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If $n$ and $m$ are positive and integer numbers prove that

$$\int_0^1 x^m \cdot(1-x)^ndx = \int _0^1 x^n \cdot(1-x)^mdx.$$

I have tried to substitute $u=1-x$ but I got nowhere.

farruhota
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G.nino
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    The substitution $u=1-x$ suffices. – Mark Viola Jun 16 '17 at 03:23
  • Does the star denote a convolution or a multiplication? –  Jun 16 '17 at 04:50
  • Use $$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx,$$ See also : https://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3xdx/439856#439856 https://math.stackexchange.com/questions/1073120/integral-int-12011-frac-sqrtx-sqrt2012-x-sqrtxdx/1073121#1073121 https://math.stackexchange.com/questions/578957/definite-integral-int-24-frac-sqrt-log9-x-sqrt-log9-x-sqrt-log3/578960#578960 – lab bhattacharjee Jun 16 '17 at 05:12
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    Where did you actually get stuck? The substitution $u=1-x$ works perfectly. – Jack D'Aurizio Jun 16 '17 at 08:43
  • I substituted but totally forgot to evaluate the integration limits in u – G.nino Jun 16 '17 at 16:29

3 Answers3

6

The substitution is correct.

Let $u=1-x$. Then

\begin{align} \int_0^1x^m(1-x)^ndx&=\int_1^0(1-u)^mu^n(-1)du\\ &=\int_0^1(1-u)^mu^ndu\\ &=\int_0^1x^n(1-x)^mdx\\ \end{align}

CY Aries
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Your substitution is correct:

$$\begin{align}\int\limits_0^1 x^m (1-x)^n \,\mathrm d\,x ~&=~ \int\limits_{1}^{0} (1-u)^m u^n (-\mathrm d u) &&[x\gets 1-u, \mathrm dx\gets (-\mathrm du)] \\ &= \int\limits_0^1 (1-u)^mu^n\,\mathrm d\,u \\ &=\int\limits_0^1 (1-x)^mx^n\,\mathrm d\,x && \text{alpha replacement}\end{align}$$

Graham Kemp
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Max
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Alternatively: Note the Beta and Gamma functions are related as follows: $$B(x,y)=\int_0^1t^{x-1}(1-t)^{y-1}dt=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}=\frac{(x-1)!(y-1)!}{(x+y-1)!}.$$

Hence: $$B(m+1,n+1)=\int_0^1x^m(1-x)^ndx=\frac{m!\cdot n!}{(m+n+1)!}=$$ $$\frac{n!\cdot m!}{(n+m+1)!}=\int_0^1x^n(1-x)^mdx=B(n+1,m+1).$$

farruhota
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