Finding $\int_{0}^{2\pi} \frac {(1+2\cos\theta)^n\cos(n\theta)}{3+2\cos\theta} \operatorname{d}\theta$

Question

If $n$ is a positive integer find $$ \int_{0}^{2\pi} \frac {(1+2\cos\theta)^n\cos(n\theta)}{3+2\cos\theta} \operatorname{d}\theta $$

I know that I have to use contour integral with a circle of radius 1 centered at the origin, but I am having trouble converting the integral into the form $\int_{|z|}$

$$\int_{|z| = 1} \frac{(1+z+1/z)^n\cos(n\theta)}{3+(z+1/z)} \frac{1}{iz} \operatorname{d}z$$

I cant seem to find a way to expand $\cos(n\theta)$ into a function of $z$.

From the above equation, I can get that the poles of is at $z = -1.5 \pm \frac{\sqrt{5}}{2}$ and only the residual of $z = -1.5 + \frac{\sqrt{5}}{2}$ should be included.

Answer 1

Because I can't submit a comment, I submit an answer. Maple calculates it for concrete values of $n$. Hope, the following would be useful: the Maple command $$[seq(int((1+2*cos(theta))^n*cos(n*theta)/(3+2*cos(theta)), theta = 0 .. 2*Pi), n = [5, 10, 15, 30])] $$ produces $$ [-1760\,\pi +{\frac {3936}{5}}\,\sqrt {5}\pi ,-6927360\,\pi +{\frac { 15490048}{5}}\,\sqrt {5}\pi ,-27264286720\,\pi +{\frac {60964798464}{5 }}\,\sqrt {5}\pi ,-1662161745149511598080\,\pi +{\frac { 3716706651753989275648}{5}}\,\sqrt {5}\pi ]. $$

Answer 2

Note that

$$\cos{n \theta} = \frac12 \left ( z^n + z^{-n}\right )$$

so upon your substitution, you get

$$-\frac{i}{2} \oint_{|z|=1} \frac{dz}{z} \frac{(1+z+z^{-1})^n (z^n+z^{-n})}{3+z+z^{-1}}$$

This can be rearranged to get

$$-\frac{i}{2} \oint_{|z|=1} \frac{dz}{z^{2 n}} \frac{(1+z+z^2)^n (1+z^{2 n})}{1+3 z+z^2} $$

There are poles where $z^2+3 z+1=0$, or

$$z_{\pm} = \frac{-3\pm \sqrt{5}}{2}$$

i.e., $z_- = -\phi^2$, $z_+=-1/\phi^2$, where $\phi=(\sqrt{5}+1)/2$. Thus only $z_+$ is within the unit circle and contributes to the integral.

For the pole at $z=0$, it may be easiest to find the coefficient of $z^{2 n-1}$ in the Maclurin expansion of

$$\frac{(1+z+z^2)^n (1+z^{2 n})}{1+3 z+z^2} $$

Not an easy task. Good luck.

Answer 3

I don't know if it will be too late for the answer. What you need is to add $i\sin(n\theta)$ to the integral. Clearly $$ \int_{0}^{2\pi} \frac {(1+2\cos\theta)^n\sin(n\theta)}{3+2\cos\theta} \operatorname{d}\theta=0. $$ Noting for $z=e^{i\theta}$, one had $$ 1+2\cos\theta=|1+z|^2-1=(1+z)(1+\frac1z)-1=1+z+\frac1z, $$ and $$3+2\cos\theta=1+(1+z)(1+\frac1z) $$ and hence \begin{eqnarray} &&\int_{0}^{2\pi} \frac {(1+2\cos\theta)^n\cos(n\theta)}{3+2\cos\theta} \operatorname{d}\theta\\ &=&\int_{0}^{2\pi} \frac {(1+2\cos\theta)^ne^{ni\theta }}{3+2\cos\theta} \operatorname{d}\theta\\ &=&-i\int_{|z|=1} \frac {(1+z+\frac1z)^{n}z^n}{1+(1+z)(1+\frac1z)}\frac{\operatorname{d}z}{iz}\\ &=&-i\int_{|z|=1} \frac {(z^2+z+1)^{n}}{z^2+3z+1}\operatorname{d}z\\ &=&-i\int_{|z|=1} \frac {(z^2+z+1)^{n}}{(z-z_1)(z-z_2)}\operatorname{d}z\\ &=&-i\cdot 2\pi i \cdot\text{Res}(\frac {(z^2+z+1)^{n}}{z-z_2},z=z_1)\\ &=&2\pi \frac {(z_1^2+z_1+1))^{n}}{z_1-z_2}\\ &=&\frac{2\pi(3-\sqrt5)^{n}}{\sqrt5}. \end{eqnarray} Here $z_1,z_2$ are two of $z^2+3z+1=0$, $$ z_1=\frac{-3+\sqrt5}{2}, z_2=\frac{-3-\sqrt5}{2}. $$