This is the final part of a longer question on Fourier transforms in $\mathbb{R}^N$.
Previous Questions
We define the Fourier transform here as: $\mathcal{F}[u](\xi) = \int_{\mathbb{R}^N} u(x) e^{-2 \pi i x \cdot \xi} \text{d}x$.
Up until now, we have proven that, for $u:\mathbb{R}^N \rightarrow \mathbb{R} $, $u \in \mathscr{S}(\mathbb{R}^N)$, if $u$ is radially symmetric then so is $\mathcal{F}[u]$. (Here, $\mathscr{S}$ denotes the Schwartz class).
We have also found the effect of 'rescaling' on the Fourier transform in the following sense:
Define $u_{\lambda}(x) = \lambda^{-N}u(\lambda^{-1} x) $. Then $\mathcal{F}[u_\lambda](\xi) = \mathcal{F}[u](\lambda \xi) $.
Current Problem
We are somehow expected to use these two previous results in order to calculate $\mathcal{F}[|x|^{-\mu}]$, up to a constant multiple. Here, $\mu \in (0,N.)$
Clearly $|x|^{-\mu}$ is radially symmetric, so we can expect the Fourier transform to be so as well. I'm thinking we could somehow use this fact to rewrite the integral in polar coordinates, and then consider only the integral over the radius:
$\mathcal{F}[|x|^{-\mu}] = \int_{\mathbb{R}^N} |x|^{-\mu} e^{-2 \pi i x \cdot \xi} \text{d}x = K \int^{\infty}_{0} r^{N - 1 -\mu} e^{-2 \pi i r |\xi|} \text{d}r $
So $K$ is just the surface area of the $N-1$-Sphere in $N$ dimensions, and we have written $r = |x|$. However, I fail to see how the result about rescaling can be used here.
Am I completely wrong about how this integral is to be solved? Any help appreciated.
We can't directly calculate the FT with a Lebesgue integral. instead, we need to completely rely on the two properties already proven, as mentioned in the question.
Let $u(x) := |x|^{-\mu} $, for $x \in \mathbb{R}^n$, some $\mu \in (0,n)$.
Define $ u_{\lambda} (x) := \lambda^{-n}u(\lambda^{-1}x) $, with $\lambda > 0$.
Then we know that $ \mathcal{F}[u_\lambda](\xi) = \hat{u}(\lambda \xi) $.
But also, by definition of $u$, we can see that:
$u_\lambda(x) = \lambda^{-n} |\frac{x}{\lambda}|^{-\mu} = \lambda^{\mu - n}|x|^{-\mu} = \lambda^{\mu - n}u(x) $
Thus, $\hat{u}(\lambda \xi) = \lambda^{\mu -n} \hat{u}(\xi) $.
We also know that, since $u$ is radially symmetric, so is $\hat{u}$. Thus, for $r := |\xi|$:
$\hat{u}(\xi) = \hat{u}(r)$.
Then we have $ (\lambda r)^{n - \mu} \hat{u}(\lambda r) = r^{n-\mu} \hat{u}(r) $.
Since this holds for any $\lambda >0$, we must have that $r^{n-\mu}\hat{u}(r) = C$, a constant.
Therefore, $\hat{u}(r) = \hat{u}(\xi) = Cr^{\mu - n}$.