Show F is the Fourier transform of an integrable function

Question

I've come across this exercise:

Show that for $\epsilon>0$,

$$F(\xi)=\frac{1}{(1+\lvert\xi\rvert^2)^\epsilon}$$ is the Fourier transform of $f\in L^1$.

I know 2 different ways of dealing with this kind of questions: showing that $F$ is a Schwartz function (hence the Fourier transform is bijective) or showing that it's square integrable (making the transform an isomorphism).

However both fail in this case and I don't know what else to try. What key property of the Fourier transform am I missing?

Thank you in advance.

Answer 1

You are looking for the Bessel potential. Stein's "Singular integrals and differentiability properties of functions" Chapter 5 Section 3 covers all the details.

Answer 2

$$ \lim_{R\rightarrow\infty}\int_{-R}^{R}e^{ix\xi}\frac{1}{(1+\xi^2)^{\epsilon}}d\xi \\ = \lim_{R\rightarrow\infty}2\int_{0}^{R}\cos(x\xi)\frac{1}{(1+\xi^2)^{\epsilon}}d\xi \\ = \left.2\frac{1}{x}\sin(x\xi)\frac{1}{(1+\xi^2)^{\epsilon}}\right|_{\xi=0}^{\infty}+\frac{2}{x}\int_{0}^{\infty}\sin(x\xi)\frac{2\epsilon\xi}{(1+\xi^2)^{1+\epsilon}}d\xi \\ = \frac{4\epsilon}{x}\int_{0}^{\infty}\sin(x\xi)\frac{\xi}{(1+\xi^2)^{1+\epsilon}}d\xi \\ = \frac{4\epsilon}{x^2}\int_{\xi=0}^{\infty}\cos(x\xi)\frac{d}{d\xi}\frac{\xi}{(1+\xi^2)^{1+\epsilon}}d\xi \\ = \frac{4\epsilon}{x^2}\int_{\xi=0}^{\infty}\cos(x\xi)\frac{(1+\xi^2)^{1+\epsilon}-2(1+\epsilon)\xi^2(1+\xi^2)^{\epsilon}}{(1+\xi^2)^{2+2\epsilon}}d\xi $$ It appears that this function is integrable in $x$ near $x=\infty$.