Newtonian off-diagonal kernel of the transverse projector

For $\epsilon >0$, $m_{\epsilon }(\mathbf{k})=e^{-\epsilon \mathrm{\mid }\mathbf{k}\mathrm{\mid }}\mathrm{/}\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^2$ is integrable: [F3] and $\sigma (S^2)=4\pi$ give $\int \mathrm{\mid }{m}_{\epsilon }\mathrm{\mid }{d}^3\mathbf{k}=4\pi {\int }_0^{\mathrm{\infty }}{e}^{-\epsilon s}ds<\mathrm{\infty }$. Its inverse Fourier integral is radial. Rotate the polar axis to $\mathbf{x}$ using [F4], and apply the zonal identity separately to the real and imaginary parts of $u\mapsto e^{is\mathrm{\mid }\mathbf{x}\mathrm{\mid }u}$. The polar formula [F3] is stated for nonnegative Borel functions; since this complex integrand is absolutely integrable, apply it separately to the positive and negative parts of its real and imaginary parts and recombine by [F5]. The angular integral is $4\pi \sin (sr)\mathrm{/}(sr)$, where $r=\mathrm{\mid }\mathbf{x}\mathrm{\mid }$. Thus $G_{\epsilon }(r)=(2\pi {)}^{-3}\int e^{i\mathbf{k}\cdot \mathbf{x}}{m}_{\epsilon }(\mathbf{k})d^3\mathbf{k}=\frac{1}{2{\pi }^{2}r}{\int }_0^{\mathrm{\infty }}{e}^{-\epsilon s}\frac{\sin (sr)}{s}ds\mathrm{.}$ Since $\sin (sr)\mathrm{/}s={\int }_0^r\cos (st)dt$ and ${\int }_0^r{\int }_0^{\mathrm{\infty }}{e}^{-\epsilon s}\mathrm{\mid }\cos (st)\mathrm{\mid }dsdt\le r\mathrm{/}\epsilon$, Fubini [F7] applies. The real part of ${\int }_0^{\mathrm{\infty }}{e}^{-(\epsilon -it)s}ds=(\epsilon -it{)}^{-1}$ is $\epsilon \mathrm{/}({\epsilon }^2+t^2)$; integrating it over $t\in [0,r]$ gives $\arctan (r\mathrm{/}\epsilon )$. Therefore $G_{\epsilon }(r)=\frac{\arctan (r\mathrm{/}\epsilon )}{2{\pi }^{2}r}\mathrm{.}$

For every test $\phi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^3)$ [F8], the bound $\mathrm{\mid }{G}_{\epsilon }(\mathbf{x})\mathrm{\mid }\le (4\pi \mathrm{\mid }\mathbf{x}\mathrm{\mid }{)}^{-1}$ and [F3] make $G_{\epsilon }\phi$ dominated by an integrable function; [F5] gives the distributional limit $G_{\epsilon }\to G$. To identify the same limit in Fourier variables, repeated integration by parts in a coordinate with $\mathrm{\mid }{k}_j\mathrm{\mid }\ge \mathrm{\mid }\mathbf{k}\mathrm{\mid }\mathrm{/}\sqrt{3}$ gives $\mathrm{\mid }\widehat{\phi }(\mathbf{k})\mathrm{\mid }\le C_N(1+\mathrm{\mid }\mathbf{k}\mathrm{\mid }{)}^{-N}$ for every integer $N$, while $\mathrm{\mid }\widehat{\phi }\mathrm{\mid }\le \mathrm{\parallel }\phi {\mathrm{\parallel }}_1$ near zero. Hence $\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^{-2}\widehat{\phi }(-\mathbf{k})$ is integrable by [F3]. Fubini [F7] identifies $\int G_{\epsilon }\phi$ with $(2\pi {)}^{-3}\int m_{\epsilon }(\mathbf{k})\widehat{\phi }(-\mathbf{k})d^3\mathbf{k}$. Since $\mathrm{\mid }{m}_{\epsilon }\mathrm{\mid }\le \mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^{-2}$, [F5] shows this tends to the inverse Fourier pairing with $\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^{-2}$. Both limits are therefore the same distribution $G$, proving ${\mathcal{F}}^{-1}(\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^{-2})=G$. By [F1] this is the published normalized fundamental solution.

Integration by parts against compact smooth tests gives Fourier multiplier $-k_i{k}_j$ for ${\mathrm{\partial }}_i{\mathrm{\partial }}_{j}G$. Step 2.1 therefore gives ${\mathcal{F}}^{-1}\text{\hspace{0.17em}⁣}({\delta }_{ij}-\frac{k_i{k}_j}{\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^2})={\delta }_{ij}{\delta }^3+{\mathrm{\partial }}_i{\mathrm{\partial }}_{j}G\mathrm{.}$ For $\mathbf{x}\mathrm{\ne }0$, put $s=\mathrm{\mid }\mathbf{x}{\mathrm{\mid }}^2$ by [F6], so $G=(4\pi {)}^{-1}{s}^{-1\mathrm{/}2}$. The power, chain and sum rules [F7] give ${\mathrm{\partial }}_{i}G=-\frac{x_i}{4\pi s^{3\mathrm{/}2}},{\mathrm{\partial }}_i{\mathrm{\partial }}_{j}G=-\frac{{\delta }_{ij}}{4\pi r^3}+\frac{3x_i{x}_j}{4\pi r^5}\mathrm{.}$ Substitution into [F2] proves the stated equal-time commutator kernel.