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If $a, b, \in R^+$, then prove that: $$\int_{0}^{1} x^a (1-x)^b dx = \int_{0}^{1} x^b(1-x)^a dx$$

My solution:

Choose u = 1 - x $\Leftrightarrow$ x = u - 1, then -du = dx.

$$-\int_{1}^{0} u^b (1-u)^a du = \int_{0}^{1} u^b (1-u)^a du$$

So I got the u value to match but idk if I can let u = x.

2 Answers 2

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Solution 1:

Let $u = 1-x \implies du = -dx$

Our integral becomes $$\int_{1}^{0} (1-u)^a (u)^b (-du)$$ The negative differential flips the integral back and we are done.

This is exactly what you did, but to answer

idk if I can let u = x.

We first start by noting that we can let $x=1-u$, we note that this is a bijective substitution on the interval $[0,1]$. We also note that integrals allow dummy variable substitutions, so there is nothing special about the symbols $x$ or $u$, and so we can change them to be whatever we like.

Solution 2:

Our function is the integral definition of the Beta function, which is symmetric in $a,b$ because $$\operatorname{B}(a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} = \frac{\Gamma(b)\Gamma(a)}{\Gamma(b+a)} = \operatorname{B}(b,a)$$

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    Whats the difference from what I did and what you did – 2017-02-11
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    @user349557 Absolutely nothing. That is the point - you did it all correctly :) The reason you can substitute $x = 1-u$ is because this is a bijective relationship on $[0,1]$ – 2017-02-11
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    Oh okay. So the u is the same as the x? – 2017-02-11
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    @user349557 Oh, I see your concern. Yeah, we are just renaming $u$ back to $x$ – 2017-02-11
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    @user349557 I included a bit about this in my post. Sorry for the confusion. – 2017-02-11
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As the result is that $$\int_{0}^{1} x^a (1-x)^b dx=\int_{0}^{1} u^b (1-u)^a du$$ and the variables on both sides of the equal sign will be replaced by excuting the integration, it does not matter if you rename $u \to x$.