How to calculate $c_a$ where $\left(f\mapsto\int_{\mathbb{R}}\frac{f(t)-f(0)}{|t|^{a}}dt\right)=c_a\mathcal{F}_x(|x|^{-1+a})$ From this question I know that for every $a\in\mathbb{R}$ there exists a unique radial positive homogeneous tempered distribution of degree $a$, up to a multiplicative constant.
Also, it is easily proven that if $g$ is a radial positive homogeneous tempered distribution of degree $a$, then its Fourier transform $\mathcal{F}(g)$ is a radial positive homogeneous tempered distribution of degree $-1-a$.
Now, it is clear that if $n\in\{1,2,...\}$ and $a\in\left(n,n+1\right)$, the following:
$$f\mapsto\int_\mathbb{R} \frac{f(t)-\sum_{k=0}^{n-1}f^{(k)}(0)t^k}{|t|^{a}}\operatorname{d}t$$
is a non-null radial positive homogeneous tempered distribution of degree $-a$, while:
$$x\mapsto|x|^{-1+a}$$
is locally integrable and of moderate growth, so it represents a tempered distribution that is also non-null, radial and positive homogeneous of degree $-1+a$ and then its Fourier transform is a radial positive homogeneous distribution of degree $-a$. So, if $\mathcal{S}$ is the space of Schwartz test functions we have that:

$$\forall n\in\{1,2,...\}, \forall a\in(n,n+1),\exists c_a\in\mathbb{C},\forall f\in\mathcal{S},\\  \left(f\mapsto\int_\mathbb{R} \frac{f(t)-\sum_{k=0}^{n-1}f^{(k)}(0)t^k}{|t|^{a}}\operatorname{d}t\right)=c_a\mathcal{F}_x(|x|^{-1+a}),$$
  How can we explicitly calculate $c_a$?

 A: The following is from (informal) lecture notes of a course I taught on distributions. The calculation corresponds to the easy case $0<a<1$ which was left out from the OP's question.
I would try to adapt the proof below to see what happens for $1<a<2$.
The notation below is as follows:
For a Schwartz function $f$ a multiindex $\beta$ and an integer $k\ge 0$, we use the seminorm $||f||_{\beta,k}=\sup_{x\in\mathbb{R}^d} \langle x\rangle^k |\partial^{\beta} f(x)|$. Here $\langle x\rangle=\sqrt{1+\sum_{1\le i\le d} x_i^2}$ is the so-called "Japanese bracket". The fourier transform is $\widehat{f}(\xi)=\mathcal{F}[f](\xi)=\int_{\mathbb{R}^d} e^{-i\xi x}f(x)\ d^dx$. 

Example:
Let $0<\alpha <d$. Then
$``\displaystyle \int _{\mathbb{R}^d}e^{-i\xi x}\frac{1}{|x|^\alpha}\ d^d x=\frac{\Gamma(\frac{d-\alpha}{2})}{\Gamma(\frac{d}{2})}2^{d-\alpha} \pi^{\frac{d}{2}}\frac{1}{|\xi|^{d-\alpha}}"
$
Proof:
Note that L.H.S. does not make sense as Lebesgue integral as 
$$
\int_{\mathbb{R}^d}|x|^{-\alpha} d^dx=\infty\ .
$$ 
But we will work in the language of distributions. Define 
$\displaystyle \phi(x)=
\begin{cases}
\frac{1}{|x|^\alpha} & x\neq 0,\\
0 & x=0
\end{cases}$
Define the distribution associated to $\phi$, say $T$, i.e. given $f\in S'$, we have $\displaystyle \langle T,f \rangle =\int _{\mathbb{R}^d}\phi(x) \, f(x) \, d^dx$. 
$$
\int |\phi f|=\int \frac{1}{|x|^\alpha}\frac{\langle x \rangle^{d+1}}{\langle x \rangle^{d+1}}|f(x)|d^dx\leq ||f||_{0,d+1}\int_{\mathbb{R}^d}\frac{d^dx}{|x|^\alpha \langle x \rangle^{d+1}}<\infty
$$
Thus $T$ is well-defined and continuous. So, $T\in S'$. 
$\displaystyle \forall f\in S'$,
\begin{align*}
\langle \widehat{T},f \rangle := \langle T,\widehat{f} \rangle &=\int_{\mathbb{R}^d\setminus\{0\}}d^dx\frac{1}{|x|^\alpha}\widehat{f}(x)\\
&=\int_{x\neq0}d^dx\Big(\frac{1}{\Gamma(\frac{\alpha}{2})}\int_0^\infty\frac{dt}{t}t^{\frac{\alpha}{2}}e^{-t|x|^2}\Big)\widehat{f}(x)\\
& \overset{\text{Fubini}}{=}\frac{1}{\Gamma(\frac{\alpha}{2})}\int_0^\infty \frac{dt}{t}t^{\frac{\alpha}{2}}\int_{\mathbb{R}^d\setminus\{0\}}d^dx\ \overline{e^{-t|x|^2}}\widehat{f}(x)\quad (*)
\end{align*}
For $a>0$, 
\begin{align*}
 \mathcal{F}[\xi \to e^{-a\xi^2}](x) &=\int_{\mathbb{R}^d}e^{-ix\xi}e^{-a\xi^2}d^d\xi\\
 &=(2a)^{-\frac{d}{2}}\int e^{-\frac{\eta^2}{2}-i\frac{x}{\sqrt{2a}\eta}}d^d\eta , ~~\eta=\sqrt{2a}\xi\\
 &=(\frac{\pi}{a})^{\frac{d}{2}}e^{-\frac{x^2}{4a}}
\end{align*}
Take $\frac{1}{4a}=t$; $\displaystyle e^{-t|x|^2}=\mathcal{F}[\xi\to (4\pi t)^{-\frac{d}{2}}e^{-\frac{\xi^2}{4t}}](x)$
Substitute into (*) and use Plancherel, 
\begin{align*}
    \langle T,\hat{f} \rangle & =\frac{1}{\Gamma(\frac{d}{2})}\int_0^\infty \frac{dt}{t} t^{\frac{\alpha}{2}}(2\pi)^d\int_{\mathbb{R}^d\setminus\{0\}}d^d\xi\ (4\pi)^{-\frac{d}{2}}e^{-\frac{\xi^2}{4t}}f(\xi) \\
    &  \overset{Fubini}{=}\frac{1}{\Gamma(\frac{\alpha}{2})}\int_{\xi \neq 0}d^d \xi \ f(\xi) \pi^{\frac{d}{2}}\int_0^\infty \frac{dt}{t}t^{\frac{\alpha-d}{2}}e^{-\frac{\xi^2}{4t}} \\
    & =\frac{1}{\Gamma(\frac{\alpha}{2})}\int_{\xi \neq 0}d^d \xi \ f(\xi) \pi^{\frac{d}{2}}(\frac{\xi^2}{4})^{\frac{\alpha-d}{2}}\int_0^\infty \frac{ds}{s}s^{\frac{d-\alpha}{2}}e^{-s},\quad s=\frac{\xi^2}{4t}
\end{align*}
Thus,
$$
\langle \widehat{T},f \rangle =\frac{\Gamma(\frac{d-\alpha}{2})}{\Gamma(\frac{\alpha}{2})}
2^{d-\alpha}\pi^{\frac{d}{2}}
\int_{\xi \neq 0}d^d\xi\ \frac{1}{|\xi|^{d-\alpha}}f(\xi)
$$
The last part does define an element in $S'$ as $d-\alpha<d\iff \alpha>0$.
