Fourier transform of $|x|^{-m}$ for $m\geq n$ in $\mathbb R^n$ According to this question the Fourier transform of $|x|^{-m}$ for $x\in\mathbb R^n$ and $0<m<n$ is given by 
$$
\mathcal F[|x|^{-m}](\xi)=C(n,m)|\xi|^{m-n} \tag{1}
$$
for some constant depending on $m$ and $n$.
My question: How do you calculate the Fourier transform of $|x|^{-m}$ for $m\geq n$?
For $m\in\mathbb R^+\setminus\mathbb N$ the formula (1) should still hold by analytic continuation. How do we get the case $m\in\mathbb N_{\geq n}$?
Edit: Definition: Let $\mathcal F[f(x)](\xi)=(2\pi)^{-n/2}\int_{\mathbb R^n} f(x)\exp(ix\cdot\xi) dx$ be the Fourier transform of a Schwartz function $f$.
 A: By definition of Fourier transform (with your normaliztion)
$$
F(\xi):=\mathcal{F}[|x|^{-m}](\xi)=\frac{1}{(2\pi)^{n/2}}\int |x|^{-m }e^{ix\cdot \xi} d^nx
$$
This integral, however, has a lot of problems. So instead let's work with a closely related (convergent) integral:
$$
\begin{aligned}
G(\xi;K,\epsilon)=\frac{1}{(2\pi)^{\frac{n}{2}}} \int_{|x|<\frac{1}{K}} |x|^{-m }(e^{ix\cdot \xi}-1) d^nx\\+
\frac{1}{(2\pi)^{\frac{n}{2}}} \int_{|x|\geq \frac{1}{K}} |x|^{-m }e^{ix\cdot \xi}e^{-\epsilon |x|} d^nx
\end{aligned}
$$
where $K$ is very large and $\epsilon$ is very small. This is designed to avoid the following problems: 1) $F(\xi=0)$ is infinite for $n>m$, because of the behavior near infinity, 2) if $m>n$ the integral $F(\xi)$ is divergent altogether because of the behavior near zero. Note that naively,
$$
\begin{aligned}
F(\xi)&=G(\xi,K,0)+\frac{1}{(2\pi)^{\frac{n}{2}}} \int_{|x|<\frac{1}{K}} |x|^{-m } d^nx \\
&=G(\xi;K,0)+\frac{S_n}{(2\pi)^n}\left.\frac{x^{n-m}}{n-m}\right|_0^{1/K}
\end{aligned}
$$
with $S_n=2\pi^{n/2}/\Gamma(n/2)$ and $\Gamma$ the Gamma function. The second summand is infinity if $m>n$ (in a sense there is "an infinity" embeded in $F(\xi)$ that you cannot get rid of). However, if $n>m$ (note that I'm avoiding $n=m$ case like the plague! All hell breaks loose in that situation! And I personally don't know how to deal wih it), the second summand grows like $\sim K^{-(n-m)}$. In fact, if $n>m$ then one can safely take $K=\infty$. I hope this explains to you why people always only talk about the case $m<n$, because the $m>n$ is badly divergent. Even if $m<n$ the raw Fourier transform is still divergent for $\xi=0$, but that's much more managable! So let's study $G$ now.
Observation 1. Let $R$ be a rotation around the origin. Since $R$ is an isometry, then $G(R\xi;K,\epsilon)=G(\xi;K,\epsilon)$. So $G(\xi;K,\epsilon)=G'(|\xi|;K,\epsilon)$.
