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Consider the integral $$A=\int_{0}^{1} x^n(1-x)^ndx$$ Prove that $A^{-1}$ is a natural number.
I have no idea how to compute the integral. But by putting $n=1$ I got $A^{-1}=6$ a natural number. Is there any simple method to show that $A^{-1}$ is a natural number?

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  • $\begingroup$ A rather simple way to evaluate the integral is performing integration by parts $n$ times. Each time the primitive part vanishes $\endgroup$
    – b00n heT
    Commented Jun 5, 2019 at 20:32

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Expanding on my comment: \begin{align}\int_0^1x^n(1-x)^n\,dx&=\frac{x^{n+1}}{n+1}(1-x)^n\big|^{1}_{0}+\frac{n}{n+1}\int_0^1x^{n+1}(1-x)^{n-1}\,dx\\ &=\frac{n}{n+1}\int_0^1x^{n+1}(1-x)^{n-1}\,dx \end{align} now you repeat the process $n$ times so to get $$\int_0^1x^n(1-x)^n\,dx=\frac{n\cdot (n-1)\cdots 1}{(n+1)\cdot (n+2) \cdots 2 n}\int_0^1x^{2n}\,dx=\frac{n\cdot (n-1)\cdots 1}{(n+1)\cdot (n+2) \cdots 2n\cdot (2n+1)}$$ thus $$\int_0^1x^n(1-x)^n\,dx=\frac{n!n!}{(2n+1)!}$$ you shall conclude.

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There is also a way to interpret this integral as a probability and therefore obtain a closed formula instantly. Imagine choosing $n$ blue, $n$ red and one black point uniformly at random on $[0,1]$. What is the probability that there will be $n$ red, one black and $n$ blue points appearing on this segment in this order? On one hand it is our integral - if $x$ is the coordinate of the black point, $x^n(1-x)^n$ is the probability that $n$ red points will be on the left and $n$ blue points on the right. On the other hand all possible arrangements are equally likely - there are $(2n+1)!$ of them in total and $(n!)^2$ of them result in the required order.
It also generalizes nicely and allows to compute values of a Beta function for all positive integer values.

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  • $\begingroup$ Very nice, was trying to find a probability argument. $\endgroup$ Commented Jun 5, 2019 at 20:46
  • $\begingroup$ So, $A^{-1}$ is equal to the number of sequences in $\{ \mathrm{red}, \mathrm{blue}, \mathrm{black} \}^{2n+1}$ in which red and blue each appear $n$ times, and black appears 1 time. $\endgroup$ Commented Jun 5, 2019 at 21:47
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You can use this to prove $$A^{-1}=\frac{(2n+1)!}{n!^2}=(2n+1)\binom{2n}{n}.$$

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  • $\begingroup$ And to cut to the chase, so to speak, one possible proof of that involves evaluating the double integral $\int_0^\infty \int_0^\infty x^n y^n e^{-(x+y)} \,dy \,dx$ in two different ways. $\endgroup$ Commented Jun 5, 2019 at 20:37
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$$A=\int_0^1x^n(1-x)^n\mathrm dx=B(n+1,n+1)=\frac{\Gamma^2(n+1)}{\Gamma(2n+2)}=\frac1{(2n+1)\binom{2n}n}$$ Thus, $$A=\frac1{(2n+1)\binom{2n}n}~\implies~A^{-1}=(2n+1)\binom{2n}n$$

$$\therefore~A^{-1}\in\Bbb N$$

Here we have applied the integral representation of the Beta Function combined with its quite simple relation to the Gamma Function. What remains is to show that the binomial coefficient is always an integer; which is quite fundamental knowledge.

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  • $\begingroup$ Careful: $\Gamma(n)=(n-1)!$. $\endgroup$
    – J.G.
    Commented Jun 5, 2019 at 20:51
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    $\begingroup$ @J.G. I haven't done enought math for to long; sorry, I will fix this. $\endgroup$
    – mrtaurho
    Commented Jun 5, 2019 at 20:53

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