# Exponential decay leeds to analytic Fourier transform - Paley-Wiener Theorem

I stumbled (in this paper by Ingrid Daubechies ) across an application of a Paley-Wiener theorem which uses the decay condition

$\int_{\mathbb{R}^d} |f(x)|e^{\alpha|x|} dx <\infty$.

to infer that $\hat f$ is analytic.

Unfortunately I cannot find a reference for this.

Can someone explain this or point me to an explanation?

Let $d =1$. The condition $|f(x)| e^{|\alpha|x} \in L^1$ implies $F(\omega)=\int_{-\infty}^\infty f(x) e^{-i \omega x}dx$ and all its derivatives $F^k(\omega)=\int_{-\infty}^\infty f(x) (-ix)^k e^{-i \omega x}dx$ converge absolutely and uniformly for $|\Im(\omega)| \le \alpha-\epsilon$. Thus $F(\omega)$ is holomorphic (and hence analytic) for $|\Im(\omega)| < \alpha$.
You could also show directly its Taylor series $$F(\omega) = \sum_{k=0}^\infty \frac{F^{(k)}(\omega_0)}{k!} (\omega-\omega_0)^k = \sum_{k=0}^\infty \frac{(\omega-\omega_0)^k}{k!} \int_{-\infty}^\infty f(x) (-ix)^k e^{-i \omega_0 x}dx$$ $$\le \sum_{k=0}^\infty \frac{|\omega-\omega_0|^k}{k!} \int_{-\infty}^\infty |f(x)x^k e^{-i \omega_0 x}|dx$$ converges absolutely for $|\Im(\omega_0)| + |\omega-\omega_0| < |\alpha|$.