Compactly supported $f \in L^1$. Then $\hat f$ can be extended to an holomorphic function on $\mathbb C$, where $\hat f$ is the fourier transform of f.
I tried the following
$\hat f(x)=\frac{1}{\sqrt{2 \pi}}\int_{\mathbb{R}}f(t)\exp(-itx) d\lambda(t)=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}f(t)\exp(-itx) dt=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}f(t)\sum_{n=0}^{\infty}\frac{(-itx)^n}{n!}dt=\frac{1}{\sqrt{2 \pi}}\sum_{n=0}^{\infty}x^n\int_{-\infty}^{\infty}f(t)\frac{(-it)^n}{n!}dt=\sum_{n=0}^{\infty}\frac{1}{\sqrt{2 \pi}}x^na_ndt$
with $a_n:=\int_{-\infty}^{\infty}f(t)\frac{(-it)^n}{n!}$
so $\hat f(x)$ can be written as a power series.
For interchanging integration and summation I thought:
For $x \in \mathbb C$ with $|x|\le T$ and $\mathrm{supp} f \subseteq [-R,R]$ we have
$\int \sum_{n=0}^{\infty}|f(t)||\frac{(-itx)^n}{n!}|\le \sum_{n=0}^{\infty}\frac{(RT)^n}{n!}\|f\|_1=e^{RT}\|f\|_1<\infty$
and then using dominated convergence theorem
Yes, this works.
More generally, suppose $(T, \Sigma, \mu)$ is a $\sigma$-finite measure space, $g: T \times U \to \mathbb C$ is a measurable function (where $U \subseteq \mathbb C$ is open) such that $g(t, \cdot)$ is analytic in $U$ for each $t$ and uniformly bounded on $T \times K$ for all compact subsets $K$ of $U$, and $f \in L^1(\mu)$. Then $F(z) = \int_T f(t) g(t,z)\; d\mu(t)$ is analytic in $U$. For this you can use Morera's theorem, as for any closed contour $C$ in $U$ whose inside is in $U$,
$$ \oint_C F(z)\; dz = \oint_C \int_T f(t) g(t,z)\; d\mu(t) \; dz = \int_T \oint_C f(t) g(t,z)\; dz \; d\mu(t) = 0 $$