# Factorials of Non-Natural

$$n!=\int_{0}^{\infty}{e^{-x}x^{n}\:dx}$$

(a) I want to find a formula for the above and then prove it by induction. The answer according to Wolfram is $n!$, however I have no idea how to get there. Any hints or ideas on how I should tackle this one?

(b) Also, I want to understand why it is called is Gamma Function!

(c) Finally, it is difficult to understand what is factorial of non natural number? (For example Wolfram tells me that $3.5! = 11.6317$.)
So how to compute $3.5!=\int_{0}^{\infty}{e^{-x}x^{3.5}\:dx}$ integral?

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Either do induction, and/or integrate by parts $n$ times. The boundary terms of the integration always dissapear – Alex R. Oct 28 '12 at 15:43

Let's see what happens when we integrate $\displaystyle \int_0^\infty e^{-x}x^{n+1}dx$ by parts.

$$\int_0^\infty e^{-x}x^{n+1}dx = -e^{-x}x^{n+1}\mid_0^\infty + (n+1)\int_0^\infty e^{-x}x^n dx = (n+1)\int_0^\infty e^{-x}x^n dx$$

This gives you most of an inductive argument for showing what you want.

By the way, it turns out that that the integral you are doing has a name: it's called the Gamma Function. And it manages to come up a lot.

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The following website has some good insight on the subject: http://www.sosmath.com/calculus/improper/gamma/gamma.html

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