Using induction to prove integral

[rheighton]

I'm rather confused on this problem.. it asks me to use mathematical induction to prove that for each positive integer 'n':

(0^∞)∫(x^n)(e^-x)dx = n!

It also asks to solve the integral when n = 0, which logically seems to be 0! = 1

If somebody could help me out with this problem I'd really appreciate it!

 

[royhaas]

Use integration by parts.

 

[galactus]

Try starting with:

\(\displaystyle \L\\\int_{0}^{\infty}e^{-ax}dx\)

\(\displaystyle \L\\=\frac{-1}{a}e^{-ax}={\infty}-(-\frac{1}{a})=\frac{1}{a}\)

Now, differentiate both sides, with respect to a, over and over to see what goes on.......Leibniz Rule, Ever heard of it?.:

\(\displaystyle \L\\\int_{0}^{\infty}-xe^{-ax}dx=\frac{-1}{a^{2}}\)

\(\displaystyle \L\\\rightarrow\int_{0}^{\infty}xe^{-ax}dx=\frac{1}{a^{2}}\)

Again:

\(\displaystyle \L\\\int_{0}^{\infty}x^{2}e^{-ax}dx=\frac{2}{a^{3}}\)

Again:

\(\displaystyle \L\\\int_{0}^{\infty}x^{3}e^{-ax}dx=\frac{6}{a^{4}}=\frac{3!}{a^{4}}\)

Now, generally we have:

\(\displaystyle \L\\\int_{0}^{\infty}x^{n}e^{-ax}dx=\frac{n!}{a^{n+1}}\)

Put a=1:

\(\displaystyle \L\\\int_{0}^{\infty}x^{n}e^{-x}dx=n!\)

Therefore, we have an definite integral whose value is n!. Put n=0 and you get......?. What?. See it?.

 

[rheighton]

Ahh thanks so much for your help!