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Let $\alpha,\beta \in \mathbb{R}$. I was wondering if there is a systematic way to solve integrals of the following form:

\begin{equation} \int^1_{0} x^\alpha(1-x)^\beta dx \end{equation}

I have seen similar kind of integrals many times. Any help/hints would be really appreciated. Thanks!

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Using the definition of the Beta Function: $$B(x,y)=\int_0^1t^{x-1}(1-t)^{1-y}dt$$ By matching coefficients,we would conclude that: $$\int_0^1x^\alpha(1-x)^\beta dx=B(\alpha+1,1-\beta)$$

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Begin with the definition of the Gamma function:

$$\Gamma(\alpha+1) = \int_0^{\infty} du\; u^{\alpha} e^{-u} $$

Similarly,

$$\Gamma(\beta+1) = \int_0^{\infty} dv\; v^{\beta} e^{-v} $$

Multiply these together...

$$\begin{align}\Gamma(\alpha+1) \Gamma(\beta+1) &= \int_0^{\infty} du\; \int_0^{\infty} dv\; u^{\alpha} v^{\beta} e^{-u-v}\\ &= \frac1{2^{\alpha+\beta+1}} \int_0^{\infty} dp \; \int_{-p}^p dq \, (p+q)^{\alpha} (p-q)^{\beta} e^{-p} \\ &= \frac1{2^{\alpha+\beta+1}} \int_0^{\infty} dp \; p^{\alpha+\beta+1} e^{-p} \, \int_{-1}^1 dr \, (1+r)^{\alpha} (1-r)^{\beta}\\ &= \Gamma(\alpha+\beta+2) \frac1{2^{\alpha+\beta+1}} \int_0^2 dr \, r^{\alpha} (2-r)^{\beta} \\ &= \Gamma(\alpha+\beta+2) \int_0^1 dx \, x^{\alpha} (1-x)^{\beta}\end{align}$$

In the second line, I used the transformation $p=u+v$, $q=u-v$, which has Jacobian $1/2$. Thus,

$$\int_0^1 dx \, x^{\alpha} (1-x)^{\beta} = \frac{\Gamma(\alpha+1) \Gamma(\beta+1)}{\Gamma(\alpha+\beta+2)} $$

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