Let $n\in\mathbb{N}$ and let $k\in\mathbb{Z}^+$ such that $0\leq k\leq n$.
Evaluate the following integral:
$$I=\oint\limits_{|z|=1}\frac{e^{z}}{z^{n+1}}\sum_{k=0}^{n}k!z^{n-k}\ dz$$
My Attempt
The integrand is not analytic inside $|z|=1$, thus it is not necessarily true that $I=0$. For that reason my best guess would be to use Cauchy's Integral Formula for derivatives:
$$f^{(n)}(z_0)=\frac{n!}{2\pi i}\oint\limits_{\Gamma}\frac{f(z)}{(z-z_0)^{n+1}}dz$$
If I substitute $z_0=0$:
$$2\pi if^{(n)}(0)=\oint\limits_{|z|=1}\frac{n!f(z)}{z^{n+1}}\ dz$$
Equating the RHS with the desired $I$ we receive:
$$2\pi if^{(n)}(0)=\oint\limits_{|z|=1}\frac{n!f(z)}{z^{n+1}}\ dz=\oint\limits_{|z|=1}\frac{e^{z}}{z^{n+1}}\sum_{k=0}^{n}k!z^{n-k}\ dz=I$$
So actually, the analytic function $f(z)$ I am looking for is:
$$f(z)=\frac{e^z}{n!}\sum_{k=0}^{n}k!z^{n-k}$$
Since if I can prove $f(z)$ is analytic, I would get that:
$$I=2\pi if^{(n)}(0)$$
Which seems like a much easier thing to calculate. However, I am still not sure how to calculate that. I didn't put much effort to it, because I have a more basic problem - I don't know how to prove that $f(z)$ I found is anayltic inside $|z|=1$.
I would be glad to hear your thoughts, or maybe a more clever solution that you thought of yourself.
Important: I haven't learned Laurent Series yet.
Thank you!
This is $$\sum_{k=0}^n k!\oint_{|z|=1}\frac{e^z}{z^{k+1}}\,dz.$$ But $$\oint_{|z|=1}\frac{e^z}{z^{k+1}}\,dz=\frac{2\pi i}{k!} f^{(k)}(0)$$ by Cauchy's integral formula for derivatives, where $f(z)=e^z$. Then $f^{(k)}(z)=e^z$ and so $$\sum_{k=0}^n k!\oint_{|z|=1}\frac{e^z}{z^{k+1}}\,dz =\sum_{k=0}^n2\pi i\frac{k!}{k!}e^0=2\pi i(n+1).$$