# Paley-Wiener theorem for a sector $\{\zeta:-\epsilon<\arg(\zeta)<\pi+\epsilon\}$

One of the variations of the Paley-Wiener theorem yields: If $f\in L^2(\mathbb R_+)$, then the Fourier transform $F$, defined by $$F(\zeta)=\int_0^\infty f(x)e^{ix\zeta}dx$$ is a holomorphic function in the upper half-plane $\{\zeta:0<\arg(\zeta)<\pi\}$.

What I'd like to state is that there exists $\epsilon>0$ such that the function $F$ is analytic in the sector $S_\epsilon=\{\zeta:-\epsilon<\arg(\zeta)<\pi+\epsilon\}$.

Is there any restrictions, I can make on $f$ in order to say that $F$ is holomorphic in $S_\epsilon$ for some $\epsilon>0$?

If $f$ is supported on $[-N,N]$, then the other Paley-Wiener theorem says $F(\zeta) = \int_{-N}^N f(z)\; e^{ix\zeta}\; dx$ is an entire function. So that's certainly one restriction you can put on.
An obstruction to the kind of conclusion you want: if $f(x) = \exp(-bx)$ with $\text{Re}(b) > 0$, then $\int_0^\infty f(x)e^{ix\zeta}\; dx$ diverges for $\text{Im}(\zeta) = - \text{Re}(b)$, with a pole at $\zeta = -ib$, so even if you allow analytic continuation rather than requiring convergence of the integral, these functions can't satisfy your property if $\text{Im}(b)/\text{Re}(b)$ is large enough.
• Thank you for the comment. I like your example but if the analytic continuation of $\int_0^\infty f(x)e^{ix\zeta}dx$ has only one pole at $\zeta=-ib$ then I can always find $\epsilon>0$ small enough so that it is holomorphic in $S_\epsilon$. What I am worried about is the case when $F(\zeta)$ has infinitely many poles. For instance, $F(\zeta)=\tan(\zeta+i)$ – Aleksandr Aug 14 '14 at 0:58