Fundamental solution to the Poisson equation by Fourier transform

The following argument works for d>3. From Fourier transform of $1/|x|^{\alpha}$. we know that if $f(x) = 1/|x|^{d-2}$ then $$\widehat f(x) = \frac{\pi^{(d-2)/2}}{\pi \Gamma((d-2)/2)}\frac{1}{x^2}.$$

Since $\Gamma(d/2) = ((d-2)/2)\Gamma((d-2)/2)$, we have $$\widehat f(x) = \frac{(d-2)\pi^{d/2}}{\pi^2 2\Gamma(d/2)x^2} = \frac{(d-2)\Omega_d}{4\pi^2 x^2 }.$$ Therefore, if we take the fundamental solution $$u(x) =\frac{1}{(2-d)\Omega_d |x|^{d-2}}, $$ we obtain $$\widehat u (\xi) = -\frac{1}{4\pi^2 \xi^2}.$$