Use residue theory to establish the following integral

[Volt]

Homework Statement

Use residue theory to establish the result:

$$\int^{\pi}_{0}$$$$\frac{dx}{A + Bcosx}$$ = $$\frac{\pi}{\sqrt{A^2 - B^2}}$$

The Attempt at a Solution

So I've gotten to the point that the above integral =

$$\frac{1}{2}$$ $$\oint^{2\pi}_{0}$$ $$\frac{-2i}{Bz^{2} + 2Az + B}$$ dz

which I know is correct. The problem I'm having is that the problem doesn't state any restrictions on A and B, but obviously the result I'm asked to prove can't hold for any A and B, but I'm just not sure what the restrictions are.

For example, can't A and B be chosen such that there are no poles inside the unit circle, and therefore the integral would just equal 0? Or that you have only one pole in the circle, or two poles, etc.

I factored the denominator as $$\frac{-A \pm \sqrt{A^{2} - B^{2}}}{B}$$ but I have no idea what to do from here. I could now rewrite the denominator of the integral using the two factors but without knowing the values of A and B I don't know how I can possibly continue (i.e. determine if there are even any poles inside the unit circle).

 

[clamtrox]

I don't understand what you are doing here. Let's take the special case of A = B, where the integral clearly should diverge. Then your formula is just

$$\oint^{2\pi}_{0} \frac{-i}{B(z+1)^2} dz$$

which only diverges if z=-1 is in your integration path.

 

[Volt]

I received clarification that for this question A > B > 0. After getting that information I *almost* managed to get the result that they give, except there was a B in the numerator
(i.e. (B*pi) / sqrt(A^2 - B^2)).

Also, I don't know what you're confused about for the integral I got in the original post. It's simply using the fact that the integral from 0 to pi is equal to 1/2 the integral from 0 to 2pi because of the cosine term, then converting it using dx = dz/iz and cosx = (z + z^-1) / 2.

 

[Count Iblis]

Well, you know that the substitution w = 1/z leaves the form of the integral the same:

z + 1/z = w + 1/w

1/z dz -----> w d(1/w) = -1/w dw

The contour in the w-plane is now traversed clockwise, if you make that anticlockwise you get an additional minus sign which cancels the minus sign from -1/w dw

So, in terms of w the integrand is the same as what you had before in terms of z. But the poles inside and outside the contour are interchanged. You know that you have two poles and that the sum of the residues at the poles inside the unit circle gives you the integral. It thus follows that you have to have one pole inside te contour and one outside the contour.

It also follows that the residue at the pole inside the contour is minus the residue at the pole outside the contour. Yoiu can see that in two ways. If you consider the contour integral over the coircle with radius and let R go to infinity, then the contour integral tends to zero. So, obviously, the sum of the residues is zero, which means that the the integral is equal to zero whenever the two poles are inside it.

Ypou can also consider a small contour that encircles one pole and not the origin. Then transforming the the w-plane goes as above, but now without the minus sign from the direction of the contour. The contour in the w-plane is traversed in the same way as in the z-plane, sinply because near either pole the transformation w = 1/z is analytic, so it is approximately a linear function. Under a linear map a counterclockwise contour will be mapped to another counterclockwise contour.

So, the residue at a pole in the z-plane is mapped to minus the residue at the corresponding point in the w-plane. Because of the symmetry, this latter residue is the same as the residue in the z plane at the other pole.

 

[Volt]

Can anyone manage to prove the result in my original post? I get very close but end up with a B in the numerator and I can't figure out where the mistake is. (it comes from factoring the quadratic, with B in the denominator).

 

[Count Iblis]

You have the contour integral over the unit circle of:

1/i 1/[B z^2 + 2 A z + B]

The residue at a pole z_p is the limit:

1/i Lim z to z_p of (z-z_p)/[B z^2 + 2 A z + B]

Using L'Hôpital's rule we get:

1/(2i) 1/[A + B z_p]

If you solve the quadratic equation, you see that the pole that is inside the unit circle is:

z_p = -A/B + sqrt[A^2/B^2 - 1]

(You can see that immediately, because certainly the pole with the minus sign in front of the square root is outside the unit circle and from the quadratic equation you see that the product of the two roots equals 1.)

The integral is thus:

2 pi i/(2 i) 1/[B sqrt(A^2/B^2 - 1)]=

pi/sqrt[A^2 - B^2]

 

[Count Iblis]

This is a good practice problem. Compute the integral of

cos(n theta)/[1-2 r cos(theta) + r^2] d theta

from zero to 2 pi, where r < 1.

This gives you the coefficients of the Fourier series of the function

1/[1-2 r cos(theta) + r^2]

You can then verify the answer by summing the entire Fourier series.