TITLE: Integral kernel for the resolvent of the laplace operator
QUESTION [7 upvotes]: 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.

REPLY [2 votes]: I finally found the solution in Evans' book on Partial Differential equations,
Chapt. 4.3. Using that
$$ \frac{1}{\xi^2 - z} = \int_0^{\infty} e^{- t ( \xi^2 - z)} d t, $$
and substituting into
$$ \int_{\mathbb{R}^2}  \frac{e^{ix \cdot \xi}}{\xi^2 - z} d \xi $$
then using Fubini, completing the square in the exponential function and
evaluating a complex integral along a line $ \{ Im = const. \} $, one arrives at the solution.
Thanks everybody for the fine answers.