Compute Fourier transforms of homogeneous functions of the form,
$$
\frac{1}{|x|^{n+d}}P\_d(x)
$$
where $P\_d$ is a homogenous harmonic polynomial of degree $d$ in $n+1$ variables.



Fourier transforms of homogeneous functions
Your function is, with $P\_d$ homogeneous harmonic polynomial of degree $d$ in $n$ variables,
$$
u(x)=\frac{P\_d(x)}{\vert x\vert^{n+d}}.
\tag{1}
$$
This is an homogeneous distribution of degree $-n$ on $\mathbb R^n\backslash\{0\}$. The first thing to do is to extend that distribution to a distribution on $\mathbb R^n$. Let us first check for a test function $\phi\in C\_c^\infty(\mathbb R^n\backslash\{0\})$, the absolutely converging integral
\begin{multline}
\langle T, \phi\rangle=
\int\_{\mathbb R^n}\frac{P\_d(x)}{\vert x\vert^{n+d}}\phi(x) dx 
=\int\_0^{+\infty} r^{n-1}\int\_{\mathbb S^{n-1}}
\frac{P\_d(r\omega)}{r^{n+d}}\phi(r\omega)d\sigma(\omega) dr
\\=
\int\_{0}^{+\infty}\int\_{\mathbb S^{n-1}}
{P\_d(\omega)}\phi(r\omega)d\sigma(\omega)\frac{dr}r
\end{multline}


$\mathbf {\text{Claim:}}$ Let $\psi$ be a function in the Schwartz class of $\mathbb R$ and let $H=\mathbf 1\_{\mathbb R^+}$.
We have
$$\int\_0^{1}\frac{\psi(r)-\psi(0)}{r} dr
+\int\_1^{+\infty}\frac{\psi(r)}{r} dr
=
\langle \frac{d}{dr}\bigl(
H(r) \ln r
\bigr),\psi(r)\rangle\_{\mathscr S'(\mathbb R),\mathscr S(\mathbb R) }.
$$
Indeed we have,
\begin{align}
\langle \frac{d}{dr}\bigl(
H(r) \ln r
\bigr),\psi(r)\rangle\_{\mathscr S'(\mathbb R),\mathscr S(\mathbb R) } 
&=-\int\_0^{+\infty}\psi'(r)
\ln r\ dr
=\lim\_{\epsilon \rightarrow 0\_+}
-\int\_\epsilon^{+\infty} \psi'(r)
\ln r\ dr,
\\
&\text{and}\hskip25pt
\\
-\int\_\epsilon^{+\infty} \psi'(r)
\ln r\ dr&=\bigl[\psi(r)
\ln r\bigr]^{\epsilon}\_{+\infty}
+\int\_{\epsilon}^{+\infty}\psi(r) r^{-1} dr
\\&=\psi(\epsilon)
\ln \epsilon
+\int\_{\epsilon}^{+\infty}
\psi(r) r^{-1} dr
\\
&=
\int\_{1}^{+\infty}\psi(r) r^{-1} dr
+\int\_{\epsilon}^{1}\psi(r) r^{-1} dr
-\int\_{\epsilon}^{1}\psi(\epsilon)r^{-1} dr
\\
&=
\int\_{1}^{+\infty}\psi(r) r^{-1} dr
+\int\_{\epsilon}^{1}
\bigl[\psi(r) -\psi(\epsilon)\bigr]r^{-1} 
dr,
\end{align}
proving the claim.$\ \mathbf{\square}$


We may thus extend $T$ as a temperate distribution $\tilde T$
on $\mathbb R^n$ with the definition
for $\phi\in\mathscr S(\mathbb R^n
)$,
$$
\langle \tilde T, \phi\rangle\_{\mathscr S'(\mathbb R^n),\mathscr S(\mathbb R^n) } =
\int\_{\mathbb S^{n-1}}
{P\_d(\omega)}
\langle \frac{d}{dr}\bigl(
H(r) \ln r
\bigr),\phi(r\omega)\rangle\_{\mathscr S'(\mathbb R\_r),\mathscr S(\mathbb R\_r) } 
d\sigma(\omega).
\tag{2}$$
However the distribution $\tilde T$ fails to be homogeneous on $\mathbb R^n$, as it is seen when you calculate
$
\langle \tilde T(\lambda x), \phi(x)\rangle.
$
Note that an homogeneous distribution of degree $\mu$
is temperate and its Fourier transform is also homogeneous with degree $-\mu-n$.
That does not mean that you cannot calculate the Fourier transform of $\tilde T$. In particular Theorem 7.1.18 in the ALPDO of Lars Hörmander (Grundlehren 256) shows that
the Fourier transform of ${\tilde T}$ is smooth outside of the origin. It is also possible to calculate directly
$$
\langle \widehat{\tilde T}, \phi\rangle=\langle \tilde T, \hat \phi\rangle,
$$
when $\hat \phi(0)=\int \phi(x) dx =0$,
but in general you will need to use the Fourier transform of
the one-dimensional distribution
$\frac{d}{dr}\bigl(
H(r) \ln r
\bigr)$.


