# Fourier transform of $\log|x|$ in $\mathbb{R}^2$ and the solution to $\Delta u = \delta_0$

In my analysis class, we are covering tempered distributions now. I was given this two-part question in Fourier transforms of distributions.

a. We are asked to compute the Fourier transform of $$\log |x|$$ as a tempered distribution in $$\mathbb{R}^2$$. Here, $$|x|$$ is the 2d Euclidean norm of the vector $$x$$.

b. We are asked to find the fundamental solution to the Laplace equation in $$\mathbb{R}^2$$, $$\Delta u = \delta_0$$, where $$u$$ is a distribution and understood in the sense of weak solutions.

Here, we take the Laplacian $$\Delta u = -\frac{\partial^2 u}{\partial x^2}-\frac{\partial^2 u}{\partial y^2}$$. We take the Schwartz space of functions $$S(\mathbb{R}^2)$$ and its continuous dual space, $$S'(\mathbb{R}^2)$$, the space of tempered distributions. For $$T \in S'(\mathbb{R}^2)$$ and $$\phi \in S(\mathbb{R}^2)$$, we use the notation $$\langle T,\phi \rangle = T(\phi)$$ and when $$T(x)$$ is a function, we define $$\langle T,\phi \rangle = \int_{\mathbb{R}^2} T(x)\phi(x)dx$$ For the Foruier transform on tempered distributions, we define $$\langle \mathcal{F}T,\phi \rangle = \langle T,\mathcal{F}\phi \rangle.$$ To be honest, I have no idea how to compute the Fourier transform of $$\log|x|$$ in $$\mathbb{R}^2$$ and how to use it to find the fundamental solution in part b, I do think I need to move to the Fourier domain in b but other than that I am lost. I thank anyone who can help with parts A and B.

The most coherent/memorable approach I know is not the most direct: first, compute that $$\Delta |x|^s=s(s+n-1)\cdot |x|^{s-2}$$ on $$\mathbb R^n$$. One can also check that the residue of the meromorphic family of tempered distributions $$s\to |x|^s$$ at $$s=-n$$ is an explicit constant multiple of $$\delta$$. Thus, for $$n\not=2$$, the limit as $$s\to -n$$ shows that $$\Delta |x|^{2-n}$$ is a constant multiple of $$\delta$$.
The squared factor $$s^2$$ disrupts this, moving the residue to the degree-one term in the (distribution-valued) Laurent expansion, but differentiating in $$s$$ gives reduces the degree. Thus, $$\Delta(\log|x|\cdot |x|^s) \;=\; \Delta {\partial\over \partial s}|x|^s \;=\; {\partial\over \partial s}\Delta |x|^s \;=\; {\partial\over \partial s} s^2|x|^{s-2}$$ Taking the limit as $$s\to 0$$ gives $$\Delta \log|x|$$ in the first term, and captures the residue of $$|x|^s$$ in the last. Namely, $$\Delta \log|x|=2\pi\cdot \delta$$.
Taking Fourier transform of this shows that $$|x|^2\cdot \widehat{\log|x|}$$ is a constant multiple of $$1$$. Since $$1/|x|^2$$ is not locally integrable in $$\mathbb R^2$$, we cannot say that the Fourier transform of $$\log|x|$$ is (a constant multiple of) $$1/|x|^2$$...