Observation 2. Let $\lambda> 0$ be a real number, then 
$$
\begin{aligned}
G(\lambda\xi;\lambda K,\lambda\epsilon)&=G'(\lambda|\xi|;\lambda K,\lambda \epsilon)\\
&=
\lambda^{m-n}G'(|\xi|;K,\epsilon)=
\lambda^{m-n}G(\xi;K,\epsilon)
\end{aligned}
$$
Therefore $G'$ is a homogeneous function of degree $m-n$. In case $m<n$, $G'$ is equal to $1/p(|\xi|;K, \epsilon)$ where $p$ is a homogeneous polynomial of degree $n-m$. With assuption $K=0$, and $\xi=0$ you can find that $p(0,0,\epsilon)=\epsilon^{n-m}$, which shows the divergent behavior I was talking about at $\xi=0$. At the same time, since $F(\xi)$ is not identically infinity (i.e. there exist $\xi$ such that $1/F(\xi)\neq 0$), $p(|\xi|,0,0)=|\xi|^{n-m}$.
So up until now, we have shown that away from $\xi=0$  we have $F(\xi)=C(m,n)|\xi|^{m-n}$ for some $C(m,n)\in \mathbb{C}$ if $n>m$. Also $F(\xi)$ is ill-defined if $m>n$. Can we calculate what $C(m,n)$ is? Sure...
Observation 3. Let $f,g$ be two function, $F,G$ their Fourier transform. Note that
$$
(f,g):=\int f(x)g(x)d^nx = \int F(\xi)G(-\xi)d^n\xi
$$
Moreover, $\mathcal{F}(\exp(-|x|^2/2)=\exp(-|\xi|^2/2)$. Now comes the trick
$$
\begin{aligned}
(|x|^{-m},\exp(-|x|^2/2)&=\int |x|^{-m}\exp(-|x|^2/2)d^nx \\
&= C(m,n)\int 
|\xi|^{m-n}\exp(-|\xi|^2/2)d^n\xi\\
\Longrightarrow C(m,n)&=\frac{\int |x|^{-m}\exp(-|x|^2/2)d^nx}{
\int |x|^{m-n}\exp(-|x|^2/2)d^nx}=\boxed{\frac{2^{\frac{n-m}{2}}\Gamma(\frac{n-m}{2})}{2^{\frac{m}{2}}\Gamma(\frac{m}{2})}}
\end{aligned}
$$
This completes the discussion.
A: You can define $r_+^{-m}$ as
$$(r_+^{-m}, \phi) =
\int_0^\infty r^{-m} \left(
 \phi(r) - \sum_{k = 0}^{m - 2} \frac {\phi^{(k)}(0)} {k!} r^k -
  \frac {\phi^{(m - 1)}(0)} {(m - 1)!} r^{m - 1} H(1 - r) \right) dr$$
and then define $r^{-m}$ in $\mathbb R^n$ as the integral in spherical coordinates:
$$(r^{-m}, \phi) =
\left( r_+^{-m + n - 1}, \int_{\mathcal S_r} \phi dS \right),$$
where the inner integral is taken over the surface of an $(n - 1)$-sphere. Then the formula
$$\mathcal F[r^\lambda] =
\int_{\mathbb R^n}
 r^\lambda e^{i \boldsymbol x \cdot \boldsymbol \xi} d \boldsymbol x =
\frac {2^{\lambda + n} \pi^{n/2} \Gamma \!\left( \frac {\lambda + n} 2 \right)}
 {\Gamma \!\left( -\frac \lambda 2 \right)} \rho^{-\lambda - n}$$
still holds for $\lambda = -n -2 k - 1$, while for $\lambda = -n - 2k$ the result will be the regular part of $\mathcal F[r^\lambda]$, which isn't a homogeneous function of $\rho$:
$$\mathcal F[r^{-n - 2k}] =
[(\lambda + n + 2k)^0] \mathcal F[r^\lambda] = \\
\frac {\pi^{n/2}} {\Gamma(k + 1) \Gamma \!\left( k + \frac n 2 \right)}
 \left( -\frac {\rho^2} 4 \right)^k
 \left( -\ln \frac {\rho^2} 4 +
  \psi \!\left( k + \frac n 2 \right) + \psi(k + 1) \right), \\
k \in \mathbb N^0.$$
