Let $f\in L_2(\mathbb R_+)$ and consider its Fourier transform $$F(\zeta)=\int_0^\infty f(x)e^{ix\zeta}dx$$
Is it true that analytic continuation of $F(\zeta)$ has at most finitely many poles in a half-plane $\{\mathrm{Im}(\zeta)>-a\}$ for any $a>0$? Any counterexample or proof would be much helpful.
Let $f_{n}(x)=e^{inx-x}$ on $[0,\infty)$ for $n=0,\pm 1,\pm 2,\pm 3,\cdots$. The Fourier transform of $f_{n}$ is $$ \int_{0}^{\infty}e^{ix\zeta}f_{n}(x)\,dx =\int_{0}^{\infty}e^{inx-x}e^{ix\zeta}\,dx = \frac{1}{1-in-i\zeta}=\frac{i}{\zeta+i+n}. $$ The function $f_{n}$ is in $L^{2}[0,\infty)$ with $\|f_{n}\|=1$ for all $n$. Choose any absolutely convergent sequence $\{ a_{n}\}$ and let $F(x)=\sum_{n}a_{n}f_{n}$. $F$ converges in $L^{2}[0,\infty)$, and its Fourier transform is $$ \sum_{n=-\infty}^{\infty} \frac{ia_{n}}{\zeta+i+n}. $$