I have never seen an exercise like this one and I'm really uncertain about the correct way to solve it...
Evaluate $$I= \sum_{n=0}^{\inf}\oint_{\gamma} \frac{(4s)^n}{s-4}cot(s)ds$$
where $\gamma$ is the cirle of radius |2|, centered in s=0.
Clearly we have only one singularity (s=0) inside $\gamma$, but I don't know how to manage it... I can't use the Cauchy formula
$$f(z)=\frac{1}{2\pi j} \oint \frac {f(s)}{s-z}ds$$ because clearly the function is not in this form....
I can't do a Taylor expansion around the singularity...
Many thanks for your help
Following the sunggestion in the comments, I have tried 2 'different' ways, but in one case I have to impose a restriction, in the other one no restriction was needed... Which is the correct one?
WAY 1:
$$I= \sum_{n=0}^{\inf}\oint_{\gamma} \frac{(4s)^n}{s-4}cot(s)ds= \oint_{\gamma} \sum_{n=0}^{\inf} \frac{(4s)^n}{s-4}cot(s)ds = \oint_{\gamma} \sum_{n=0}^{\inf} {(4s)^n} \frac{1}{s-4}\frac{\cos (s)}{\sin(s)}ds=$$
$\displaystyle =\oint \frac{1}{1-4s}\frac{1}{s-4} \frac{\cos s}{\sin s}$ provided that |s|<1
Keeping in mind the $\gamma$ contour and the restriction, there is only one singularity, s=0. Applying the Theorems of residuals:
$$=2\pi j (\lim _{s->0} s \cdot \frac{1}{1-4s}\frac{1}{s-4} \frac{\cos s}{\sin s})=(-1/4)\cdot 2\pi j$$
WAY 2:
$$I= \sum_{n=0}^{\inf}\oint_{\gamma} \frac{(4s)^n}{s-4}cot(s)ds= \oint_{\gamma} \sum_{n=0}^{\inf} \frac{(4s)^n}{s-4}cot(s)ds =\oint_{\gamma} \frac{1}{s-4}\frac{\cos s}{\sin s} +\oint_{\gamma}\sum_{n=1}^{\inf} \frac{(4s)^n}{s-4}cot(s)ds $$
but
$$\lim_{s->0} \frac{s^n}{\sin s}=\lim_{s->0} \frac{n s^{n-1}}{\cos s}=0$$
so s=0 is a removable discontinuity, the second integrand function is analytic and the value of the corresponding integral is zero (Cauchy's Theorem). So:
$$I= \oint_{\gamma} \frac{1}{s-4}\frac{\cos s}{\sin s} ds= 2\pi j \cdot (lim_{s->0} s \frac{1}{s-4} \frac{\cos s}{\sin s})=(-1/4)\cdot 2 \pi j$$
First, $$ \frac{1}{2\pi i}\int_{|z|=2}\frac{(4z)^n\cot z\,dz}{z-4}=\mathrm{Res}\left(\frac{(4z)^n\cot z\,dz}{z-4},z=\frac{\pi}{2}\right)+\mathrm{Res}\left(\frac{(4z)^n\cot z\,dz}{z-4},z=-\frac{\pi}{2}\right)=-\frac{(2\pi)^n}{\frac{\pi}{2}-4}-\frac{(-2\pi)^n}{-\frac{\pi}{2}-4}, $$ since $$ \mathrm{Res}\left(\cot z,\pm\frac{\pi}{2}\right)=-1. $$ Hence $$ \sum_{n=0}^\infty \frac{1}{2\pi i}\int_{|z|=2}\frac{(4z)^n\cot z\,dz}{z-4} $$ diverges.