Fundamental solution to the Poisson equation by Fourier transform 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?
 A: 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
$$\widehat u (\xi) = -\frac{1}{4\pi^2 \xi^2}.$$
A: 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}
