The equivalence of the two definitions of fractional Laplacian Using the Fourier transform we can easily define the fractional Laplacian by
$$(-\Delta)^{s/2}f(x)=(|\xi|^s\hat f(\xi))^\vee(x), \ \ f\in C_0^\infty.  $$
However, I learned that there is another definition using the principal value of singular integral
$$(-\Delta)^{s/2}f(x)=C_{n,s}P.V.\int_{\mathbb{R}^n}{\frac{f(x)-f(y)}{|x-y|^{n+s}}dy}, \ \ 0<s<2,  $$
where $C_{n,s}$ is some constant.
I am curious about the proof that these two definitions are equivalent, but I can not find the proof on the internet or textbooks. By the way, I found that one may symmetrize the integral above to regularize the integral 
$$ P.V.\int_{\mathbb{R}^n}{\frac{f(x)-f(y)}{|x-y|^{n+s}}dy}= P.V.\int_{\mathbb{R}^n}{\frac{f(x)-f(x-y)}{|y|^{n+s}}dy}\\=\frac12\int_{|y|\le h}{\frac{2f(x)-f(x+y)-f(x-y)}{|y|^{n+s}}dy}+\int_{|y|\ge h}{\frac{f(x)-f(x-y)}{|y|^{n+s}}dy}.$$ 
 A: As indicated by the now deleted answer, a full proof is available in this wonderful paper  by Nezza-Palatucci-Valdinoci (Arxiv link here, published version here). I'll paraphrase it here to answer your question.
The equality is proven for Schwartz functions $\mathscr S$. Call the Fourier definition $Fu = \mathscr F^{-1}|\xi|^s \mathscr F u $ (I used $u$ instead of $f$ to avoid too many "f"s and to follow the above paper) and call the principal value definition $Pu$. The goal is to show $F=P$ on $\mathscr S$. Notice also that your second integral for $Pu$ can also be symmetrized to obtain
$$ Pu = \frac12C_{n,s}\int_{\mathbb R^n} \frac{2u(x) - u(x+y)-u(x-y)}{|y|^{n+s}} \,\mathrm dy.$$
With $Pu$ rewritten in this way, it is enough to prove that their Fourier transforms are equal, namely $\mathscr F Pu = \mathscr F  Fu = |\xi|^s\mathscr F u$.
By the decay of $u\in \mathscr S$, and the cancellation at $y=0$ from the numerator, Fubini on $\mathbb R^{2n}$ applies to give us $ \mathscr F P u = P\mathscr F u$, which is in full,
\begin{align} (\mathscr FPu)(\xi) 
&
= \frac{1}{2}C_{n,s}\int_{\mathbb R^n} \frac{2\mathscr F u(\xi) - \mathscr F_x [u(x+y)](\xi)-\mathscr F_x [u(x-y)](\xi)}{|y|^{n+s}} \,\mathrm dy \\
&
= \frac{1}{2}C_{n,s}\int_{\mathbb R^n } \frac{2 -e^{i\xi\cdot y}-e^{-i\xi\cdot y}}{|y|^{n+s}} \,\mathrm dy\, \mathscr F u (\xi)\\
&
= \left[ C_{n,s} \int_{\mathbb R^n} \frac{1-\cos(\xi\cdot y)}{|y|^{n+s}} \,\mathrm dy  \right] \mathscr F u (\xi).
\end{align}
Here, we used that translations correspond to modulations in the Fourier transform, i.e. 
$$ \mathscr F [u(\cdot +y) ](\xi) = e^{-i\xi\cdot y} \mathscr Fu(\xi).$$
This is already in the form of a multiplier. Moreover you don't seem to care what the constant is, so we might be satisfied by computing the dependence of our symbol $S(\xi):=\int_{\mathbb R^n} \frac{1-\cos(\xi\cdot y)}{|y|^{n+s}} dy$ on $|\xi|$. This is easy to see by performing the scaling
$$ z = |\xi|y ⟹ dz = |\xi|^n \,\mathrm dy,$$ 
which gives
$$ S(\xi) = |\xi|^s S\left(\frac{\xi}{|\xi|}\right). $$
Actually, it should be clear that $S$ is a radial function. The proof is identical in its steps to the proof that a radial function has radial Fourier transform. Hence, $S\left(\frac{\xi}{|\xi|}\right)  = S\big((1,0,\dots ,0)\big)$ and depending on who you ask, we can even take $C_{n,s} := S\big((1,0,\dots ,0)\big)^{-1}$ which gives precisely the result.
