Integration Problem

Question

how can I prove this identity:

$$ n \in N_0, m \in N $$

$$ \int_0^1 x^{n}(1-x)^{m}dx\ = \frac{n!m!}{(n+m+1)!} $$

I thought about the geometric series but I am not sure

Thanks in advance!

Answer 1

Let $I_{n,m}$ be the integral in question.

Integration by parts with $u=x^n$, $dv=(1-x)^m\,dx$, $du=nx^{n-1}\,dx$, $v=-\frac{(1-x)^{m+1}}{m+1}$ yields $$ \begin{align*} I_{n,m}&=\int_0^1x^n(1-x)^m\,dx\\ &=\left.-\frac{x^n(1-x)^{m+1}}{m+1}\right\vert_{x=0}^{x=1}+\frac{n}{m+1}\int_0^1x^{n-1}(1-x)^{m+1}\,dx\\ &=\frac{n}{m+1}I_{n-1,m+1}. \end{align*} $$ Applying this iteratively, we see that $$ \begin{align*} I_{n,m}&=\frac{n}{m+1}\cdot I_{n-1,m+1}\\ &=\frac{n}{m+1}\cdot\frac{n-1}{m+2}\cdot I_{n-2,m+2}\\ &=\ldots\\ &=\frac{n}{m+1}\cdot\frac{n-1}{m+2}\cdots\frac{1}{m+n}\cdot I_{0,m+n}\\ &=\frac{n!\cdot m!}{(m+n)!}\cdot I_{0,m+n} \end{align*} $$ Now, can we compute $I_{0,m+n}$? $$ \begin{align*} I_{0,m+n}&=\int_0^1(1-x)^{m+n}\,dx\\ &=\left.-\frac{(1-x)^{m+n+1}}{m+n+1}\right\rvert_{x=0}^{x=1}\\ &=\frac{1}{m+n+1}. \end{align*} $$ So, in all, we get $$ I_{n,m}=\frac{n!\cdot m!}{(n+m)!}\cdot\frac{1}{m+n+1}=\frac{n!\cdot m!}{(n+m+1)!}, $$ as desired.

Answer 2

Note that the beta function can be defines as $$B(x,y) = \int_{0}^{1}t^{x-1}(1-t)^{y-1}dy = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)} = \frac{(x-1)!(y-1)!}{(x+y -1)!}$$ Note that $$\Gamma(n) = (n-1)!$$ and $$\Gamma(n+1) = n\Gamma(n)$$ In your case we have $$\int_{0}^{1}x^n (1-x)^{m}dx$$ Thus applying the formula above we have $$\int_{0}^{1}x^n (1-x)^{m}dx = \frac{\Gamma(n+1)\Gamma(m+1)}{\Gamma(n+m+1)}$$

You should be able to continue from here.