When is a function a Fourier transform of an integrable function? Specifically, in the case $f(\xi)=\frac{1}{(1+\xi ^2)^\epsilon}$ where $0<\epsilon<1$.
I wish to prove this is a Fourier transform of a $L_1$ function. Any insight into the manner would be appreciated.
 A: Let
$$ g_R(x) = \int_{-R}^R \frac1{(1+\xi^2)^\epsilon} e^{i\xi x} \, d\xi .$$
If you integrate by parts, you get
$$ g_R(x) = \frac1{ix} \left[\frac1{(1+\xi^2)^\epsilon} e^{i\xi x}\right]_{-R}^R + \frac1{ix} \int_{-R}^R \frac{2\epsilon\xi}{(1+\xi^2)^{1+\epsilon}} e^{i\xi x} \, d\xi .$$
Therefore $\check f(x) = g(x) = \lim_{R\to \infty} g_R(x)$ is defined for $x \ne 0$, and $|g(x)| \le C_1/|x|$.  Integrate by parts again, and you will see that $|g(x)| \le C_2/|x|^2$.  (Here $C_1$ and $C_2$ are positive constants, where for example $C_1 = \int_{-\infty}^\infty 2 \epsilon |\xi|/(1+\xi^2)^{1+\epsilon} \, d\xi$.)  Hence $g(x)$ decays fast enough as $x\to \pm\infty$ to be in $L^1$.
Now we need to show that $g(x)$ blows up slowly enough as $x\to 0$ to be in $L^1$.  See that
$$ g(x) = \frac2x \int_0^\infty \frac{2\epsilon\xi}{(1+\xi^2)^{1+\epsilon}} \sin(\xi x) \, d\xi .$$
So use $|\sin(a)| \le \min\{|a|,1\}$ to estimate
$$ \left|\int_0^\infty \frac{2\epsilon\xi}{(1+\xi^2)^{1+\epsilon}} \sin(\xi x) \, d\xi \right|$$
$$ \le \int_0^{1/x} \frac{2\epsilon\xi^2 x}{(1+\xi^2)^{1+\epsilon}} \, d\xi + \int_{1/x}^\infty \frac{2\epsilon\xi}{(1+\xi^2)^{1+\epsilon}} d\xi $$
Elementary estimates show it decays as least as fast as $x^{2\epsilon}$ as $x \to 0$.
