Question about a proof regarding the spectral radius of a linear bounded operator I'm studying basic spectral theory from the book Elements of functional analysis by Hirsch and Lacombe, and I've encountered some difficulties in understanding the proof of the following theorem:

Theorem Let $E$ be a Banach space over $\mathbb{C}$ and suppose that $T \in L(E)$ (where $L(E)$ is the set of all bounded linear
  operators on $E$). Then the spectrum $\sigma(T)$ of $T$ is nonempty
  and $r(T) = \max \{|\lambda|: \lambda \in \sigma(T) \} $ (where $r(T) =
 \lim ||T^n||^{\frac{1}{n}}$).

I write here the beginning of the proof:
Let $\rho = \max \{|\lambda|: \lambda \in \sigma(T) \}$ and assume by contradiction $\rho<r(T)$. Take $\lambda \in \mathbb(C)$ s.t. $|\lambda|>r(T)$. So $\lambda I - T$ is invertible and
$$ (\lambda I - T) ^{-1}= R(\lambda, T)= \sum_{n=0}^{\infty} \lambda^{-(n+1)}T^n,$$
where $R(\lambda, T)$ denotes the resolvent operator.
Take $p \in \mathbb{N}$ and consider
$$ \lambda^{p+1} R(\lambda, T) =\sum_{n=0}^{\infty} \lambda^{(p-n)}T^n. $$
Then write $\lambda$ as $\lambda = te^{i \theta}$, with $t>0$ and $\theta \in [0,2\pi] $. Then
$$t^{p+1}e^{i(p+1)\theta}R(e^{i \theta}, T) = \sum_{n} t^{p-n} e^{i(p-n) \theta}T^n. $$
Here I cite:
Integrating the result (with respect to $\theta$) from $0$ to $2\pi$, we obtain, by continuity of the Riemann integral with values in $L(E)$, $$ \int_{0} ^{2\pi}t^{p+1}e^{i(p+1)\theta}R(e^{i \theta}, T)d \theta = \\\sum_{n} \int_{0}^{2\pi} t^{p-n} e^{i(p-n) \theta}T^n d \theta = 2\pi T^p. $$ (The proof continues, but my question regards only this part that I've just written). 
Now my question: In the last expression they integrate an expression in which appears an operator which takes value in an abstract Banach space, and they mention Riemann integral with values in $L(E)$. How should I interpret this phrase? Also, I would like to know why exchanging integral and summation sign is a legal operation. I know that this operation is allowed with standard Riemann integrals and absolutely convergent series, but as I said the context seems to be here more abstract. 
 A: What you need in order to understand the integral given there is the theory of the Banach space - valued Riemann integral. This is a more general integral as it is for functions $f: [a, b] \longrightarrow \mathbb{R}$. What one uses here for the construction are step functions $s: [a, b] \longrightarrow X$, which take values in a Banach space $X$. The usual construction of this integral is done by the BLT - theorem (Bounded linear transformation). See for example Reed/Simon 1 or here. It should be also noted that in the latest version of the book by Hirsch and Lacombe, there is a huge exercise on page 20, where this integral and its properties are constructed in eight steps. Note also that if $X$ is a Banach space, $L(X)$ is a Banach space and therefore the integral taken over bounded operators makes sense.
Now to your second question concerning the legitimation of interchanging the sum and the integral. Let $E$ be a Banach space, $T \in L(E)$ an operator and $\lambda \in \mathbb{K}$ such that $\lvert \lambda \rvert > r(T)$. Consider $r \in (r(T), \lvert \lambda \rvert )$. Then $r > r(T)$ and therefore there is an $n_0 \in \mathbb{N}$ such that $\|T^n \| \leq r^n $ for all $n \geq n_0$. Since $r < \lvert \lambda \rvert$, the series $\sum_{n = 0}^\infty \lambda^{- n - 1} T^n$ is absolutely convergent in the Banach space $L(E)$. Since the same assumptions are given in the proof you wrote in your question, it follows that your sum $\lambda^{p +1}R(\lambda, T)$ is absolutely convergent in $L(E)$. Now take $t \in (r(T), \infty)$ (note that $t$ is not just $>0$ as you wrote it, by the assumption on $\lambda$). It now follows easily by the Weierstraß-M-Test that the series $t^{p + 1} \mathrm e^{\mathrm i (p + 1) \theta} R(\mathrm e^{\mathrm i \theta}, T)$ is uniformly convergent (with respect to $\theta$) in $L(E)$. Here, you also have to use that the sum $\lambda^{p +1}R(\lambda, T)$ is convergent, which follows from absolute convergence in a Banach space. By the properties of the integral, interchanging the sum and the integral is then allowed.
