Integrating around contour (Cauchy)

[bwinter]

Homework Statement

The question asks that you prove that
$$\int\frac{sin^{2}x}{x^2}dx = \pi / 2$$
The integral is from zero to infinity, but I don't know how to add those in latex.

Homework Equations

Use a contour integral to get around the pole at z = 0. The problem is, I'm really really foggy on how to do that. Professor says residue theorem doesn't need to be used--and besides, isn't the residue at z = 0 just 0 anyway?

The Attempt at a Solution

Integrate around it by subtracting out a small circular contour around the pole, and then adding it back in.

First, I converted the integral to complex form, and changed the boundaries to -inf to inf, by multiplying by one half. I know this much is right.
$$Re \left\{\frac{1}{2} \int\frac{e^{i2z}-1}{z^2}dz\right\}$$

Then I integrate around a small contour: but am I doing this right? Do I just sub these in?
$$z = \epsilon e^{i\theta}$$ and $$dz = i\epsilon e^{i\theta}d\theta$$. Bounds of the integral should be from $$\pi$$ to 0 if I'm integrating above. Then what? I'm lost. Any integral I try to solve from then on just gives me 0. And what do I do about the exponential in the exponential? I've never seen e to the e before and I don't know what to do with it when integrating, which makes me question whether I'm doing this whole contour thing right. Help!

 

[Count Iblis]

Note that sin^2(x) = 1/2 [1 - cos(2x)], so you need to include a minus sign in your expression involving exp(2 i z).

Then, what you do is you expand [exp(2 i z) - 1]/z in powers of z, the expansion will start with z^(-1). If you then integrate around the small circle (which passes the point z = 0 in the upper half plane), then it will be the z^(-1) term which will make a nonvanishing contribution in the limit of epsilon to zero. The contribution of all other terms will be zero.

You ten close the contour in the upper half plane using a half circle of radius R. You can prove that the conribution of that big half corcle tends to zero in the lmit R to infinity. Since the total contour integral must be zero (because the contour does not contain the singualrity at z = 0, that's why we let the small circle pass over it in the upper haplf plane), it follows that the integral is minus the contribution of the small circle.

 

[bwinter]

I expanded the expression and took the limit of epsilon going to zero, and was left with $$a_{-2}=-1$$. If I integrate this over pi to zero, I get pi. Putting the 1/2 back in from in front of the integral, I get pi/2, which is my answer. But is this right? I didn't use any radius R or anything.

 

[Count Iblis]

I expanded the expression and took the limit of epsilon going to zero, and was left with $$a_{-2}=-1$$. If I integrate this over pi to zero, I get pi. Putting the 1/2 back in from in front of the integral, I get pi/2, which is my answer. But is this right? I didn't use any radius R or anything.

Yes, the expansion around zero will contain the answer. But to see that the value of the integral is given by that number, you need to set up the complete contour integral and show that the big half circle tends to zero. The reason for that is basically that the 1/z^2 term goes to zero rapidly enough to make the integral zero and that
exp(2 i z) also tends to zero:

exp(2 i z)

for z = R exp(i theta) can be written as:

exp[2 i R (cos(theta) + i sin(theta)]

The absolute value of this is:

exp(- 2 R sin(theta))

and sin(theta) > 0 in the upper half plane.

This is why you have to close the contour in the upper half plane. And closing the contour in the upper half plane means that you have to let the small coircle move over z = 0 in the upper half plane too in order to make sure that there are no poles in the contour. The total integral is then zero by Cauchy's theorem.

If instead you let the small circle go below the point z = 0, then the contribution of that small circle changes sign. But then the value of the contour integral is not zero but it is given by minus twice the value oif the small circle you found previously (that follows from the residue theorem). The result you find for the integral is then unchanged, of course. But this illustrates that you do need to consider the whole contour integral.