Heat operator formalism via spectral projections and Dirac measure I am currently reading very helpful notes on the Heat kernel $p(t,x,y)$ on a Riemannian manifold $M$ -- there is one aspect though that I am not sure I understand notationally, it says that we can think of the heat kernel formally as 
$$
p(t,x,y) = \int_0^\infty e^{-t\lambda} \,dE_\lambda(\delta_x,\delta_y)
$$
where the $E_\lambda$ denote spectral projection operators and $\delta_x$ is denotes the Dirac delta function based at $x \in M$ (and likewise for $\delta_y$).
I am more familiar with the notation that is used in the holomorphic functional calculus, here one defines the heat operator
$$
e^{ - t\Delta} = \frac{i}{2\pi} \int_\Gamma e^{-t\lambda}(\Delta - \lambda)^{-1} \;d\lambda
$$
where $\Gamma$ is an "infinite" contour that contains the positive half real line (thus inculding the spectrum of the Laplacian). The Schwartz kernel $k(t,x,y)$ of this operator is defined by the equation
$$
e^{-t\Delta}f(x) = \int_M k(t,x,y)\,f(y) d\mu(y)
$$
and a little reformulation shows that 
$$
k(t,x,y) = \frac{i}{2\pi} \int_\Gamma e^{-t\lambda} \kappa(x,y,\lambda) d\lambda
$$
where $\kappa(x,y,\lambda)$ denotes the Schwartz kernel of the Resolvent $(\Delta - \lambda)^{-1}$. I suspect that the latter can be connected formally with the kernel given in the first line -- though I am not sure how to handle the object (measure ?) $dE_\lambda(\delta_x,\delta_y)$ correctly. 
For example, how can I act on a function with the right hand side above so that 
$$
u(t,x) = \int_M p(t,x,y)f(y) \,d\mu(y) \overset{?}{=}  \left(\int_0^\infty e^{-t\lambda} E_\lambda(\delta_x,\delta_y) \right)(f) 
$$
Many thanks for your help!
 A: It is perfectly rigorous to think of the Spectral Integral representation of the resolvent of a semibounded operator such as $-\Delta$ as a limiting form of the holomorphic functional calculus where the contour is allowed to approach the real line:
$$
         (-\Delta-\lambda I)^{-1}x = \frac{1}{2\pi i}\int_{\Gamma}(-\Delta-z I)^{-1}xdz
$$
The procedure is the same as when you collapse a contour onto a branch cut. The convergence that takes place is only in the strong operator topology, which is the reason I inserted a vector $x$ in your equation. This is Stone's formula for the spectral measure $E$:
$$
      E[a,b]x = \lim_{\epsilon\downarrow 0}\frac{1}{2\pi i}\int_{a}^{b}\{(-\Delta-(u+i\epsilon)I)^{-1}-(-\Delta-(u-i\epsilon)I)^{-1}\}x\, du
$$
If $a$ and/or $b$ happen to be an eigenvalue of $-\Delta$, then the left side above is not fully correct, but must be replaced with an expression that picks up only half of the residue mass at $a$ and $b$, but all of the mass in $(a,b)$:
$$
           E(a,b)x + \frac{1}{2}E\{a\}x +\frac{1}{2}E\{ b\}x.
$$
If $[a,b]$ is contained in the resolvent of $-\Delta$, then the above obviously results in $E[a,b]=0$. The end result is that, if $-\Delta \ge 0$, the resolvent may be written as
$$
               (-\Delta-\lambda I)^{-1}x = \int_{0}^{\infty}\frac{1}{t-\lambda}dE(t)x.
$$
One can show that the measure $E$ is a selfadjoint projection-valued measure. From this, one can show that the $C^{0}$ semigroup with generator $-\Delta$ (which is the heat operator) may be written as
$$
             e^{-t(-\Delta)}x = \int_{0}^{\infty}e^{-t\lambda} dE(\lambda)x
$$
Using your kernel representation for the resolvent, you'll be able to find $E$, and to connect it with what you're doing in order to find a kernel representation. I prefer your approach used in conjunction with Stone's formula to get the job done through what you know about the resolvent kernel. The spectral integral formulation is one way to eliminate sloppy delta function types of arguments; to inject $\delta_{x}$ into such a formulation seems particularly strange to me in light of that.
