In the free case one can compute the resolvents of the Laplacian $-\Delta$ in many cases explicitly, in the sense that they are given by an integral operator. Often, one uses the Hille-Yosida theorem or Fourier transform to do so. 


Also, the Schwartz-kernel theorem tells us that in a distributional sense there is always an "integral kernel" associated to the resolvent of a Schrodinger operator. 


However, examples in the free case suggest that the resolvent of $H:=-\frac{d^2}{dx^2} +V$ ( I am now in dimension $1$ as I would just like to get some conceptual insight) where $V$ is a $C^{\infty}$ and bounded potential with suitable boundary conditions such that this operator is self-adjoint, gives rise to a resolvent 


$$(H-z)^{-1}\psi(x) = \int K\_z(x,y)\psi(y) dy$$ where 
$K\_z$ has, for fixed $z$ in the resolvent set, nice properties (almost always smooth) in $x$ and $y$.


See for example this calculation: [of the one-dimensional resolvent](https://math.stackexchange.com/questions/1762074/evaluation-of-an-integral-associated-with-integral-kernel-of-resolvent-of-laplac)
or this one of the three-dimensional [on math.stackexchange](https://math.stackexchange.com/questions/325418/fourier-transform-of-integral-kernel-of-the-free-resolvent?rq=1)


I would like to ask why this is the case? Which theorem guarantees us that such nice $K\_z$ kernels exist?



Resolvents of Schrodinger operators
There is a general answer for any dimension $n\geq1$. Let $\Omega\subseteq \mathbb R^n$ denote the underlying domain.



> 
> For all $z$ in the resolvent set of $H$, $(H-z)^{-1}$ is a (classical) pseudodifferential operator of order $-2$, as $-\Delta+V$ is elliptic. Hence, its kernel $K\_z$ is a distribution on $\Omega\times\Omega$ which is **conormal with respect to the diagonal** $\Delta\_\Omega = \{(x,x)\mid x\in\Omega\}$. This implies that the wave front set of $K\_z$ is contained in the conormal bundle $N^\ast \Delta\_\Omega\setminus0 = \{(\xi,-\xi)\in T\_{(x,x)}^\*(\Omega\times\Omega)\mid x\in\Omega,\,$ $\xi\in T\_x^\*\Omega\setminus0\}$. In particular, its singular support is contained in (actually, it is equal to) $\Delta\_\Omega$.
> 
> 
> 


Further properties of the kernel $K\_z$ depend on the situation under consideration. For instance, if $\Omega$ is bounded and smooth and $V$ is smooth up to the boundary, then $(H-z)^{-1}$ belongs to Boutet de Monvel's calculus, see e.g. Gerd Grubb [Functional calculus of pseudodifferential boundary problems](http://www.springer.com/de/book/9780817637385). For $\Omega=\mathbb R^n$, there are (refined) pseudodifferential calculi adapted to the behavior of the potential $V$ as $|x|\to\infty$.




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A general reference for the first part of this answer is Chapter XVIII (especially, Sections 18.1 and 18.2) of Lars Hörmander [The analysis of linear partial differential operators III](http://www.springer.com/us/book/9783540499374).


