7
$\begingroup$

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

1 Answers 1

2

The methodology and resulting solution in the OP are correct. We can easily corroborate the result using straightforward real analysis, without successive integration by parts. Rather, we exploit symmetry and the relationship between the Beta and Gamma functions. To that end we now proceed to independently evaluate the integral.

We have

$$\begin{align}\int_0^{2\pi}\cos^{2n}(\theta)\,d\theta&=4\int_0^{\pi/2}\cos^{2n}(\theta)\,d\theta\\\\ &=2B\left(1/2,n-1/2\right)\\\\ &=2\frac{\Gamma(1/2)\Gamma(n+1/2)}{\Gamma(n+1)}\\\\ &=2\frac{\sqrt\pi \left(\sqrt\pi (2n-1)!!\right)}{2^nn!}\\\\ &=\frac{2\pi}{2^n\,n!}\frac{(2n)!}{2^n\,n!}\\\\ &=\frac{\pi}{2^{2n-1}}\binom{2n}{n} \end{align}$$

as was to be shown!