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On wikipedia the proof that $$ B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} $$ puts the whole expression under a double integral.

Would it be possible to prove the fact that $$\int_0^1 t^{x-1}(1-t)^{y-1}dt = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ By expanding the integral and having a sum? I have tried to do so but have been unsuccessful many times.

Stanislas CASTELLANA
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  • Since $\operatorname{B}(x,,y)=2\int_0^{\pi/2}\sin^{2x-1}u\cos^{2y-1}udu$, $\operatorname{B}(\tfrac12,,\tfrac12)=\pi$. If you can write this in terms of Gamma functions, you get $\Gamma(\tfrac12)=\sqrt{\pi}$, which is equivalent to evaluating the Gaussian integral. Almost every well-known proof of this uses a double integral, but there are exceptions. A generalization thereof might be your best hope, but you can hopefully now see why it's difficult. – J.G. Dec 26 '20 at 14:08
  • See this , this and this and linked questions therein. – V.G Dec 26 '20 at 14:10

2 Answers2

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For any real $ x>0 $ we have : $$ \Gamma\left(x\right)=\displaystyle\int_{0}^{+\infty}{z^{x-1}\mathrm{e}^{-z}\,\mathrm{d}z} $$

Thus, if $ u \in \left[0,+\infty\right[ $, and for any $ y>0 $ we have :

\begin{aligned} \Gamma\left(x\right)u^{y-1}\mathrm{e}^{-u}&=u^{y-1}\mathrm{e}^{-u}\displaystyle\int_{0}^{+\infty}{z^{x-1}\mathrm{e}^{- z}\,\mathrm{d}z} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\lbrace\begin{matrix}z =u t\ \\ \mathrm{d}z =u\,\mathrm{d}t\end{matrix}\right.\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Gamma\left(x\right)u^{y-1}\mathrm{e}^{-u}&=u^{x+y-1}\mathrm{e}^{-u}\displaystyle\int_{0}^{+\infty}{t^{x-1}\mathrm{e}^{-u t}\,\mathrm{d}t} \\ \Longrightarrow\displaystyle\int_{0}^{+\infty}{\Gamma\left(x\right)u^{y-1}\mathrm{e}^{-u}\,\mathrm{d}u}&=\displaystyle\int_{0}^{+\infty}{u^{x+y-1}\mathrm{e}^{-u}\displaystyle\int_{0}^{+\infty}{t^{x-1}\mathrm{e}^{-u t}\,\mathrm{d}t}\,\mathrm{d}u}\\ \iff\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Gamma\left(x\right)\Gamma\left(y\right)&=\displaystyle\int_{0}^{+\infty}{t^{x-1}\displaystyle\int_{0}^{+\infty}{u^{x+y-1}\mathrm{e}^{-\left(1+t\right)u}\,\mathrm{d}u}\,\mathrm{d}t} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\lbrace\begin{matrix}\tau =\left(1+t\right)u \\ \mathrm{d}u =\frac{\mathrm{d}\tau}{1+t}\ \ \ \ \ \ \ \ \ \ \ \end{matrix}\right.\\ &=\displaystyle\int_{0}^{+\infty}{\displaystyle\frac{t^{x-1}}{\left(1+t\right)^{x+y}}\displaystyle\int_{0}^{+\infty}{\tau^{x+y-1}\mathrm{e}^{-\tau}\,\mathrm{d}\tau}\,\mathrm{d}t}\\ &=\Gamma\left(x+y\right)\displaystyle\int_{0}^{+\infty}{\displaystyle\frac{t^{x-1}}{\left(1+t\right)^{x+y}}\,\mathrm{d}t} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\lbrace\begin{matrix}\varphi =\frac{1}{1+t}\ \ \\ \mathrm{d}t =-\frac{\mathrm{d}\varphi}{\varphi^{2}}\end{matrix}\right. \\\Gamma\left(x\right)\Gamma\left(y\right)&=\Gamma\left(x+y\right)\displaystyle\int_{0}^{1}{\displaystyle\frac{1}{\varphi^{2}}\left(\displaystyle\frac{1}{\varphi}-1\right)^{x-1}\varphi^{x+y}\,\mathrm{d}\varphi}\\ \iff\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \displaystyle\frac{\Gamma\left(x\right)\Gamma\left(y\right)}{\Gamma\left(x+y\right)}&=\displaystyle\int_{0}^{1}{\varphi^{y-1}\left(1-\varphi\right)^{x-1}\,\mathrm{d}\varphi}\end{aligned}

CHAMSI
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Too long for a comment:

The integral is $\sum_{n\ge0}\frac{(-1)^n(y-1)_n}{n!}\int_0^1t^{n+x-1}dt=\sum_n\frac{(-1)^n(y-1)_n}{n!(n+x)}$ in terms of falling Pochhammer symbols. Since $(y-1)_n=\frac{\Gamma(y)}{\Gamma(y-n)}$ (if $y-n$ is an integer $\le0$ the denominator diverges so the symbol is $0$, or we could just truncate the infinite series), we want to prove $\sum_n\frac{(-1)^n}{n!(n+x)\Gamma(y-n)}=\frac{\Gamma(x)}{\Gamma(x+y)}$. In terms of the Hankel contour, the series is$$\frac{i}{2\pi}\oint_H\sum_n\frac{t^n}{n!(n+x)}(-t)^{-y}e^{-t}dt=\frac{i}{2\pi}\oint_H\Gamma(x,\,0,\,-t)(-t)^{-x-y}e^{-t}dt,$$with the last $\Gamma$ a generalized incomplete Gamma. I expect this can be evaluated to the desired result with the residue theorem.

J.G.
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