# Use calculus of residues to evlauate

Use calculus of residues to evaluate the integral $$\int_0^{2\pi}\cos^{2n}\theta d\theta$$

My Ateempt :

$$\int_0^{2\pi}\frac{(1+\cos2\theta)^n}{2^n}d\theta$$ $$=\frac{1}{2^n} \int_C [1+\frac{1}{2}(z^2+\frac{1}{z^2})]^n\frac{dz}{iz}$$ $$=\frac{1}{2^{2n}i} \int_C \frac{(z^4+2z^2+1)^n}{z^{2n+1}}dz$$

Now the integrand has a pole of order $2n+1$ at $z=0$. The residue at $z=0$ is given by $$\frac{\phi^{2n}(z)}{(2n)!}$$

How do i find this residue ?

• I think it's better to not rewrite $\cos^2 \theta = \frac{1+\cos (2\theta)}{2}$. – Daniel Fischer Jun 16 '14 at 18:56
• @DanielFischer yeah right. without that substitution i get the same result rather easily. how do i find the residues ? – Aman Mittal Jun 16 '14 at 19:01

First notice that

$$(z^4+2z^2+1)^{2n} = (x+i)^{2n} (x-i)^{2n}$$

then the residue is given by, using the product rule for differentiation,

$$\frac{1}{(2n)!} \lim_{x \to 0}\sum_{k=0}^{2n} {2n\choose k} \left((x-i)^{2n}\right)^{(k)} \left((x+i)^{2n}\right)^{(2n-k)}=\dots\,.$$

Notes:

1) the general formula for computing the residue of $f(z)$ at a point $z=z_0$ with a pole order $m$ is

$$r = \frac{1}{(m-1)!} \lim_{z\to z_0} \frac{d^{m-1}}{dz^{m-1}}\left( (z-z_0)^m f(z) \right).$$

2) You need the formula

$$\frac{d^q}{d t^q} t^m = \frac{\Gamma(m+1)}{\Gamma(m-q+1 )} t^{m-q}$$

to find the required derivatives.