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I am trying to solve the integral $\int_0^{2\pi} \cos^{2n}\theta d\theta$ using residues. I get the wrong answer so could you please say what I am doing wrong?

We start with the substitution $z = e^{i\theta}$. Then $d\theta = \frac{dz}{iz}$ and $\cos \theta = (z + z^{-1})/2$. So our integral now looks like $\frac{1}{i2^{2n}}\int_{|z|=1} \frac{(z+z^{-1})^{2n}}{z}dz$. The inner part of the integral has a singularity at $z = 0$. So we expand it as a Laurent series around the origin. $\frac{(z+z^{-1})^{2n}}{z} = \frac{1}{z}\sum_{t=0}^{2n}\binom{2n}{t}z^{2n-t}z^{-t} = \sum_{t=0}^{2n}\binom{2n}{t}z^{2n-2t-1}$. The residue of that beast is the coefficient of $1/z$, that is $\binom{2n}{n}$. Therefore the integral is equal to $\frac{2\pi i}{i2^{2n}}\binom{2n}{n} = \frac{\pi}{2^{2n-1}}\binom{2n}{n}$. But this is the wrong answer! Please help me!

I know that this can be solved using partial integration but I don't want that!

  • 0
    How do you get $\cos^{2n}=\frac{(z+z^{-1})^{2n}}{2^{2n}z}dz$?2012-09-14
  • 0
    I get that $cos^{2n}\theta = \frac{(z+z^{-1})^{2n}}{2^{2n}}$. The $z$ in the denominator comes from $d\theta = \frac{dz}{iz}$2012-09-14
  • 0
    $z=ie^{i\theta}\rightarrow dz = ie^{i\theta}d\theta$, but $e^{i\theta} = z$, then $dz = iz d\theta\rightarrow \frac{dz}{iz}=d\theta$.2012-09-14
  • 1
    Your answer is correct. What makes you think it is wrong?2012-09-14
  • 0
    I think this is correct. I also did some numerical computation using Wolfram alpha which agree with your result.2012-09-14
  • 0
    You don't solve integrals, you evaluate them.2012-09-16
  • 0
    Looks like you were right. Would you mind posting an answer?2013-05-21

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