Question Number 28442 by abdo imad last updated on 25/Jan/18
$${let}\:{give}\:{B}\left({x},{y}\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{u}^{{x}−\mathrm{1}} \left(\mathrm{1}−{u}\right)^{{y}−\mathrm{1}} {du}\:\:{and}\:\left({beta}\:{function}\right) \\ $$ $${and}\:\Gamma\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:{u}^{{x}−\mathrm{1}} \:{e}^{−{u}} \:{du}\:\:\:\:\:\left({x}>\mathrm{0}\right)\left({gamma}\:{function}\:{of}\:{euler}\right) \\ $$ $$\left.\mathrm{1}\right)\:{prove}\:{that}\:\:\:\:\Gamma\left({x}\right)=\:\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:{u}^{\mathrm{2}{x}−\mathrm{1}} \:{e}^{−{u}^{\mathrm{2}} \:} {du}\:. \\ $$ $$\left.\mathrm{2}\right)\:{prove}\:{that}\:\:{B}\left({x},{y}\right)\:=\:\frac{\Gamma\left({x}\right).\Gamma\left({y}\right)}{\Gamma\left({x}+{y}\right)}\:. \\ $$ $$ \\ $$