= Solution
Since $0<f_\alpha\leq1$, its <regular distribution> satisfies
$$
|\langle f_\alpha,\phi\rangle|\leq\|\phi\|_1
\leq\pi\bigl(p_{0,0}(\phi)+p_{2,0}(\phi)\bigr),
$$
so $f_\alpha$ is a <tempered distribution> for every $\alpha>0$.
Put $a=\alpha/2$. The defining <integral> of the <gamma function>, with $r=(1+x^2)t$, gives
$$
f_\alpha(x)=\frac1{\Gamma(a)}\int_0^\infty t^{a-1}e^{-t}e^{-tx^2}\,dt.
$$
The <Fourier transform of a Gaussian> suggests the following representative for the <Fourier transform of an inverse power of one plus x squared>:
$$
\boxed{h_\alpha(\xi)=\frac{\sqrt\pi}{\Gamma(a)}\int_0^\infty t^{a-3/2}e^{-t-\xi^2/(4t)}\,dt\quad(\xi\ne0).}
$$
All factors are positive. <Tonelli's theorem> evaluates its total <integral>, allowing an infinite value at the single point zero:
$$
\begin{aligned}
\int_{\mathbb R}h_\alpha(\xi)\,d\xi
&=\frac{\sqrt\pi}{\Gamma(a)}\int_0^\infty t^{a-3/2}e^{-t}
\left(\int_{\mathbb R}e^{-\xi^2/(4t)}\,d\xi\right)dt\\
&=\frac{2\pi}{\Gamma(a)}\int_0^\infty t^{a-1}e^{-t}\,dt=2\pi.
\end{aligned}
$$
Assign any finite value at zero if necessary. This gives an <integrable function> with $\|h_\alpha\|_1=2\pi$.
To verify equality as <tempered distributions>, take any <Schwartz function> $\phi$. Substituting the mixture into $\langle f_\alpha,\widehat\phi\rangle$ is justified by
$$
\frac1{\Gamma(a)}\int_0^\infty t^{a-1}e^{-t}\int e^{-tx^2}|\widehat\phi(x)|\,dx\,dt
\leq\|\widehat\phi\|_1.
$$
For each $t>0$, <Fubini's theorem> and the <Fourier transform of a Gaussian> identify the inner pairing with $\int\sqrt{\pi/t}\,e^{-\xi^2/(4t)}\phi(\xi)\,d\xi$. Interchanging the resulting <integrals> is valid because their absolute value is bounded by $2\pi\|\phi\|_\infty$ after integrating the positive kernel. Therefore
$$
\langle\widehat{f_\alpha},\phi\rangle=\int h_\alpha(\xi)\phi(\xi)\,d\xi.
$$
Finally, on a <compact set> with $|\xi|\geq\delta>0$, the mixture <integrand> is bounded by $t^{a-3/2}e^{-t-\delta^2/(4t)}$, which is integrable both at zero and at infinity. <Dominated convergence> proves continuity there. Hence $\boxed{\widehat{f_\alpha}=h_\alpha\in L^1(\mathbb R),\ h_\alpha(\xi)>0\text{ and continuous for }\xi\ne0}$.
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