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I've noticed that some complex analysis textbooks discuss evaluating real-valued integrals like $\int_{0}^{\infty} \frac{\sqrt{x}}{1+x^{2}} \, dx $ using a keyhole contour before they have defined the Cauchy principal value of an integral (first definition).

But isn't using a keyhole contour a principal value approach in the sense that the contour approaches the singularity at the origin in a symmetrical way?

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The reason to use a keyhole contour is to do with the fact that $z^{1/2}$ has a branch point at the origin.

You don't really need to know the concept of Cauchy Principal Value to be able to take the limit of the inner radius going to $0$.

But, in a way you are right that it is a principal value approach.

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    Using the example I gave, it would seem to me that by letting the radius of the inner circle go to 0, what you end up calculating is $\text{PV} \int_{0}^{\infty} \frac{\sqrt{x}}{1+x^{2}} \ dx$. Then you can argue that you can drop the PV sign. But some textbooks don't seem to look at it that way, and it's quite confusing.2012-03-29
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    @Random: I was talking about the integral on the inner "circle". As a real/complex integral, the integral you gave is well defined and PV does not appear, that looks right.2012-03-29