One motivation for studying contour integrals in complex analysis is that they can be used to give elegant evaluations of certain real integrals, which tend to have more direct physical and geometrical interpretations (and are therefore easier to motivate). The methods of complex analysis often seem to give a more elegant way to compute real integrals than if real analysis alone was used. However, in many cases where an integral is computable using complex analysis there also seems to exist a way to evaluate it using only real analysis (possibly in several variables). I therefore ask the following vague question (see below for an attempt to make it more precise):

Is there a real integral for which it can be proved that it cannot be evaluated without using methods of complex analysis?

To make the question a bit more precise, first of all assume that all the functions involved are elementary functions (over $\mathbb{Q}$, say). A real integral is meant to be an integral of a real valued function along (some, possibly unbounded, part of) the real line. For the sake of this question, let's say that 'methods of complex analysis' means anything involving Cauchy's theorem or any of its consequences such as Cauchy's Residue Theorem. It's a bit tricky to define precisely what 'evaluate' should mean, but let's say it means that the integral is assigned the value of an explicit elementary function (over $\mathbb{Q}$) at a rational number.

Unless the answer to the question is yes, perhaps there is some general way to show that the above question has a negative answer. For example, I can imagine the possibility of results in logic saying that any sentence expressing the evaluation of a real integral in some language incorporating Cauchy's theorem is equivalent to a sentence in a language including only real analysis. Is anything in this direction known?

  • $\begingroup$ I guess there is some. How could we stand cold in front of complex analysis when it is so much helpful to number theory, or when we read the characterization by Loewner of operator monotone functions? $\endgroup$ Feb 3, 2012 at 10:46
  • $\begingroup$ I vaguely remember in Feynman's autobiography, he had a bet with a mathematician whose name I don't recall, let's say X, that if X could do an integral with residues then Feynman could do it without it. It was Feynman's book, so of course he won... $\endgroup$ Feb 3, 2012 at 13:51
  • $\begingroup$ Denis Serre: I don't think anybody stands cold in front of complex analysis. $\endgroup$ Feb 3, 2012 at 21:38
  • $\begingroup$ @DonuArapura even though it was Feynman's book, he actually lost. See the question math.stackexchange.com/questions/253910/…. $\endgroup$
    – KConrad
    Mar 5, 2019 at 19:41

1 Answer 1


I'm quite sure one could camuflage any complex analysis technique into a real calculus computation, leaving no appeal even to the notion of complex numbers. For instance, the Cauchy's formula on a circle may be translated into a formula for Fourier coefficients; the homotopy invariance of a path integral may be recovered by means of the Stokes theorem and so on. It is even possible that in some case this way one gets a simpler or more elementary computation, and sometimes, why not, it could be a reasonable and welcome operation to do (e.g. I'm thinking to the needs of an elementary course where a certain computation has to be done but the audience is supposed to have no or little Complex Analysis).

But, I'd say that in any case this would be by no means a piece of evidence against the utility of Complex Analysis techniques, nor would suggest that one can renounce to them. I think this is an a general issue about big theories. Big theories are important not only beacuse they provide useful tools to solve problems, but, even before, because they show us the way, like the Polar star. To make an example, you do not need the Open Mapping theorem of Functional Analysis to know that a certain concrete operator is invertible, if you are able to prove the convenient particular bounds, that is, just an inequality. But it's Functional Analysis who told you in advance that an inequality as the wanted one could be proved, and addressed you there.

  • $\begingroup$ Just to be clear: There has been no suggestion that complex analysis is redundant. There is no doubt about its conceptual clarity, elegant results and utility. $\endgroup$ Feb 3, 2012 at 21:36

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