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Things to remark: $\Gamma(x) = \int_0^{\infty} t^{x-1}e^{-t}dt.$

$\int_0^1 t^{x-1}(1-t)^{y-1}dt $ = $\frac {\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ = $B(x,y)$. enter image description here

I don't understand from where do $(102)$ follow?

There is the $8.19$ theorem: enter image description here

Any help would be appreciated.

  • https://math.stackexchange.com/q/846626/321264, https://math.stackexchange.com/q/2098859/321264, https://proofwiki.org/wiki/Legendre's_Duplication_Formula. – StubbornAtom Nov 27 '21 at 14:20
  • The idea is to define $f(x)$ as $\frac{2^{x-1}}{\sqrt{\pi}} \Gamma \left( \frac{x}{2} \right) \Gamma \left( \frac{x+1}{2} \right)$ and then show that $f(x)$ satisfies the three conditions of Theorem 8.19. – awkward Nov 27 '21 at 15:30
  • ^ Property (a) follows immediately from the definition once use the fact that $\Gamma(z+1)=z\Gamma(z)$ for all $z\in (0,\infty)$ (in particular when $z=\frac{x}{2}$). Property (b) follows from $2^0=1$, $\Gamma(\frac{1}{2})=\sqrt{\pi}$ (which is property (99)), and that $\Gamma(1)=1$ (clear from integral definition). Property (c) is also easy to verify; just take the $\log$ of the RHS and show it is strictly convex (for example show the second derivative is strictly positive) using the known fact that $\log\circ \Gamma$ is a smooth strictly convex function. – peek-a-boo Nov 27 '21 at 15:39

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