Given the equation $(1-\Delta)u=f$ for $f \in S(\mathbb{R}^n)$ (rapidly decreasing functions) we get by taking the Fourier transform that 


$u = \left(\frac{1}{2\pi}\right)^{\frac{n}{2}}\mathcal{F}^{-1}\left(\frac{1}{1+|.|^2} \right)\*f$


This is all well-defined in the sense of $S'(\mathbb{R}^n).$


Obviously, for $n \le 3$ the function $\frac{1}{1+|.|^2}$ is in $L^2$ so it is meaningful to write 


$u$ as $u(x) = \int\_{\mathbb{R}^n } K(x-y)f(y) dy$ for some integral kernel defined via the inverse Fourier transform, but is there also a way to make sense out of this integral kernel representation for $n>3$?- Or if not, how can I see that there is no such kernel in higher dimensions?



Solution to inhomogenous PDE
This kernel is called the [Bessel potential](https://en.wikipedia.org/wiki/Bessel_potential#Representation_in_Fourier_space). It is smooth away from $0$, and in your case, this kernel is locally integrable.


