The fundamental solution (or Green function) for the Laplace operator in $d$ space dimensions $$\Delta u(x)=\delta(x),$$ where $\Delta \equiv \sum_{i=1}^d \partial^2_i$, is given by $$ u(x)=\begin{cases} \dfrac{1}{(2-d)\Omega_d}|x|^{2-d}\text{ for } d=1,3,4,5,\ldots\\ \dfrac{1}{2\pi}\log|x| \ \ \ \ \ \ \ \ \ \ \ \ \text{for } d=2, \end{cases} $$ where $\Omega_d$ is the $d$-dimensional solid angle $ \Omega_d=2\pi^{d/2}/\Gamma(d/2). $ Indeed, it can be shown directly that for a test function $\varphi$: $$ \int d^dx\,u(x)\Delta\varphi(x)=\varphi(0). $$ I tried to obtain the same result by Fourier transforming the distributional equation $ \Delta u(x)=\delta(x) $ thus $$ -k^2 \widehat u(k)=1\implies u(y)=-\frac{1}{(2\pi)^d}\int\frac{e^{ik\cdot y}}{k^2}d^dk. $$ Now, using $d$ dimensional polar coordinates, we can write $$ u(y)=-\frac{1}{(2\pi)^d}\int_0^{2\pi}d\phi\int_0^{\pi}d\theta_1\int_0^{\pi}d\theta_2\ldots\int_0^{\pi}d\theta_{d-2}\\ \sin\theta_1\sin^2\theta_2\ldots\sin^{d-2}\theta_{d-2}\int_0^\infty{dq}\,q^{d-1}\frac{e^{iq|y|\cos\theta_1}}{q^2} $$ by an appropriate choice of the axis along $y$. Since the integrand depends only on the angle $\theta_1$, we can use: $$ \Omega_d=\int_0^{2\pi}d\phi\int_0^{\pi}d\theta_1\int_0^{\pi}d\theta_2\ldots\int_0^{\pi}d\theta_{d-2} \sin\theta_1\sin^2\theta_2\ldots\sin^{d-2}\theta_{d-2} $$ $$ \implies \int_0^{2\pi}d\phi\int_0^{\pi}d\theta_2\ldots\int_0^{\pi}d\theta_{d-2} \sin^2\theta_2\ldots\sin^{d-2}\theta_{d-2}=\frac{\Omega_d}{2}; $$ $$ u(y)=-\frac{\Omega_d|y|^{2-d}}{2(2\pi)^d}\int_0^{\infty}dl\,l^{d-3}\int_0^{\pi}d\theta\, \sin\theta\, e^{il\cos\theta}=-\frac{\Omega_d |y|^{2-d}}{(2\pi)^{d}}\int_0^\infty dl\,l^{d-4}\, \sin l. $$ The integral in the last line can be manipulated as follows: for $2<\Re e(d)<4$ Wick rotation yields $$ \int_0^\infty dx\, x^{d-4}\, \sin x = \Im m \int_0^\infty dx\, x^{d-4}\, e^{ix}=\Im m \int_0^\infty idy\, (iy)^{d-4}\, e^{-y}=\Im m \left(i e^{id\pi/2}\right)\Gamma(d-3)=\Gamma(d-3)\cos\frac{\pi d}{2}. $$ Of course this must be justified via an appropriate contour integration: see Proof that $ \int_0^\infty x^{d-4}\sin x\, dx = \cos \frac{\pi d}{2} \Gamma(d-3)$, for $2<\Re(d)<4 $?
Now for $2<\Re e(d)<4$ we get: $$ u(x)=-\frac{\Omega_d|x|^{2-d}}{(2\pi)^d}\Gamma(d-3)\cos\frac{\pi d}{2}, $$ which reduces to the familiar $$ -\frac{1}{4\pi|x|} $$ for $d=3+\varepsilon$ as $\varepsilon\to0.$ It even yields $$ \frac{1}{2\pi}\ln|x| $$ for $d=2+\varepsilon$ as $\varepsilon\to0$ if one neglects an $x$-independent pole in $\varepsilon$.
The two expressions we have for $u(x)$ are not trivially coincident in the strip where the latter is defined, and in fact Wolframalpha shows that the coefficients are equal $$ \frac{\Gamma(d/2)}{2 \pi^{d/2}(2-d)}=-\frac{2^{1-d} \Gamma(d-3) \cos(\pi d /2)}{\pi^{d/2}\Gamma(d/2)} $$ in the interval $2\le d \le 4$ only for $d=2$ and $d=3$. So there is no point in trying to show that their expressions coincide in the whole strip. However, is there a way to recover the above expression, which holds for any dimension $d=1,2,3,\ldots$ from the Fourier transform calculation?
The really tricky thing is just the coefficient, though: If $\Delta u(x)=\delta(x)$ then $u(x)=C |x|^{2-d}$ via Fourier transform?
