Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

(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?

share|improve this question
    
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

2 Answers 2

up vote 5 down vote accepted

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.

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.