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 $f$.

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 $\rho$: $$\mathcal F[r^{-n - 2k}] = [(\lambda + n + 2k)^0] \mathcal F[r^\lambda] = \\ \frac {\pi^{n/2}} {\Gamma(k + 1) \Gamma \!\left( k + \frac n 2 \right)} \left( -\frac {\rho^2} 4 \right)^k \left( -\ln \frac {\rho^2} 4 + \psi \!\left( k + \frac n 2 \right) + \psi(k + 1) \right), \\ k \in \mathbb N^0.$$

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.