Decide if the improper integral of a Fourier transform converges

Question

I have the function: $$f(x)=\left\{\begin{matrix} e^{-x^{10}} & ,x>0\\ -e^{-x^{10}} & ,x<0 \end{matrix}\right.$$

I need to answer:

• $\int_{-\infty}^{\infty}|\widehat{f}(\omega)|d\omega < \infty$
  

and I'm not quite sure how to do it.

Answer 1

If an integrable function has an integrable Fourier transform, Fourier inversion yields $f(x)=\int \hat{f}(\xi)e^{2\pi i x \xi}\,d\xi $ for (almost all) $x$. In particular, $f$ has a continuous "version" (with respect to a.e. equality). -- PhoemueX

... which is not the case for this $f$:

Assume $\int_{-\infty}^{\infty}|\widehat{f}(\omega)|d\omega < \infty$. by the inversion theorem and by the fact that $f$ is absolutely integrable we get that we can take the transform of $\widehat{f}$ and to have a continuous version of $f$ which is impossible because $f$ has a jump discontinuity at $0$ and thus $\int_{-\infty}^{\infty}|\widehat{f}(\omega)|d\omega$ does not converge. -- YaG32