1. ## Integration proof

How would i prove this integral
for all integrals m,n >0

2. $\displaystyle m,n\in\mathbb{Z} \ \mbox{or} \ m,n\in\mathbb{R}$

If integers, I would think by induction.

3. for all integers m,n greater or equal to 0

4. Proof: by induction
Show P(1) is true
Assume P(k) is true
Prove P(k+1) is true.

5. Originally Posted by Monster32432421
How would i prove this integral
for all integrals m,n >0
Use repeated integration by parts:

$\displaystyle u = (1 - x)^n \Rightarrow du = -n (1 - x)^{n-1} dx$

$\displaystyle dv = x^m dx \Rightarrow v = \frac{1}{m+1} x^{m+1}$

Then

$\displaystyle \displaystyle I(m, n) = \frac{n}{m+1} \int_0^1 x^{m+1} (1 - x)^{n-1} \, dx$
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$\displaystyle \displaystyle = \frac{n(n-1)(n-2) .... (n - [t-1])}{(m+1)(m+2)(m+3) .... (m+t)}\int_0^1 x^{m+t}(1 - x)^{n-t} \, dx$

Substitute $\displaystyle t = n$ and it's left for you to finish.

6. Since

$\displaystyle \displaystyle \int_0^1 x^{\alpha - 1} \;(1-x)^{\beta -1} dx$ is a Beta function with $\displaystyle B(\alpha,\beta) = \dfrac{\Gamma(\alpha) \; \Gamma(\beta)}{\Gamma(\alpha + \beta)}$

Also, $\displaystyle \Gamma(\alpha) = (\alpha-1)!$

therefore,

you can write:

$\displaystyle \displaystyle \int_0^1 x^m \; (1-x)^n \;dx = \int_0^1 x^{(m+1)-1} \; (1-x)^{(n+1)-1} \;dx$,

which is a beta function $\displaystyle B(m+1, n+1)$

$\displaystyle B(m+1, n+1) = \dfrac{\Gamma(m+1) \; \Gamma(n+1)}{\Gamma(m+n+2)}$

Now finish it off!

7. Originally Posted by harish21
Since

$\displaystyle \displaystyle \int_0^1 x^{\alpha - 1} \;(1-x)^{\beta -1} dx$ is a Beta function with $\displaystyle B(\alpha,\beta) = \dfrac{\Gamma(\alpha) \; \Gamma(\beta)}{\Gamma(\alpha + \beta)}$

Also, $\displaystyle \Gamma(\alpha) = (\alpha-1)!$

therefore,

you can write:

$\displaystyle \displaystyle \int_0^1 x^m \; (1-x)^n \;dx = \int_0^1 x^{(m+1)-1} \; (1-x)^{(n+1)-1} \;dx$,

which is a beta function $\displaystyle B(m+1, n+1)$

$\displaystyle B(m+1, n+1) = \dfrac{\Gamma(m+1) \; \Gamma(n+1)}{\Gamma(m+n+2)}$

Now finish it off!
I don't think that quoting, more or less, a standard result constitutes a proof in the sense that the OP has requested.