Let $f$ be indefinitely differentiable on $\mathbb R$ that has compact support.
$\implies f$ belongs to the Schwartz space.
Consider:
$$I(\xi) = \frac1{\Gamma(\xi)} \int_0^\infty f(x)x^{-1+\xi}dx$$
Where $\xi \in \mathbb C$
The idea is to prove that $I$ has an analytic continuation in the entire complex plane. (It is clear that this holds for $Re(\xi) > 0$).
Using this, the second point is to prove that:
$$I(-n) = (-1)^nf^{(n)}(0) \space \space \space , \space \space \space \forall n \ge0$$
$$\implies I(0) = f(0)$$
Anyone have any ideas?