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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.

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2 Answers 2

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

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  • $\begingroup$ Thank you, I'll try to go on. $\endgroup$
    – Doesbaddel
    Dec 10, 2019 at 19:13
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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 $$I_{n,m}=\int_0^1 t^n(1-t)^{m}dt=\frac{n!m!}{(n+m+1)!}.$$

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  • $\begingroup$ 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. $\endgroup$
    – Doesbaddel
    Dec 10, 2019 at 19:12
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    $\begingroup$ @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. $\endgroup$
    – clathratus
    Dec 10, 2019 at 19:23
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    $\begingroup$ Yeah you're right, thank for showing a different solution to me! $\endgroup$
    – Doesbaddel
    Dec 10, 2019 at 19:30

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