# Confusion on contour integration

1. Sep 7, 2010

### quasar_4

Hi everyone,

I'm terribly confused and my advisor is nowhere to be found. Can anyone help with the following scenario?

I am trying to integrate a series of ugly functions of two variables over a solid angle. For most of them, I can get things into a single integral over theta (0 to Pi). The problem is that all of these functions blow up at pi. So, first I integrated and took a limit to try and see if the area underneath the function is finite:

$$\lim_{a \rightarrow \pi} \int_0^a f(\theta) d\theta$$

And when I took these limits they all came out to be infinite (not good). But for kicks and because I'd done the algebra already, I tried doing the integrals as contour integrals - made a substitution z=exp(i*theta), etc. and I got nice numerical (finite!!) answers.

I guess this is similar to the integral of $$\int_0^1 \frac{1}{x} dx$$.

Apparently, if you do this as a contour integral, you get a scalar multiple of pi*i from the residue theorem. But I don't understand this! How can doing the integration one way give you an infinite answer, and the other method give you something finite? What gives? Am I totally wrong on the contour integration, or do you actually get two different results this way? :grumpy:

2. Sep 7, 2010

### snipez90

See http://en.wikipedia.org/wiki/Line_integral#Complex_line_integral".

In particular, read the example that illustrates the use of the definition of the complex line integral. It should be evident that what's happening in the example is much different from the intuitive notion of determining the area under the curve 1/x over (0,1).

Last edited by a moderator: Apr 25, 2017
3. Sep 7, 2010

### quasar_4

That helps a bit. I suppose it's just not as intuitive to work in the complex plane. And that the point of contour integration and the residue theorem is that you can sometimes take advantage of the analyticity of regions around singularities to evaluate integrals... I still have a lot to learn

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