# Real integral which cannot be evaluated without complex analysis?

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?

• 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? Feb 3, 2012 at 10:46
• 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... Feb 3, 2012 at 13:51
• Denis Serre: I don't think anybody stands cold in front of complex analysis. Feb 3, 2012 at 21:38
• @DonuArapura even though it was Feynman's book, he actually lost. See the question math.stackexchange.com/questions/253910/…. Mar 5, 2019 at 19:41