Trouble with fundamental solution of Laplace equation via Fourier transform I'm trying to find fundamental solution of Laplace equation, a.k.a. function $\Phi:\mathbb{R}^n\to\mathbb{R}$ (or, more properly, a distribution on $\mathbb{R}^n$) such that $-\Delta\Phi=\delta_0$ in the sense of distribution. (I'm doing this as part of preparation for my final exams next week, but I didn't manage to find any material that explained what I need to do properly.) What I tried to do was to simply to take a Fourier transform. Then I get $$1=\hat{\delta_0}=\widehat{-\Delta\Phi}=\sum x_i^2\hat{\Phi},$$ or $\hat{\Phi}(x)=\frac1{|x|^2}$. Than from the inverse Fourier transformation I get $$\Phi(x)=\int_{\mathbb{R}^n}\hat{\Phi}(t)e^{i t\cdot x}\,\mathrm{d}t=\int_{\mathbb{R}^n}\frac{e^{i t\cdot x}}{|t|^2}\,\mathrm{d}t.$$ But the last integral doesn't converge. And even if it did (which I'm fairly certain is not the case), I'd have no idea how to transform it into "reasonable" form.
I'm aware that the answer I'm looking for is $\Phi(x)=C_n|x|^{2-n}$ for $n\ne2$ and $\Phi(x)=C_2\ln|x|$ for $n=2$. And I did find other ways to prove that. My problem is, that I don't understand where is the problem in this approach - and that probably means there is some fundamental detail that I just don't understand properly. Maybe it's the proper definiton of fundamental solution, maybe some operation is not allowed, maybe it's actually all correct and I just don't understand that it is actually correct. I don't know. Please, help me find what it is.
 A: The Fourier transform of tempered distributions is defined by duality. For any test function $\varphi$
$$
\langle \hat{f},\varphi\rangle = \langle f,\hat{\varphi}\rangle
$$
Here, $f(x) = 1/|x|^2$ is a tempered distribution (if $n≥3$), so it has a Fourier transform in this sense (and actually, since it is locally integrable, $\langle f,\hat{\varphi}\rangle = \int_{\mathbb{R}^d} f\,\hat{\varphi}$). If you just know distribution theory but not tempered distributions, the latter are just a subspace of distributions: the space of test functions is the space of $C^\infty$ functions but decaying faster to $0$ than any power function at infinity. Most of the formulas concerning the Fourier transform still work with this definition, but not of course the definition as an integral.
With this definition, you get the result you want, see for example there $f(x) = 1 / \lvert x \rvert^2$, $x\in \mathbb{R}^3$ , for the Fourier transform F, prove by scaling: $ F(f) (y) = C \frac{1}{\lvert y\rvert}. $ in the case of the dimension $3$ (or the Book by Lieb and Loss, Functional Analysis). The same methods works for $n>3$.
In the case of dimension $2$, the situation is more difficult since the function there is not locally integrable, and this is the critical case where a $\log$ appears, the computation is done in my answer here The Fourier transform of $1/p^3$ (just take $d=2$).
In the case of the dimension $1$, see What is the Fourier transform of $|x|$?. In this case, as in dimension $2$, one has to first define what is the tempered distribution associated to $1/|x|^2$ since it is not locally integrable.
