Evaluate Integral: \frac{16cos^{2}x}{5-4cosx}

[elimenohpee]

Homework Statement

Use the residue theorem to evaluate: $$\int^{\pi}_{0}\frac{16cos^{2}xdx}{5-4cosx}$$

The Attempt at a Solution

I rewrote the integral with the substitutions
$$z=e^{ix}$$
$$dx = \frac{dz}{iz}$$
$$cosx = 0.5(z+z^{-1})$$
$$cos^{2}x = 0.25(z+z^{-1})^{2}$$

I throw all that in, convert the integral then evaluate the poles. There are two simple poles at z=2 and z=0.5. I discard the z=2 pole since it lies outside the contour. I calculate the residue at z=0.5
$$\int_{C}\frac{4(z+z^{-1})^{2}dz}{-i(z-2)(2z-1)}$$

$$res_{z->0.5}= ... = 25/3i$$

therefore, it makes the integral = $$\frac{50\pi}{3}$$

when I type this into wolfram alpha, I get 10 pi/ 3 ... any idea where I messed up?

 

[Lavabug]

Homework Statement

Use the residue theorem to evaluate: $$\int^{\pi}_{0}\frac{16cos^{2}xdx}{5-4cosx}$$

The Attempt at a Solution

I rewrote the integral with the substitutions
$$z=e^{ix}$$
$$dx = \frac{dz}{iz}$$
$$cosx = 0.5(z+z^{-1})$$
$$cos^{2}x = 0.25(z+z^{-1})^{2}$$

I throw all that in, convert the integral then evaluate the poles. There are two simple poles at z=2 and z=0.5. I discard the z=2 pole since it lies outside the contour. I calculate the residue at z=0.5
$$\int_{C}\frac{4(z+z^{-1})^{2}dz}{-i(z-2)(2z-1)}$$

$$res_{z->0.5}= ... = 25/3i$$

therefore, it makes the integral = $$\frac{50\pi}{3}$$

when I type this into wolfram alpha, I get 10 pi/ 3 ... any idea where I messed up?

I don't know what's going on in your denominator after the variable change. Shouldn't you have 5 -2(z + z^-1) downstairs?

 

[elimenohpee]

I don't know what's going on in your denominator after the variable change. Shouldn't you have 5 -2(z + z^-1) downstairs?

I do, but there is also the factor of i*z from the change of variable in the differential term. So I factor that through the denominator, and it gives me (5z - 2z^2 -2) which factors to (z-2)(2z-1)

The problem has to be with the residue, as that is the only residue which makes up the contour integral. But I don't see where I went wrong with calculating the residue at z=0.5

 

[elimenohpee]

Some how the residue should be 5/3i but I'm getting 25/3i

 

[Lavabug]

Yeah I was just working out the residue for 2, I think you're missing a minus sign. I got a 25i/3 (moving the -i to the enumerator making it -(-i) ). I guess that's what you meant

What is the contour exactly? The semicircle of what radius?

 

[elimenohpee]

semi circle of radius 1. If you get i in the numerator, when you evaluate the contour integral, wouldn't that give you a negative area? I = 2*i*pi*residue = 2i pi*(25i/3)= -50pi/3

 

[Lavabug]

I'm a bit baffled, I'll bookmark this and will check back if I figure it out/someone else does.

 

[elimenohpee]

Here is a calculation of the residue at z=0.5
http://www.wolframalpha.com/input/?i=residue+4%28z%2Bz^-1%29^2%2F%28-i%28z-2%29%282z-1%29%29+at+z%3D0.5

Here is the integral evaluated
http://www.wolframalpha.com/input/?i=integrate+16cos^2+x+dx%2F%285-4cosx%29+from+0+to+pi

 

[elimenohpee]

can anyone else offer any insight?

 

[Sethric]

By leaving the z^-1 in the numerator, you have missed the pole at z=0, an order 2 pole. The residue at that pole is 5i. Also, your contour runs over these poles, so you only take half the residue value. Therefore, the sum of the residues is (5i/2) +(-25i/6) = (-10i/6). That comes out to an integral evaluation of $$\frac{10 \pi}{3}$$

 

[Lavabug]

By leaving the z^-1 in the numerator, you have missed the pole at z=0, an order 2 pole. The residue at that pole is 5i. Also, your contour runs over these poles, so you only take half the residue value. Therefore, the sum of the residues is (5i/2) +(-25i/6) = (-10i/6). That comes out to an integral evaluation of $$\frac{10 \pi}{3}$$

Thanks for the response. Could you please expand this a little bit? (or refer me to a site or text that sums up what residues I have to include in these problems) Is this valid only for contours of the type |z| = r (like the problem or |z| < or > r? Or greater/less than or equal to r? I feel like I'm missing some important details about the residue theorem.

 

[Sethric]

Any time you have a contour that runs over a pole, the contour has to "wrap" around it, instead of actually crossing over. The contour will run in a semicircle around the residue at an infinitely small radius. In doing so, only half of the residue value will play a part in the integral. This can actually be generalized to smaller or larger fractions as well, depending on the path the contour takes around the pole.

 

[doubleaxel195]

I'm not sure if this will help, but what exactly are you integrating around? It seems like you're integrating around a closed path, correct? But the original problem asked from 0 to pi, which is half a circle. Note too that your integrand is even.

 

[Lavabug]

Any time you have a contour that runs over a pole, the contour has to "wrap" around it, instead of actually crossing over. The contour will run in a semicircle around the residue at an infinitely small radius. In doing so, only half of the residue value will play a part in the integral. This can actually be generalized to smaller or larger fractions as well, depending on the path the contour takes around the pole.

Thanks. Just want to make sure I'm getting it: do I take the half value of the residue if I have pole at the point 3 in a contour of the type |z| $$\leq$$3 ? Do I ignore the pole if the contour is |z| < 3? And take the full value if the pole is between 0 and 3?

 

[Sethric]

Thanks. Just want to make sure I'm getting it: do I take the half value of the residue if I have pole at the point 3 in a contour of the type |z| LaTeX Code: \\leq 3 ? Do I ignore the pole if the contour is |z| < 3? And take the full value if the pole is between 0 and 3?

In order:
I would say yes, as the boundary appears to be a straight line at infinitesimal values, implying that the arc is approaching half of a circle.

Honestly, for that one, I don't know. Intuition tells me I would still need to include the residue of that pole, but I would wait for someone more experienced in complex analysis to answer that. My reasoning is that the original theorem deals with an integral over an arc that encroaches on said pole. Since you are strictly less than, your arc would have to be on the inside, implying your orientation might be negative. So it could be negative half of the residue. But that's all speculation. Anyone else know the answer?

And yes. Full value.

 

[elimenohpee]

By leaving the z^-1 in the numerator, you have missed the pole at z=0, an order 2 pole. The residue at that pole is 5i. Also, your contour runs over these poles, so you only take half the residue value. Therefore, the sum of the residues is (5i/2) +(-25i/6) = (-10i/6). That comes out to an integral evaluation of $$\frac{10 \pi}{3}$$

GAHHHHHH. Why didn't I see that! That is perfect, thank you so much. I understand completely where I went wrong. Have a great night!