# Integral via complex analysis. Integral via hypercomplex analysis

If I remember rightly there are some integrals of real functions which are easier to compute by using complex analysis.

Is this because of properties of the particular function or because of a lack of a known real analysis technique?

Are there functions which would require hypercomplex analysis to integrate?

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This question is extremely vague – Casebash Jul 27 '10 at 23:21
My understanding of the question is: "There are definite integrals over parts of the real line that are easier to compute using techniques from complex analysis. What properties of the function are responsible for this? Are there any integrals that cannot be computed without further extensions with hypercomplex numbers?" This sounds fine to me. – Larry Wang Jul 27 '10 at 23:25

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There is a famous anecdote of Feynman in which he recalls competing with other people to evaluate integrals. Feynman would use differentiation under the integral sign, and he claimed that this technique (which his colleagues never used) worked on any integral. But one of his colleagues showed him an integral he could not integrate; the only method his colleague knew that worked was complex analysis. Unfortunately I don't know the integral in question. – Qiaochu Yuan Jul 27 '10 at 23:54

## 2 Answers

I'm pretty sure you mean the evaluation of definite integrals. For instance, $e^{-x^2}$ has no elementary antiderivative, but it's definite integral over the real line can be computed explicitly (it's $\sqrt{\pi}$).

You ask what complex analysis has to do with this. Well, the idea is that complex integrals around closed curves (of holomorphic functions) are usually comparatively easy to evaluate: it's because we have the residue theorem as a tool.

However, these real integrals are usually over the whole real line, which doesn't satisfy the hypothesis of a path under the residue theorem. For instance, it's not a finite closed path. However, we can replace the real number by a large semicircle (to pick one common example) that goes from $-R$ to $R$, then around the upper-half-plane from $R$ to $-R$. This is a closed path and the residue theorem can apply to this. When you let $R$ to $\infty$, the integral around semicircular path often tends to zero (by direct bounding arguments). So what you're left with is the integral along the real line, and it's equal to the residues.

Examples of this will be in any complex analysis textbook, e.g. Ahlfors's. The Wikipedia article on the residue theorem (link above) also has examples.

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A generic real-valued function carries no additional information other than its values. But many of the common functions encountered in analysis are just the restriction to $\mathbb{R}$ of holomorphic functions, and these functions carry a lot of additional information, namely, their values in $\mathbb{C}$. Unlike real-analytic functions, holomorphic functions are extremely rigid: for example, if two holomorphic functions agree on a set of complex numbers with a limit point, they must be equal everywhere. That means it's possible to deduce information about how a holomorphic function behaves on part of its domain (say, $\mathbb{R}$) from information about how it behaves on other parts of its domain, since the two have to determine each other. This information is extracted using contour integrals.

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