# Show, that $I_{m,n}:=\int\limits_{0}^{1}x^m(1-x)^n dx \text{ holds }I_{m,n}=\frac{m!\,n!}{(m+n+1)!}$

Show, that for $$m,n\in\mathbb{N}$$ holds: $$\displaystyle I_{m,n}:=\int\limits_{0}^{1}x^m(1-x)^n dx \text{ holds }I_{m,n}=\frac{m!\,n!}{(m+n+1)!}$$

I tried to do integration by parts and got $$\frac { ( 1 - x ) ^ { n + 1 } } { n + 1 } \cdot x ^ { m } - \int \frac { ( 1 - x ) ^ { n + 1 } } { n +1 }\cdot m\cdot x ^ { m - 1 } d x = \frac{(1-x)^{n+1}}{n+1}\cdot x^m-\left(\frac{1}{n+1}\cdot m\underbrace{\int (1-x)^{n+1}\cdot x^{m-1}dx}_{*}\right)\\ *=\left(\frac{(1-x)^{n+2}}{n+2}\cdot x^{m-1}-\left(\frac{1}{n+2}\cdot (m-1)\int (1-x)^{n+2}\cdot x^{m-2}dx\right)\right)=\frac{(1-x)^{n+1}}{n+1}\cdot x^m-\left(\frac{1}{n+1}\cdot m\left(\frac{(1-x)^{n+2}}{n+2}\cdot x^{m-1}-\left(\frac{1}{n+2}\cdot (m-1)\int (1-x)^{n+2}\cdot x^{m-2}dx\right)\right)\right)$$

Do I see a pattern after integrating this a few times, because I don't really come to the conclusion, that this would equal $$I_{m,n}=\frac{m!\,n!}{(m+n+1)!}$$. I've integrated two times now.

$$I_{m+1,n} = \frac{m+1}{n+1}\int_0^1 x^m(1-x)^{n+1}dx$$ by IBP.

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

Thus $$I_{m+1,n} = \frac{m+1}{m+n+2}I_{m,n}$$. We may proceed by induction on $$m$$.

By symmetry we apply the same argument in $$n$$ and thus get the result.

• Thank you, I'll try to go on. Dec 10, 2019 at 19:13

An alternate method.

Define the Gamma function: $$\Gamma(s)=\int_0^\infty t^{s-1}e^{-t}dt\qquad s>0.$$ We can show from IBP that $$\Gamma(s+1)=s\Gamma(s)$$, and it is easy to show that $$\Gamma(1)=1$$. Hence $$\Gamma(n)=(n-1)!$$ for integers $$n\ge1$$. We then see that $$\Gamma(s)=2\int_0^\infty x^{2s-1}e^{-x^2}dx.$$ Thus $$\Gamma(a)\Gamma(b)=4\int_0^\infty \int_0^\infty x^{2a-1}y^{2b-1}e^{-(x^2+y^2)}dxdy.$$ Then convert to polar coordinates to get \begin{align} \Gamma(a)\Gamma(b)&=4\int_0^{\pi/2}\int_0^\infty (r\cos\theta)^{2a-1}(r\sin\theta)^{2b-1}e^{-r^2}rdrd\theta\\ &=4\int_0^{\pi/2}\int_0^\infty \cos(\theta)^{2a-1}\sin(\theta)^{2b-1}r^{2(a+b)-1}e^{-r^2}drd\theta\\ &=\left(2\int_0^\infty r^{2(a+b)-1}e^{-r^2}dr\right)\left(2\int_0^{\pi/2}\cos(\theta)^{2a-1}\sin(\theta)^{2b-1}d\theta\right)\\ &=2\Gamma(a+b)\int_0^{\pi/2}\cos(\theta)^{2a-1}\sin(\theta)^{2b-1}d\theta. \end{align} In other words, we have $$2\int_0^{\pi/2}\cos(\theta)^{2a-1}\sin(\theta)^{2b-1}d\theta=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=2\int_0^{\pi/2}\cos(\theta)^{2b-1}\sin(\theta)^{2a-1}d\theta.$$ This is actually very convenient, because we can set $$t=\sin^2\theta$$ so that $$dt=2\sin\theta\cos\theta d\theta$$ and we get $$\int_0^1 t^{a-1}(1-t)^{b-1}dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\qquad a,b>0.$$ we then take $$a=n+1$$ and $$b=m+1$$ for integers $$n,m>0$$ and get $$I_{n,m}=\int_0^1 t^n(1-t)^{m}dt=\frac{n!m!}{(n+m+1)!}.$$

• That's cool and I really appreciate your work, but we did not define the Gamma-Function in our lecture (yet) and that's why I'm not sure if I can use this fact in order to prove the equation. Dec 10, 2019 at 19:12
• @Doesbaddel I'm glad you like it :). Since there was already an answer with an induction proof I decided to take this route. If you ask me, I think you should be allowed to solve the problem whichever way you want. Math is all about seeing how different solutions to the same problem relate to each other. Dec 10, 2019 at 19:23
• Yeah you're right, thank for showing a different solution to me! Dec 10, 2019 at 19:30