Differentiation of Fourier Transform using Morera's Theorem

Question

What is wrong with my proof?
Let Fourier transform of $f(x)$ exists and be $\mathcal{F}\{f\}(\omega)=\int_{-\infty}^{+\infty}f(x)e^{-i\omega x}dx$
We will find the condition of $\frac{d}{d\omega}\mathcal{F}\{f\}(\omega)$ exists
If $\int_{-\infty}^{+\infty}f(x)dx$ exists then
Using Morera's theorem: $\int_\gamma\int_{-\infty}^{+\infty}f(x)e^{-i\omega x}dxd\omega$=$\int_{-\infty}^{+\infty}f(x)dx\int_\gamma e^{-i\omega x}d\omega$=$\int_{-\infty}^{+\infty}f(x)dx\cdot 0$=$0$
Thus the condition of differentiation of Fourier transform is $\int_{-\infty}^{+\infty}f(x)dx$ exists
However, according to 2 following results:

1. Show that the Fourier Transform is differentiable
2. Derivative of Fourier transform: $F[f]'=F[-ixf(x)]$
   We have the condition of differentiation of Fourier transform is $\int_{-\infty}^{+\infty}xf(x)dx$ exists
   Which is true?

Answer 1

The problem with this idea to show that the Fourier transform is even holomorphic is that for Fubini's theorem you would need that $|f(x)e^{i\omega x}|$ is integrable in $\mathbb R\times \Omega$ for an open set $\Omega \subseteq\mathbb C$. You then have to estimate $$|e^{i\omega x}|=e^{\Re(i\omega x)} =e^{-\Im(\omega)x}.$$ Your idea works if $f$ has compact support. This case is treated in the Paley-Wiener-Schwartz theorem of distribution theory.