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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?

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  • $\begingroup$ Try integrating by parts for the first part. $\endgroup$ – Antonio Vargas Apr 5 '14 at 0:13
  • $\begingroup$ Actually the first part is pretty okay; it's the second part I'm having trouble proving. $\endgroup$ – zhn11tau Apr 5 '14 at 13:36

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