I have a question regarding this post that I wrote some time ago:
Question about a proof regarding the spectral radius of a linear bounded operator
At the end of the question I wrote that:
by continuity of the Riemann integral with values in $L(E)$, $$ \int_{0} ^{2\pi}t^{p+1}e^{i(p+1)\theta}R(e^{i \theta}, T)d \theta = \\\sum_{n} \int_{0}^{2\pi} t^{p-n} e^{i(p-n) \theta}T^n d \theta = 2\pi T^p. $$
Rereading the post, I really don't get why the last equality holds. Why does the sum of those integrals equals $2\pi T^p$?
The integral itself is $$ \int_0^{2\pi} e^{i(p-n)\theta}\,d\theta=\begin{cases}2\pi,&\ n=p\\[0.3cm] 0,&\ n\ne p\end{cases} $$ as the rest does not depend on $\theta$. So the expression is $$ 2\pi\,t^{0}\,T^p=2\pi\,T^p. $$