I do not follow the proof of this Theorem
Theorem Suppose
$R(\lambda) = P(\lambda) + \sum_{m,k}c_{m,k}(\lambda - \alpha_m)^{-k}$
is a rational function with poles at the points $\alpha_m$. ($P$ is a polynomial, and the given sum has only finitely many terms). If $x \in A$ and if $\sigma(x)$ contains no pole of $R$, define
$R(x) = P(x) + \sum_{m,k}c_{m,k}(\lambda - \alpha_m e)^{-k}$.
If $\Omega$ is an open set in $\mathbb{C}$ that contains $\sigma(x)$ and in which $R$ is holomorphic, and if $\Gamma$ surrounds $\sigma(x)$ in $\Omega$, then
$R(x) = \frac{1}{2\pi i}\int_\Gamma R(\lambda)(\lambda e - x)^{-1}d\lambda$.
The proof simple states that we only have to apply the following lemma:
Lemma Suppose $A$ is a Banach algebra, $x \in A$, $\alpha \in \mathbb{C}$, $\alpha \not\in \sigma(x)$, $\Omega$ is the complement of $\alpha$ in $\mathbb{C}$ and $\Gamma$ surrounds $\sigma(x)$. Then
$\frac{1}{2\pi i}\int_\Gamma (\alpha - \lambda)^n(\lambda e - x)^{-1}d\lambda = (\alpha e - x)^{-1}$ for $n = 0, \pm 1, \pm 2, \dots$
Hence it should be quite simple, but I can't remember the necessary theory from complex analysis. We want
$P(x) + \sum_{m,k}c_{m,k} (x-\alpha_m e)^{-k} = \frac{1}{2\pi i}\int_\Gamma\left(P(\lambda)+\sum_{m,k}c_{m,k}(\lambda - \alpha_m)^{-k}\right)(\lambda e-x)^{-1}d\lambda$.
The RHS should be something like
$\frac{1}{2\pi i}\int_\Gamma \left(\beta\prod_{j=1}^n(\lambda-\alpha_j)\right)(\lambda e - x)^{-1}d\lambda + \sum_{m,k}c_{m,k}\frac{1}{2\pi i}\int_\Gamma (\lambda-\alpha_m)^{-k}(\lambda e - x)^{-1}d\lambda$
For the latter term, I am not sure how to use the assumption stated in the theorem to justify the use of the lemma for each $\alpha_m$. I am guessing that a correct argument should yield
$\sum_{m,k}c_{m,k}\frac{1}{2\pi i}\int_\Gamma (\lambda-\alpha_m)^{-k}(\lambda e - x)^{-1}d\lambda = \sum_{m,k}c_{m,k}(x-\alpha_m e)^{-k}$
For the first term, I am assuming something like the following
$\frac{1}{2\pi i}\int_\Gamma \left(\beta\prod_{j=1}^n(\lambda-\alpha_j)\right)(\lambda e - x)^{-1}d\lambda = \beta\prod_{j=1}^n \frac{1}{2\pi i} \int_\Gamma (\lambda-\alpha_j)(\lambda e - x)^{-1}d\lambda = \beta\prod_{j=1}^n(x-\alpha_j) =P(x)$.
I am guessing that the arguments have something to do with the poles and the holomorphism of $R$ but it's been a couple of years since I dealt with complex analysis and I do not have any books on the subject at hand. I would appreciate greatly if something could tell me if my reasoning is in the right direction, and if so, how one should use the given assumptions to argue for the application of the Lemma.