To generate the gamma function by integration....is a MacLaurin series legitimate?

28 Views Asked by At

I've been trying to make sense of the gamma function and why some values pop up, like $\operatorname{\Gamma}\left (\dfrac{3}{2} \right) =\dfrac{\sqrt{\pi}}{2}$

I'm still trying.

But here's a thought: what if you used the MacLaurin series for $e^{-x}$ instead. Here's what I came up with:

$n! = \displaystyle \int_0^\infty x^n \displaystyle \sum_{k=0}^\infty \dfrac{(-1)^k \cdot x^k}{k!} = \displaystyle \int_0^\infty \displaystyle \sum_{k=0}^\infty \dfrac{(-1)^k \cdot x^{k+n}}{k!}$

$=\left [\displaystyle \sum_{k=0}^\infty \dfrac{(-1)^k \cdot x^{k+n+1}}{(k+n+1) \cdot k!} \right]_0^\infty$

$= \left [\displaystyle \sum_{k=0}^\infty \dfrac{(-1)^k \cdot x^{k+n+1}}{(k+1)! + nk!} \right]_0^\infty$

Evaluating that at $0$ is just $0$, but it's the upper bound I'm worried about.

Is this reasoning:

  1. Open to improvement or
  2. Totally wrong?