How is obtained the following representation for the heat semigroup? How is obtained the following representation for the heat semigroup?
$$e^{t \Delta}=\dfrac{1}{2 \pi i} \int_{\Gamma} e^{t \lambda} (\lambda I - \Delta)^{-1} d \lambda$$
where $\Gamma$ is a contour around the spectrum (the negative real axis),
which is equivalent with
$$k_t(x,y)=\dfrac{1}{2 \pi i} \int_{\Gamma} e^{t \lambda} G(x,y,\lambda) d \lambda,$$
where $k_t$ is the heat kernel and $G$ is the Green function, the kernel of the resolvent $(\lambda I - \Delta)^{-1}$.
They say that by operational calculus, but I don't know how. 
I found in Dunford and Schwartz, Linear operators, Part I, page 601, Th.4 (for unbounded closed operators) that
$$f(T)=\dfrac{1}{2 \pi i} \int_{\Gamma}f(\lambda)R(\lambda;T) d \lambda + f(\infty)I,$$
if f is analytic at infinity, which is not in my case. 
So I think that this is not the result used. What result is used? I couldn't figure it out.
Thank you.
 A: Though this was raised long time ago I still try to answer it...
I suppose $\Delta=\partial^2/\partial x^2$ is the Laplacian, say in one dimension for simplicity. The heat equation reads
$$
\partial_t u(x,t) = \partial_x^2 u(x,t),\qquad x\in\mathbb{R},\quad t>0.
$$
The $C_0$-semigroup associated with this PDE is $T=\{T(t)=e^{t\Delta}\mid t\in[0,\infty)\}$. The resolvent operator is the Laplace transform of the time evolution operator $T(t)=e^{t\Delta}$:
$$
(\lambda-\Delta)^{-1}=\int_0^\infty dt\,e^{-\lambda t}e^{t\Delta},
$$
where $\lambda$ is from the resolvent set $\rho(\Delta)$, that is any complex number excluding the negative reals (i.e. the spectrum of the Laplacian, $\sigma(\Delta)$). The representation formula in your question is just the inverse Laplace transform of the above
$$
e^{t\Delta}=\int_\Gamma\frac{d\lambda}{2\pi i}\frac{e^{\lambda t}}{\lambda-\Delta},\quad t\ge0,
$$
where $\Gamma$ is the Bromwich contour which we can deform such that it encircles the negative real axis in a counter-clockwise fashion. This is an operator equation. If we apply this equation (both sides) on the delta function we get the representation of the heat kernel (fundamental solution of the heat equation) with the Green's function
$$
k_t(x,x_0)\equiv e^{t\Delta}\delta(x-x_0)=
\int_\Gamma\frac{d\lambda}{2\pi i}\frac{e^{\lambda t}}{\lambda-\Delta}
\delta(x-x_0)
=\int_\Gamma\frac{d\lambda}{2\pi i}e^{\lambda t}G(x,x_0,\lambda).
$$
For the one dimensional heat equation you can indeed show that the heat kernel
$$
k_t(x,x_0)=\frac{1}{\sqrt{4\pi t}}e^{-(x-x_0)^2/(4t)},\quad t>0
$$
and the Green's function
$$
G(x,x_0,\lambda)=\frac{1}{2\sqrt{\lambda}}e^{-\sqrt{\lambda}|x-x_0|},
\quad\lambda\in\rho(\Delta)
$$
are indeed the Laplace transforms of each other.
