Cauchy integral type formula for self-adjoint operator Let $\Gamma$ be a differentiable Jordan curve in the resolvent set $\rho(A)$ of the self-adjoint operator $A$. How does one show
$\chi_\Omega(A) = \int_\Gamma R_A(z) dz$, where $\Omega = Int \Gamma \cap \mathbb R$ and $R_A(z)$ is the resolvent of $A$ at $z$.
Of course, this looks like Cauchy's integral formula (isnt there a $1/2\pi i$-factor missing?). The RHS is a Banach-space valued integral (the operators $R_A(z)$ are bounded operators) and somewhere i've read that function theory easily extends to the setting of general Banach algebras, but i dont feel confident enough to argue why the above formula holds. How could one use the functional calculus from the spectral theorem (for self-adjoint operators) here?
 A: The spectral theorem gives
$$
          (\lambda I-A)^{-1} =\int_{\sigma(A)}\frac{1}{\lambda -t}dE(t),
  \;\;\; \lambda\in\rho(A).
$$
If $C$ is a simple closed piecewise smooth positively oriented contour in $\rho(A)$ that encloses part of the spectrum, then $C$ crosses the real axis where the spectral measure is $0$ in a neighborhood of that crossing. So the following integrals make sense
\begin{align}
     \left(\frac{1}{2\pi i}\oint_{C}(\lambda I-A)^{-1} d\lambda\,x,y\right)
      & = \frac{1}{2\pi i}\oint_{C}((\lambda I-A)^{-1}x,y)d\lambda \\
      & = \frac{1}{2\pi i}\oint_{C}\int_{\sigma}\frac{1}{\lambda -t}d(E(t)x,y)d\lambda \\
      & = \int_{\sigma}\left(\frac{1}{2\pi i}\oint_{C}\frac{1}{\lambda-t}d\lambda \right)d(E(t)x,y).
\end{align}
The complex measure $d(E(t)x,y)$ is supported in $\sigma$, which is a positive distance from where $C$ intersects the real axis; and the inner integral is $1$ for $t\in\sigma$ inside $C$ and is $0$ for $t\in\sigma$ and outside $C$. If $\mbox{Int}(C)$ denotes the interior of $C$, then the above becomes
$$
         \left(\frac{1}{2\pi i}\oint_{C}(\lambda I-A)^{-1} d\lambda\,x,y\right)
    = (E(\mbox{Int}(C))x,y),\;\;\; x,y \in X,
$$
Because this holds for all $x,y \in X$,
$$
      \frac{1}{2\pi i}\oint_{C}(\lambda I-A)^{-1}d\lambda = E(\mbox{Int}(C)).
$$
