Consider the Laplace operator defined in the biggest possible subset of $L^2(\mathbb{R}^2)$ and let $z \in \mathbb{C}\backslash\mathbb{R}$. Therefore $z \notin \sigma (\Delta)$ the spectrum of $\Delta$, and the resolvent $R=(-\Delta - zI)^{-1}$ is well defined and bounded in all of $L^2(\mathbb{R}^2)$.


I'm trying to find the integral kernel for this operator, that is a function $K(x,y)$ such that almost everywhere in $\mathbb{R}^2$:


$$(Ru)(x)=\int\_{\mathbb{R}^2} u(y)K(x,y) dy$$


Let $f := ( - \Delta - z I)^{- 1} u$ and let $\mathcal{F}$ be the Fourier
transform. Now


$$ ( - \Delta - z I) f = u \Rightarrow \mathcal{F} ( ( - \Delta - z I) f)
 =\mathcal{F} ( u) \Rightarrow ( \xi^2 - z) \hat{f} ( \xi) = \hat{u} ( \xi)
 . $$


Solving for $\hat{f}$ and applying $\mathcal{F}^{- 1}$ we arrive at (modulo
some constant depending on your favourite definition of $\mathcal{F}$)


$$ f ( x) =\mathcal{F}^{- 1} \left( \frac{1}{\xi^2 - z} \hat{u} ( \xi)
 \right) =\mathcal{F}^{- 1} \left( \frac{1}{\xi^2 - z} \right) \ast u ( x)
$$


So I need to calculate


$$ \int\_{\mathbb{R}^2} \frac{e^{ix \cdot \xi}}{\xi^2 - z} d \xi $$


and I'm completely stuck. Does anybody have any ideas or references? Thanks.



Integral kernel for the resolvent of the laplace operator
At least when $\Re z < 0$, and assuming that you're taking $\Delta$ to be a negative operator (i.e., $-\Delta \geq 0$), you can write $(s-z)^{-1} = \int\_0^\infty e^{zt} e^{-st} dt$, so that by the functional calculus, you should be able to write your resolvent as $Ru = \int\_0^\infty e^{zt} e^{\Delta t}u dt$, and hence your integral kernel as
$$ K(x,y) = \int\_0^\infty e^{zt} K\_H(x,y;t) dt, $$ where $$K\_H(x,y;t) = \frac{1}{4\pi t}e^{-|x-y|^2/4t}$$ is the heat kernel for $\mathbb{R}^2$. Even if this sketch is complete nonsense, I still suspect that heat kernel methods might nonetheless be helpful, at least in the regime $\Re z < 0$.


EDIT: Helpful perhaps with regards to any estimates you might want--not so helpful with regards to evaluating your perfectly concrete integral...


