frac{1}{2\pi i} \frac{-e^{-ix(a-b)}}{x^2 - c^2} dx$$ where $a, b, c > 0$ are real constants.

The integrand has poles along the real axis when viewed as a complex function. My approach to solving this integral was to create a contour with two semicircles of radius $\epsilon$ and close the contour by creating a big semicircle. In other words, we would have a contour like the one in the figure below:

enter image description here

but instead of having a semicircle cutout near the origin there would be two at $\pm c$. One can then use the residue theorem.

I made a post about this integral on physics SE: https://physics.stackexchange.com/q/800770/288281. In that post I was surprised to find out that the choice of contour does matter and will give different answers, in contrast to something like $$\int_{-\infty}^\infty \frac{\sin(x)}{x}dx$$ which does not depend on the choice of contour. This is briefly discussed in this question: When does the value of a complex contour integral depend on the choice of the contour of integration?.

I have two questions:

  1. When does the value of an integral over the real line, when computed using complex contour integration, depend on the choice of contour and why?
  2. The approach used to solve this integral in physics textbooks is also to use the residue theorem, but they first push the poles slightly above or below the real axis and then take a limit. So the integral becomes $$\lim_{\epsilon \rightarrow 0} \int_{-\infty}^\infty \frac{1}{2\pi i} \frac{-e^{-ix(a-b)}}{x^2 - c^2 + i\epsilon} dx.$$ Why is it necessary to push the poles above/below the real axis? What is wrong with the contour approach I described above which keeps the poles on the real axis? Is this because the integral is not well defined?
  • $\begingroup$ In brief, two correct ways of evaluating an integral will not give different answers. So, yes, the issue is understanding the precise details about fooling around with the contour. :) $\endgroup$ Feb 9 at 2:29
  • $\begingroup$ @paulgarrett This is what I am confused about as I had thought the same. Assuming we use the residue theorem correctly why should different contours give different answers? For example in physics, different choices of contours give distributions with support lying in different areas for the integral I have in my post. $\endgroup$
    – CBBAM
    Feb 9 at 3:21
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    $\begingroup$ That integral is divergent as it stands. Do you interpret it as a principal value integral, or what? $\endgroup$ Feb 9 at 5:00
  • $\begingroup$ @HansLundmark To be honest I am not quite sure. In a lot of physics books they do not go into mathematical details and "compute" it using various contour integrals. In the mathematical treatments I've seen they interpret it as a tempered distribution. Is the reason you get different answers depending on the choice of contour due to the fact that the integral is divergent? $\endgroup$
    – CBBAM
    Feb 9 at 5:31
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    $\begingroup$ is that really the integral you want? or should there be an $i$ in the exponential? anyway, this function cannot be integrated in any sense. In general, the idea is if you have some function with some poles/singularities, then you consider a contour $\Gamma$ which avoids that bad point(s), and such that at the endpoints of the contour, the function behaves nicely. Then, for that contour, and any similar such homotopic contour, you’ll get the same answer for the integral by Cauchy’s theorem. But if you now choose a non homotopic contour, then you’ll (as expected) get a different answer. $\endgroup$
    – peek-a-boo
    Feb 9 at 5:44

1 Answer 1


Ah, with your correction of the exponent, your integral becomes a standard thing, namely a Fourier transform, and issues surrounding "principal value integrals" arise. These have been carefully considered, and understood for quite a while, though perennially lend themselves to (understandable) confusion, as in your case.

Ok, the $1/(x^2-c^2)$ (for real $c$) is not locally integrable near $\pm c$, so its Fourier transform (as distribution!?!) cannot be the integral that usually defines Fourier transforms. That's not at all necessarily fatal, because we can hope to consider it (in some reasonable fashion) as a tempered distribution, so that the Fourier transform need not be computed via that literal integral.

For the same reason, whether or not one thinks of the integral as potentially a Fourier transform, it cannot be a literal integral over the real line. Ok. So we have to make a choice about its meaning. In some larger context, this choice might be dictated. But just as "evaluate this integral", it is not clear. By partial fractions, it suffices to consider ${1\over x-c}$ and ${1\over x+c}$ somehow as tempered distributions (which will also excuse us from worrying about convergence of the integral(s) at infinity). By translating, without loss of generality consider $1/x$.

That is, consider a functional which is "trying to be" $\lambda(f)=\int_{-\infty}^\infty {f(x)\over x}\,dx$, for (for example) test functions $f$. (That's also how we avoid worry about convergence at infinity of the literal Fourier transform...) If we think that we have choices about how to make this integral behave better, there are at least three: first, the Principal Value integral $\lim_{h\to 0^+} \int_{|x|>h} {f(x)\over x}dx$. Second and third involve pushing the contour up/down very slightly, and taking a limit as the push goes to $0$. In a simpler world, these would all produce the same outcome... but they do not. As in the Sokhotski-Plemelj formula, the up/down pushes include an extra $\pm \delta$ at $0$, with opposite signs!

One way to "choose" among these possibilities is to ask what properties the "integrate (not quite literally) against $1/x$" should have. One useful property would be the obvious degree of homogeneity, and parity with respect to $x\to -x$. One can show that this uniquely determines a distribution, up to scalar multiplies.

One can also show that Fourier transform on $\mathbb R$ converts things with homogeneity degree $-s$ to things with homogeneity degree $-(1-s)$... with suitable specification of what homogeneity exactly means for distributions. Fourier transform preserves parity, to the Fourier transform of (principal value integral) $1/x$ is a multiple of the sign function! From this one can easily see the Fourier transform of (principle value) $1/(x^2-c^2)$.

(The push up/down functionals will have an extra constant, due to the Fourier transform of $\delta_{\pm c}$).

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    $\begingroup$ Thank you for your answer! To view this as a Fourier transform are we viewing $(a-b)$ as the frequency/momentum variable? If I have understood everything correctly the root of the problem is we would like to find the Fourier transform of $1/(x^2-c^2)$, which formally should be equal to $$\int_{-\infty}^\infty \frac{1}{2\pi i} \frac{-e^{-ix(a-b)}}{x^2 - c^2} dx$$ but this is ill-defined so we have to somehow give it meaning and there are a variety of ways to do it, some of which may lead to different distributions. Which method one ends up choosing depends on the context. $\endgroup$
    – CBBAM
    Feb 9 at 19:22
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    $\begingroup$ Ah, yes, I'm just thinking of your $a-b$ as the "dual variable", whatever you want to call it. :) And, yes. $\endgroup$ Feb 9 at 19:23
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    $\begingroup$ Well, it's not so much that they have several different FT's, but that the integrate-against-$1/(x^2-c^2)$ functional is ambiguous, because the function is not locally $L^1$, so it cannot literally be "integrate-against-1/(x^2-c^2)$... It's underspecified, even if the symbols look ok. :) After you tell me which disambiguation you are interested in, the FT is unambiguous. :) $\endgroup$ Feb 9 at 19:27
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    $\begingroup$ So once we "modify" $1/(x^2-c^2)$ to make it locally $L^1$ then the ambiguity goes away, that's the whole idea? $\endgroup$
    – CBBAM
    Feb 9 at 19:30
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    $\begingroup$ Yes, without context, it is absolutely not clear what distribution (integrate against) $1/(x^2-c^2)$ refers to. So, yes, FT of that "interpreted such-and-such a way" is solid. :) $\endgroup$ Feb 9 at 19:42

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