A Definite Integral Using the Residue Theorem

[AppleFritters]

Homework Statement

I'm trying to solve this definite integral using the residue theorem:

[tex]\int _0^\pi \frac{d \theta}{ (2+ \cos \theta)^2}
[/tex]​

Homework Equations

I got the residue theorem which says that
[tex] \oint_C f(z)dz = 2 \pi i \ \ \text{times the sum of the residues inside C}[/tex]

The Attempt at a Solution

[tex]\int _0^\pi \frac{d \theta}{ (2+ \cos \theta)^2} = \frac{1}{2} \int _0 ^ {2\pi} \frac{d \theta}{ (2+ \cos \theta)^2} = \frac{1}{2} \int _0 ^ {2\pi} \frac{d \theta}{ (4+ 4 \cos \theta + \cos^2 \theta)} \\
[/tex]

I can rewrite [tex] \cos \theta [/tex] as [tex]\cos \theta = \frac{1}{2} (e^{i \theta} + e^{-i \theta})[/tex]

So substituting this in, I get

[tex] \frac{1}{2} \int _0 ^ {2\pi} \frac{d \theta}{ (\frac{9}{2} + \frac{1}{4}e^{2i \theta} + \frac{1}{4}e^{-2i \theta} + 2e^{i \theta} + 2e^{-i\theta})} [/tex]

Now I make the integral into a contour integral and also make a change of variables:

[tex] z=e^{i \theta}, \hspace{2ex} dz = izd\theta , \hspace{2ex} d \theta = \frac{1}{iz} dz\\
\frac{1}{2}\oint_{\text{unit circle}} \frac{1}{iz} \left(\frac{dz}{\frac{9}{2} + \frac{z^2}{4} + \frac{1}{4z^2} + 2z + \frac{2}{z}}\right) = \frac{1}{2i} \oint \frac{1}{z} \left( \frac{dz}{\frac{1}{4z^2} (z^4 + 8z^3 + 18z^2 + 8z + 1)} \right) = \frac{1}{2i} \oint \frac{zdz}{\frac{1}{4}(z^2 + 4z + 1)^2}[/tex]

To find the singularities, I used the quadratic equation and got
[tex] z_1 = -2 + \sqrt{3} \\
z_2 = -2 - \sqrt{3} [/tex]
[itex]z_1[/itex] lies inside the unit circle but [itex]z_2[/itex] does not. It looks like I'm going to have a second order pole at [itex] z_1 = -2 + \sqrt{3} [/itex] so now I'm going to find the residue at [itex] z_1 [/itex]

[tex] \frac{1}{2i} \oint \frac{zdz}{\frac{1}{4}(z^2 + 4z + 1)^2} = \frac{1}{2i} \oint \frac{zdz}{\frac{1}{4}((z + z_1)(z+z_2))^2} [/tex]

Since this is a second order pole, to find the residue, I multiply [itex] f(z) [/itex] by [itex] (z+z_1)^2 [/itex], then differentiate the result once.

[tex] (z+z_1)^2 f(z) = \frac{z(z+z_1)^2}{((z+z_1)(z+z_2))^2} = \frac{z}{(z+z_2)^2} \hspace{3ex} \rightarrow \hspace{3ex} \frac{d}{dz} \left( \frac{z}{(z+z_2)^2} \right) = -\frac{2z}{(z+z_2)^3} + \frac{1}{(z+z_2)^2} [/tex]

Now take the limit as [itex] z \rightarrow z_1 [/itex] or just evaluating the expression at [itex] z=z_1 [/itex]

[tex] -\frac{2(-2+\sqrt{3})}{(-2 + \sqrt{3} + -2 - \sqrt{3})^3} + \frac{1}{(-2 + \sqrt{3} + -2 - \sqrt{3})^2} = \frac{-2(-2 + \sqrt{3})}{(-4)^3} + \frac{1}{(-4)^2} = \frac{\sqrt{3}}{32} [/tex]

So I get [itex] \frac{\sqrt{3}}{32} [/itex] for the residue. Now applying the residue theorem,

[tex] \frac{1}{2i} \oint \frac{zdz}{\frac{1}{4} (z^2+4z+1)^2} = \frac{1}{2i} 2 \pi i (4) \left( \frac{\sqrt{3}}{32} \right) = \boxed{\frac{\pi \sqrt{3}}{8}}[/tex]

The answer is supposed to be [itex] \frac{2 \pi}{3\sqrt{3}}[/itex]. I can't figure out what I did wrong. Can anyone please point out where I made a mistake? Thank you.

 

[vela]

Sign errors. In terms of ##z_1## and ##z_2##, ##f(z)## is ##\frac{z}{(z-z_1)^2(z-z_2)^2}##.

 

[AppleFritters]

That was such a silly mistake. I kept reworking this problem so many times because of this simple mistake. Thank you so much