I had an integral of $$\int_{0}^{2\pi}\cos^3(\theta) d\theta$$
The answer came out to be
integral over the curve $$\int_{C} \dfrac{(z^2+1)^3}{8iz^4} dz$$
$$=-i* \int_{C}\dfrac{(z^2+1)^3}{8z^4} dz$$
The only pole of the integrand is at z = 0, and the residue is given by the coefficient of $z^3$ in the numerator (by the Laurent-series computation). This is clearly zero: all coefficients of the numerator have even powers.
I don't understand starting the residue part I know if residue equals $0$ the integral must equal to $0$ according to residue theorem.
Also, from my understanding, I know this integral is equal to $\dfrac{2\pi}{3!}f^{(3)}(0)$ and since by cauchy's formula no matter where you evaluate the singular point, as long as there is an analytic function that is analytic on the singular point, the value will be same. However as you can see the 3rd derivative of $(z^2+1)^3$ is pain to calculate and I don't have time to do that during exam. I think I am missing some information. Can someone please help me out leading me to shortcut of this question?
I can conclude that since $\dfrac{2\pi}{3!}f^{(3)}(0)$ which turned out to be 0 so the whole integral became 0. Also, I can conclude that since coefficient of $\dfrac{2\pi}{3!}f^{(3)}(0)$ equals 0 and the coefficient of $z^3$ of this power series (well in this case this cannot be laurent series since you can see the residue does not exist) is also 0. Lastly, I have a question about one matter which has been bugging me through out the course. When $\dfrac{2\pi}{3!}f^{(3)}(0)$ you derive this value is there way where you can derive $\dfrac{2\pi}{2!}f^{(2)}(0)$ value solely from the value$\dfrac{2\pi}{3!}f^{(3)}(0)$ ? without having to do laurent expansion of integrand. If not what exactly does generalized cauchy's formula serve its purpose for? I mean I can see that if you have a integrand of pole order 2, you can calculate residue at that singularity and you estimate whether or not coefficient of some general index is bounded to prove questions like liouville theorem. However, I don't see any other purpose....