Understanding a theorem 10.25 from Rudin comparison with method to solve differential equations. I'm going to be very informal... so i hope i'm going to make my point anyway... otherwise tell me what i should adjust in this question.
The theorem 10.25 of rudin - functional analysis states that in a banach algebra we can define a "rational" function as the following
$$R(x) = \frac{1}{2\pi i} \int_{\Gamma} R(\lambda)(\lambda e - x)^{-1} d\lambda$$
The integral i reported is specifically defined for banach algebra value (there's another question where i've asked an example of that), and $R(\lambda)$ is defined as
$$R(\lambda) = P(\lambda) + \sum_{m,k} c_{m,k}(\lambda - \alpha_{m})^{-k}$$
where $\alpha_m$ are poles of the rational function $R(\lambda)$. I was wandering if there's some kind of relationship between such integral and the methods to solve differential equation (that usually people learn in basic analysis courses). For example... if $$A = \frac{d^2}{dx^2} + 2\frac{d}{dx} + 5$$
To solve such equation we use the polynomial
$$p(\lambda) = \lambda^2 + 2 \lambda + 5$$.
Is there some relationships?
 A: Yes, the two topics can have something to do with each other, except that the operator $\frac{d}{dx}$ is not bounded. Consider the differential $\partial = \frac{d}{dx}$ on $C[0,1]$ with domain $\mathcal{D}(\partial)$ consisting of all continuously differentiable functions $f$ on $[0,1]$ for which $f(0)=0$. Then
$$
        \partial : \mathcal{D}(\partial)\subset C[0,1]\rightarrow C[0,1]
$$
has a resolvent $(\partial-\lambda I)^{-1}$, where $g=(\partial -\lambda I)^{-1}f$ is the unique solution of
$$
                             g'-\lambda g = f \\
                                g(0) = 0.
$$
The solution is found using an integrating factor of $e^{-\lambda t}$:
$$
         (\partial -\lambda I)^{-1}f = \int_{0}^{x}f(t)e^{\lambda(x-t)}dt.
$$
This is a strange example because $\partial$ on this domain has no spectrum!
Suppose you want to now solve
$$
           \left(\frac{d^{2}}{dx^{2}}+2\frac{d}{dx}+5\right)f=g,\\
                      f(0)=0,\;\; f'(0)=0.
$$
This is the same as $p(\partial)f=g$.
Write $p(\lambda)=(\lambda^{2}+2\lambda+5)=(\lambda+1+2i)(\lambda+1-2i)$, integrate around the roots of $p$, and see what you get
$$
\begin{align}
       &\frac{1}{2\pi i}\oint_{C}\frac{1}{p(\lambda)}(\lambda I-\partial)^{-1}gd\lambda \\
& =-\frac{1}{2\pi i}\oint_{C}\frac{1}{p(\lambda)}(\partial-\lambda I)^{-1}gd\lambda\\
  & = -\frac{1}{2\pi i}\oint_{C}\frac{1}{p(\lambda)}\int_{0}^{x}e^{-\lambda t}g(t)dt d\lambda \\
  & = -\int_{0}^{x}\left[\frac{1}{2\pi}\oint_{C}\frac{e^{\lambda(x-t)}}{p(\lambda)}d\lambda\right]g(t)dt
\end{align}
$$
This is an integral solution where the integral kernel is found by evaluating
$$
     \frac{1}{2\pi i}\oint_{C}\frac{e^{\lambda(x-t)}}{(\lambda-(-1-2i))(\lambda-(-1+2i))}d\lambda \\
   = \frac{e^{(-1-2i)(x-t)}}{-1-2i-(-1+2i)}+\frac{e^{(-1+2i)(x-t)}}{-1+2i-(-1-2i)} \\
   = -\frac{1}{4i}(e^{(1+2i)(t-x)}-e^{(2-i)(t-x)}) \\
   = -\frac{e^{(t-x)}}{2}\sin(2(t-x)).
$$
The final solution obtained using this method is
$$
          f(t)=\frac{1}{2}\int_{0}^{x}g(t)e^{(t-x)}\sin(2(t-x))dt.
$$
You can probably see that this method is going to give you the same thing as playing around with the operators directly:
$$
              (\partial^{2}+2\partial+5I)f=g.
$$
A functional calculus such as this turns operator manipulations into algebraic manipulations on fractions/polynomials in $\lambda$ because manipulating the algebraic expressions in the integrand becomes equivalent to manipulating the operators outside the integral.
