1. Hard integration problem

The question:
Show that $\int_0^1 {x^m(1 - x)^n \ dx = \frac{m!n!}{(m + n + 1)!}$ for all nonnegative integers m and n.

My attempt:
Using integration by parts:
$u = x^m$
$u' = mx^{m - 1}$
$v' = (1 - x)^n$
$v = -\frac{(1 - x)^{n + 1}}{n + 1}$
$I_{m,n} = _0^1[-x^m\frac{(1 - x)^{n + 1}}{n + 1}] + m\int_0^1 {x^{m - 1}\frac{(1 - x)^{n + 1}}{n + 1}}$
= $_0^1[-x^m\frac{(1 - x)^{n + 1}}{n + 1}] + \frac{m}{n + 1}\int_0^1 {x^{m - 1}(1 - x)^{n + 1}}$
= $\frac{m}{n + 1}\int_0^1 {x^{m - 1}(1 - x)^{n + 1}}$
= $\frac{m}{n + 1} I_{m-1, n+1}$

I think that's right to far. I'm not sure what to do from here. Any assistance would be most appreciated!

2. I don't think that's going to work.

What you have is a special case of the Beta function. Try looking at the Wiki page on that; it shows how the beta function relates to the gamma function.

3. I have a different method, but I don't know if it will help you.

Here we go

The convolution of two functions $f(t),g(t)$ is defined to be

$\displaystyle f*g=\int_{0}^{t}f(t-\tau)g(\tau)d\tau$

using this

$\displaystyle t^n*t^m=\int_{0}^{t}(t-\tau)^n\tau^m d\tau=t^n\int_{0}^{t}(1-\frac{\tau}{t})^n\tau^md\tau$

now sub $\displaystyle u=\frac{\tau}{t} \implies tdu=d\tau$

$\displaystyle t^n\int_{0}^{1}(1-u)^n(tu)^mtdu= t^{n+m+1}\int_{0}^{1}(1-u)^nu^mdu$

Now taking the Laplace transform of both sides gives

$\displaystyle \mathcal{L}(t^n*t^m)=\mathcal{L}\left((t^{n+m+1}\i nt_{0}^{1}(1-u)^nu^mdu\right)=\left( \int_{0}^{1}(1-u)^nu^mdu\right)\mathcal{L}(t^{n+m+1})$

$\displaystyle \frac{n!}{s^{n+1}} \cdot \frac{m!}{s^{m+1}}=\left( \int_{0}^{1}(1-u)^nu^mdu\right)\frac{(n+m+1)!}{s^{n+m+2}}$

Now solving for the integral gives

$\displaystyle \int_{0}^{1}(1-u)^nu^mdu=\frac{n!m!}{(n+m+1)!}$

4. Thanks. I'm not familiar with those techniques, I think your method is a bit beyond my level. I'm sure there's a way to do this with more basic methodology.

\begin{aligned}
I_{m,n} & =\frac{m}{n+1}I_{m-1,n+1}\\
& =\frac{m(m-1)(m-2)...1}{(n+1)(n+2)....(n+m)}I_{0,n+m}\\
& =\frac{m!n!}{(n+m)!}I_{0,n+m}\end{aligned}

Then you just have to work out $I_{0,n+m}$ and you have the answer...

6. Originally Posted by Glitch
The question:
Show that $\int_0^1 {x^m(1 - x)^n \ dx = \frac{m!n!}{(m + n + 1)!}$ for all nonnegative integers m and n...
In...

Beta Function -- from Wolfram MathWorld

... You can verify that...

$\displaystyle \int_{0}^{1} u^{m}\ (1-u)^{n}\ du = B(m+1,n+1) = \frac{m!\ n!}{(m+n+1)!}$ (1)

Merry Christmas from Italy

$\chi$ $\sigma$