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 ?
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.
$$ \int_0^{2\pi} \cos ^{2n} \theta \, d\theta = \oint \frac {(z+z^{-1})^{2n}}{2^{2n}} \, \frac{dz}{iz} = \frac{1}{2^{2n}i}\oint \frac {(z^2+1)^{2n}}{z^{2n+1}} \, dz$$ The integral is taken counter-clockwise around the unit circle. Now apply the binomial theorem: $$\int_0^{2\pi} \cos ^{2n} \theta \, d\theta = \frac{1}{2^{2n}i} \oint \,\frac{dz}{z^{2n+1}} \sum_{k=1}^{2n} \binom{2n}{k}z^{2k}=2\pi i\, \text{Res}_{z=0} \{ f(z)\} $$ The residue is the coefficient of the term $\displaystyle\frac{1}{z}.$ This is the term with $k=n$: $$\int_0^{2\pi} \cos ^{2n} \theta \, d\theta =\frac{2\pi}{2^{2n}}\binom{2n}{n}=\frac{2\pi}{2^{2n}} \frac{(2n)!}{n!n!}.$$
