In examples I have seen for solving an infinite integral from $-\infty$ to $\infty$ using contour integration, the real axis becomes part of the contour of integration in the complex plane, and the residue method is used. Posted here is an example of transforming the real line to a unit circle at the complex origin using Möbius transformations. The residue method is used. I have not seen this method before. Can someone point me to the literature on this method?


This method applies whenever these conditions are satisfied:

  1. $f$ is a meromorphic function on a neighborhood of the (closed) upper-half or lower-half plane, having no poles along the real axis, and
  2. viewing the domain of $f$ as an open set of $\mathbf P^1$, then $f$ extends holomorphically to $\infty$.

Then we can calculate the integral using the residue theorem. The reason for this is that the completed upper (or lower) half plane is conformal to the closed unit disc via a suitable automorphism of $\mathbf P^1$ (given by a Möbius transformation). For instance it is clear that $\left|\frac{z-i}{z+i}\right|\leq 1$ if $z$ lies in the upper-half plane, because $z$ is farther from $-i$ than from $i$; equality happens precisely along the real line (draw a triangle). Thus, by pullback, $f$ can be viewed as a meromorphic function on a neighborhood of the closed unit disc, having no poles along the unit circle. In particular, the residue theorem applies, and by the chain rule, the resulting integral along the unit circle is equal to the integral of the original $f$ along the real line (it misses the point $\infty$, but this doesn't change the integral because $\{\infty\}$ has measure zero).

In your example, the function $\frac{1}{1+z^2}$ is meromorphic on all of $\mathbf P^1$, and holomorphic at $\infty$. Thus the residue theorem applies.

  • $\begingroup$ I tried the method with $$\int_{-\infty}^{\infty}1/(1+x^4)dx$$ and the method worked. I tried the method with $$\int_{-\infty}^{\infty}cos(x)/(1+x^2)dx$$ and with the residue method I get $$\pi*cosh(1)$$, which is not correct. Is the reason for it not working in the second case because of Condition (2)? $\endgroup$ – PMay Jun 30 '14 at 17:58
  • $\begingroup$ Dear @PMay That's right: the function $\cos(x)$ has an essential singularity at $\infty$. Cheers - $\endgroup$ – Bruno Joyal Jun 30 '14 at 18:07

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