Mean of Beta distribution without using Gamma function?

Question

Is there a way to prove directly that

$$\int_0^1x^{(\alpha+1)-1}(1-x)^{\beta-1}dx=\frac{\alpha}{\alpha+\beta}\int_0^1x^{\alpha-1}(1-x)^{\beta-1}dx$$

without reference to the fact that

$$\int_0^1x^{\alpha-1}(1-x)^{\beta-1}dx=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$$

or any other Gamma function facts?

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I tried using integration by parts, but of course differentiating $x^{(\alpha+1)-1}$ meant that I had to integrate $(1-x)^{\beta-1}$, so I only managed to prove that

$$\int_0^1x^{(\alpha+1)-1}(1-x)^{\beta-1}dx=\frac{\alpha}{\beta}\int_0^1x^{\alpha-1}(1-x)^{(\beta+1)-1}dx\text{.}$$

Answer 1

Since $\frac{d}{dx}x^\alpha(1-x)^\beta = \left(\alpha-(\alpha+\beta) x\right)x^{\alpha-1}(1-x)^{b-1}$ we have $$ \int_{0}^{1}\left(\alpha-(\alpha+\beta) x\right)x^{\alpha-1}(1-x)^{\beta-1}\,dx = 0 $$ which implies $$ (\alpha+\beta)\int_{0}^{1}x^\alpha(1-x)^{\beta-1}\,dx = \alpha\int_{0}^{1}x^{\alpha-1}(1-x)^{\beta-1}\,dx $$ as wanted.

Answer 2

What you've got is enough: you can write $$ x^{\alpha-1}(1-x)^{\beta-1+1} = x^{\alpha-1}(1-x)^{\beta-1}(1-x) = x^{\alpha-1}(1-x)^{\beta-1}-x^{\alpha+1-1}(1-x)^{\beta-1}. $$ Thus combining this with the formula in the question, $$ B(\alpha+1,\beta) = \frac{\alpha}{\beta}B(\alpha,\beta+1) = \frac{\alpha}{\beta}(B(\alpha,\beta)-B(\alpha+1,\beta)) $$ and you can rearrange this to get the desired formula.