Question Number 40891 by abdo.msup.com last updated on 28/Jul/18
$${let}\:{x}>\mathrm{0}\:{and}\:{y}>\mathrm{0}\:{and} \\ $$$${B}\left({x},{y}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{x}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{{y}−\mathrm{1}} {dt} \\ $$$$\left.\mathrm{1}\right){prove}\:{that}\:{B}\left({x},{y}\right)={B}\left({y},{x}\right) \\ $$$$\left.\mathrm{2}\right){B}\left({x}+\mathrm{1},{y}\right)=\frac{{x}}{{y}}\:{B}\left({x},{y}+\mathrm{1}\right) \\ $$$$\left.\mathrm{3}\right){B}\left({x}+\mathrm{1},{y}\right)=\frac{{x}}{{x}+{y}}{B}\left({x},{y}\right) \\ $$$$\left.\mathrm{4}\right){B}\left({x},{n}+\mathrm{1}\right)=\frac{{n}!}{{x}\left({x}+\mathrm{1}\right)….\left({x}+{n}\right)} \\ $$$$\left.\mathrm{5}\right){B}\left({n},{p}\right)\:=\:\frac{\mathrm{1}}{\left({n}+{p}−\mathrm{1}\right){C}_{{n}+{p}−\mathrm{2}} ^{{p}−\mathrm{1}} } \\ $$
Commented by maxmathsup by imad last updated on 03/Aug/18
![1) B(x,y) = ∫_0 ^1 t^(x−1) (1−t)^(y−1) dt =_(1−t=u) ∫_0 ^1 (1−u)^(x−1) u^(y−1) du=B(y,x) 2)B(x+1,y) = ∫_0 ^1 t^x (1−t)^(y−1) dt by parts u=t^x and v^′ =(1−t)^(y−1) B(x+1,y) =[−(1/y)(1−t)^y t^x ]_0 ^1 +∫_0 ^1 xt^(x−1) (1/y)(1−t)^y dt =(x/y) ∫_0 ^1 t^(x−1) (1−t)^y dt = (x/y) B(x,y+1)](https://www.tinkutara.com/question/Q41208.png)
$$\left.\mathrm{1}\right)\:{B}\left({x},{y}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{{x}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{{y}−\mathrm{1}} {dt}\:=_{\mathrm{1}−{t}={u}} \:\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{u}\right)^{{x}−\mathrm{1}} {u}^{{y}−\mathrm{1}} \:{du}={B}\left({y},{x}\right) \\ $$$$\left.\mathrm{2}\right){B}\left({x}+\mathrm{1},{y}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{{x}} \left(\mathrm{1}−{t}\right)^{{y}−\mathrm{1}} {dt}\:\:{by}\:{parts}\:{u}={t}^{{x}} \:{and}\:{v}^{'} =\left(\mathrm{1}−{t}\right)^{{y}−\mathrm{1}} \\ $$$${B}\left({x}+\mathrm{1},{y}\right)\:=\left[−\frac{\mathrm{1}}{{y}}\left(\mathrm{1}−{t}\right)^{{y}} {t}^{{x}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:+\int_{\mathrm{0}} ^{\mathrm{1}} \:{xt}^{{x}−\mathrm{1}} \:\frac{\mathrm{1}}{{y}}\left(\mathrm{1}−{t}\right)^{{y}} {dt} \\ $$$$=\frac{{x}}{{y}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{t}^{{x}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{{y}} {dt}\:\:=\:\frac{{x}}{{y}}\:{B}\left({x},{y}+\mathrm{1}\right) \\ $$