My complex analysis has the following exercise in the end of the Residue Theorem chapter:
Evaluate the integral $$\int_0^{2\pi} \frac{1}{3 + 2 \cos(t)}dt$$
Because this is the first exercise on the residue theorem they gave the following tip:
consider the path $\gamma(t)=e^{it}$ with $t \in [0,2\pi]$ and $$\oint_\gamma \frac{1}{z^2+3z+1} dz$$
My approach
First I found all the singularities of the function $f(z)=\frac{1}{z^2+3z+1}$ and called them the set $A$.
So now we have:
$$\oint_\gamma \frac{1}{z^2+3z+1} dz = 2\pi i \sum_{a\in A}\text{Res}(f,a) \text{Ind}_\gamma(a)$$
With $A=\{\frac{1}{2}(\sqrt{5} – 3);-\frac{1}{2}(\sqrt{5} + 3)\}$
Because: $\text{Ind}_\gamma\left(-\frac{1}{2}(\sqrt{5} + 3)\right) = 0$ and $\text{Ind}_\gamma\left(\frac{1}{2}(\sqrt{5} – 3)\right) = 1$, we have that:
$$\oint_\gamma \frac{1}{z^2+3z+1} dz = 2\pi i \text{Res}(f,\frac{1}{2}(\sqrt{5} – 3))$$
I calculated that residue and got the following:
$$\oint_\gamma \frac{1}{z^2+3z+1} dz = \frac{2\pi i}{\sqrt{5}} $$
Because we have that $\int_\gamma f = \int_a^b f(\gamma)\gamma ' dt$ then we have:
$$\int_0^{2\pi} \frac{e^{it}}{e^{2it}+3e^{it}+1} dz = \frac{2\pi}{\sqrt{5}} $$
But now I have no idea how I can relate this integral to the original integral $\int_0^{2\pi} \frac{1}{3 + 2 \cos(t)}dt$.
My questions are:
- Did I made any mistake?
- If not, how can I relate this to the original integral I was trying to solve.
- Imagine that I was asked to evaluate this integral but I was not given any tip, how do you find the complex function that you need to integrate over?
Best Answer
Hint - note that:
$$\int _0 ^{2\pi} \frac{1}{3+2\cos(t)}dt\underset{z=e^{it}}=\int _{|z|=1} \frac{z^{-1}dz}{3+2\frac{z+z^{-1}}{2}}dt =\int _{|z|=1} \frac{dz}{z^2+3z+1}dt$$
Generally, the substitution $z=e^{it}$ is a standard way of turning rational trigonometric integrals into contour integrals, which can be solved using the residue theorem.