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 $$\widehat u (\xi) = -\frac{1}{4\pi^2 \xi^2}.$$

Just to put everything in the right place, since Hugo uses slightly different conventions. Here If $\Delta u(x)=\delta(x)$ then $u(x)=C |x|^{2-d}$ via Fourier transform? it is shown that if $$ f(x)={C_d}|x|^{1-d} $$ then $$ \widehat f({\xi})=-\frac{1}{|\xi|^2}, $$ which is what we want by the above calculation. Now, to fix the constant, one uses the identity $$ \int_{\mathbb R^d}f(x) \widehat g(x) dx = \int_{\mathbb R^d}\widehat f(\xi) g(\xi) d\xi; $$ choosing conveniently $$ g(x)=e^{-x^2/2}, $$ we know that $$ \widehat g(\xi) = \int_{\mathbb R^d} e^{-i\xi\cdot x}g(x) dx= (2\pi)^{d/2}e^{-\xi^2/2}. $$ So $$ \int_{\mathbb R^n}C_d |x|^{2-d}(2\pi)^{d/2}e^{-x^2/2} dx = -\int_{\mathbb R^d}\xi^{-2} e^{-\xi^2/2} d\xi. $$ Switching to polar coordinates \begin{align} (2\pi)^{d/2}C_d\int_{0}^{+\infty}\rho e^{-\rho^2/2}d\rho &= - \int_0^{+\infty}\rho^{d-3}e^{-\rho^2/2}d\rho \\ (2\pi)^{d/2}C_d\int_{0}^{+\infty} e^{-s/2}ds &= - \int_0^{+\infty}s^{(d-2)/2}e^{-s/2} ds\\ 2(2\pi)^{d/2}C_d &= - 2^{d/2}\int_0^{+\infty}t^{(d-4)/2}e^{-t} dt\\ 2(2\pi)^{d/2}C_d &= - 2^{d/2-1}\int_0^{+\infty}t^{(d-4)/2}e^{-t}\\ C_d &= - \frac{\Gamma\left(\frac{d-2}{2} \right)}{4\pi^{d/2}}\\ &=\frac{\Gamma(d/2)}{2\pi^{d/2}(2-d)}. \end{align}