# Integrals over an infinite range (using residue theorem)

I am taking a short course of complex variables and am trying to understand how to evaluate the integral:

$$\int\frac{dx}{1+x^{2}}$$ (1) where the integration is from -infinity to +infinity.

To do this, we must, apparently consider:

$$\oint\frac{dz}{1+z^2}$$ (2). The closed loop is a countour which is a semi circle of radius R about the origin containing ONLY the +i singularity (there are two singularities, +i and -i). .

Ofcourse we may write $$\frac{1}{1+z^2}=-\frac{1}{2i}\left(\frac{1}{z+i}-\frac{1}{z-i}\right)$$

Thus the residue at z=i is 1/2i. Therefore the integral (2) is 2pi*i*res(z=i)=pi

This implies (I cant understand why) the Integral (1) EQUALS Integral (2)=pi.

Questions:

1)why do we consider only a loop containing one of the singularities (+i)?
2)why is integral (1), with limits going from -infinity to plus infinity EQUAL to the same integral, with weird limits (-R to R along a semi circle, then along real axis from -R to R)?

## Answers and Replies

These are pretty basic contour integration questions, I think they should be explained in your course?

1) Because you are integrating over the semicircle from -R to R, (with R>1 you get $$i$$ inside of the contour). The other pole ($$z=-i$$) is not inside the contour.

2) Well you are not done yet, you still have to show that part. (Most likely using the ML inequality or Jordans lemma).

And the integral (2), has that value from the residue (as all contour integrals over a jordan curve "$$\gamma$$" in the complex plane)

$$\oint_\gamma f(z) \mathrm{d}z = 2\pi i \sum_k \text{Res}(f,a_k)$$

Where Res(f,a_k) are the residues (z=a_k) of the poles of f inside of the contour "gamma".

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thanks *-<|:-D=<-<

"2) Well you are not done yet, you still have to show that part. (Most likely using the ML inequality or Jordans lemma)."

Hmmm, rereading my notes, I get the impression that there is a trivial reason as to why integral 1 equals integral 2... as in it should be obvious from looking at both integrals, without having to resort to jordan's lemma and such like... i just dont get it though!

Hi vertices,

The complex contour integral (2) you are considering equals the original real integral (1) plus the integral over a semicircle of radius R. It turns out that, in a lot of cases, the integral over that semi circle goes to zero as R goes to infinity, so you get the equality you were seeking.

HallsofIvy
Homework Helper
I think what you are missing is that the integral over the semicircle of radius R goes to 0 as R goes to infinity.

thanks jeff and hallsofivy. Is there a non-trivial reason as to why the integral over the semi circle goes to 0 as R tends to infinity?

My lecturer wrote the first integral and literally put an "=>" sign by it before writing the second integral.

I think he put that indicator to show that the appropriate contour to choose was the closed semi-circle.

The integrand 1/(z^2+1) goes to 0 when R tends to infinity hence the integral does so too.

maybe I guess.

That said, I still don't understand why the integral (2) over a semi circle from R to -R is zero when R tends to infinity. This does not come about from Jordan's Lemma because there is no exp(iaz) term, where a>0. Are there any other techniques I could use to see why the integral is zero?

Ah you're absolutely right, i had forgot what jordans lemma was. The ML inequality or "estimation lemma" applies here however, and i think that is what you are gonna use for all these types of integrals, it is very typical for an introductory course to complex calculus.

Check out estimation lemma on wiki.

Ah you're absolutely right, i had forgot what jordans lemma was. The ML inequality or "estimation lemma" applies here however, and i think that is what you are gonna use for all these types of integrals, it is very typical for an introductory course to complex calculus.

Check out estimation lemma on wiki.
thanks, I just did - very useful:)

what about semi infinite integrals? how do we compute the arc integral in that case?