Complex Variables and Integration

[Ed Quanta]

How do I solve the following integral using complex variable techniques

The integral from 0 to infinity of [x^m/(x^2 + 1)^2]; 1<m<3

[Tide]

How do I solve the following integral using complex variable techniques

The integral from 0 to infinity of [x^m/(x^2 + 1)^2]; 1<m<3

You will have to select a contour judiciously. Since m is not generally an integer then z = 0 is a branch point. Choose the branch to extend from 0 to infinity along the postive real axis. For your contour, one part will extend from infinity along the positive real axis and loop arbitrarily close to the origin then continue along the positive real axis paying close attention to the fact that z = R e^{2 \pi i} along that portion of the contour. The rest of your contour circles around from \infty e^{0i} to \infty e^{2\pi i}.

Now just apply the Cauchy Integral Theorem! You'll pick up a residue from the small loop around the origin and the two linear segments have different phases and their sum is proportional to the integral you want to evaluate.

[Ed Quanta]

What about the singularities at z=i and at z=-i?
Do I make my contour so that these two points are not included?

[Tide]

What about the singularities at z=i and at z=-i?
Do I make my contour so that these two points are not included?

Using the contour I described, those singularities are included. You have to be aware of what is inside and outside the contour and it becomes a simple matter of proper bookkeeping!

[Ed Quanta]

So let me get this straight, we have a contour that extends from infinity along the positive real axis, loop around as it comes close to the origin, and then what does it do on the negative real axis? Sorry, just trying to get this picture in my head.

[Tide]

Okay - step by step: Start at infinity along the positive real axis and go toward the origin. Loop around the origin (tiny loop!) and go back out to infinity (on the e^{2\pi i} side!) then close your loop with a big circle at infinity. We're choosing z = R e^{2\pi i} to be the branch cut.