Integration of $x^{\alpha}(1-x)^{1-\alpha}dx$

Question

I'm trying to solve a definite integral $$\int^{1}_{0} x^{\alpha}(1-x)^{1-\alpha}dx $$ where $-1<\Re\alpha<2$ with a beta function: $$B(2-\alpha, \alpha + 1) = \frac{\Gamma(\alpha+1)\Gamma(2-\alpha)}{\Gamma(3)}$$ Is there a way to solve Gamma functions in the numerator for complex $\alpha$? Any hints would be appreciated.

Answer 1

The numerator can be further simplified assuming this is what you meant by solve Gamma functions in the numerator for complex $\alpha$. According to the functional relation of the Gamma Function combined with Euler's Reflection Formula we get

$$\begin{align} B(2-\alpha,\alpha+1)&=\frac{\Gamma(\alpha+1)\Gamma(2-\alpha)}{\Gamma(3)}\\ &=\frac12\alpha\Gamma(\alpha)(1-\alpha)\Gamma(1-\alpha)\\ &=\frac{\alpha(1-\alpha)}2\Gamma(\alpha)\Gamma(1-\alpha) \end{align}$$

$$\therefore~B(2-\alpha,\alpha+1)=\frac{\alpha(1-\alpha)}2\frac{\pi}{\sin(\pi \alpha)}$$

Note that this is only valid as long as $\alpha\notin\mathbb Z$ hence for $\alpha\in\mathbb Z$ the denominator equals $0$.

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In case you are interested in a proof of the fundamental relation between the Beta and the Gamma Function I would recommend you to take a look at J.G.'s answer.