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Question Number 1643 by 112358 last updated on 28/Aug/15
Calculate I(a,b)=∫_0 ^1 t^a (1−t)^b dt given that I(a,b)=(b/(a+1))I(a+1,b−1) (a>0,b>0). Use the fact that I(a,b)=I(a+1,b)+I(a,b+1) and I(a,b)=I(b,a) to help evaluate I(a,b).
$${Calculate}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{I}\left({a},{b}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{a}} \left(\mathrm{1}−{t}\right)^{{b}} {dt} \\ $$$${given}\:{that}\:{I}\left({a},{b}\right)=\frac{{b}}{{a}+\mathrm{1}}{I}\left({a}+\mathrm{1},{b}−\mathrm{1}\right) \\ $$$$\left({a}>\mathrm{0},{b}>\mathrm{0}\right).\:{Use}\:{the}\:{fact}\:{that} \\ $$$${I}\left({a},{b}\right)={I}\left({a}+\mathrm{1},{b}\right)+{I}\left({a},{b}+\mathrm{1}\right) \\ $$$${and}\:{I}\left({a},{b}\right)={I}\left({b},{a}\right)\: \\ $$$${to}\:{help}\:{evaluate}\:{I}\left({a},{b}\right).\: \\ $$
Answered by 123456 last updated on 29/Aug/15
I(a,b)=∫_0 ^1 t^a (1−t)^b dt=∫_0 ^1 t^(a+1−1) (1−t)^(b+1−1) dt I(a,b)=B(a+1,b+1)=((Γ(a+1)Γ(b+1))/(Γ(a+b+2)))=((a!b!)/((a+b+1)!)) −−−−−−−−−−−−− I(a+1,b)+I(a,b+1)=((Γ(a+2)Γ(b+1))/(Γ(a+b+3)))+((Γ(a+1)Γ(b+2))/(Γ(a+b+3))) =((Γ(a+2)Γ(b+1)+Γ(a+1)Γ(b+2))/(Γ(a+b+3))) =((Γ(a+1+1)Γ(b+1)+Γ(a+1)Γ(b+1+1))/(Γ(a+b+2+1))) =(((a+1)Γ(a+1)Γ(b+1)+(b+1)Γ(a+1)Γ(b+1))/((a+b+2)Γ(a+b+2))) =(((a+b+2)Γ(a+1)Γ(b+1))/((a+b+2)Γ(a+b+2))) =((Γ(a+1)Γ(b+1))/(Γ(a+b+2))) =I(a,b) (b/(a+1))I(a+1,b−1)=(b/(a+1))∙((Γ(a+2)Γ(b))/(Γ(a+b+2))) =((Γ(a+1+1)bΓ(b))/((a+1)Γ(a+b+2))) =(((a+1)Γ(a+1)Γ(b+1))/((a+1)Γ(a+b+2))) =((Γ(a+1)Γ(b+1))/(Γ(a+b+2))) =I(a,b)
$$\mathrm{I}\left({a},{b}\right)=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{t}^{{a}} \left(\mathrm{1}−{t}\right)^{{b}} {dt}=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{t}^{{a}+\mathrm{1}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{{b}+\mathrm{1}−\mathrm{1}} {dt} \\ $$$$\mathrm{I}\left({a},{b}\right)=\mathrm{B}\left({a}+\mathrm{1},{b}+\mathrm{1}\right)=\frac{\Gamma\left({a}+\mathrm{1}\right)\Gamma\left({b}+\mathrm{1}\right)}{\Gamma\left({a}+{b}+\mathrm{2}\right)}=\frac{{a}!{b}!}{\left({a}+{b}+\mathrm{1}\right)!} \\ $$$$−−−−−−−−−−−−− \\ $$$$\mathrm{I}\left({a}+\mathrm{1},{b}\right)+\mathrm{I}\left({a},{b}+\mathrm{1}\right)=\frac{\Gamma\left({a}+\mathrm{2}\right)\Gamma\left({b}+\mathrm{1}\right)}{\Gamma\left({a}+{b}+\mathrm{3}\right)}+\frac{\Gamma\left({a}+\mathrm{1}\right)\Gamma\left({b}+\mathrm{2}\right)}{\Gamma\left({a}+{b}+\mathrm{3}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\Gamma\left({a}+\mathrm{2}\right)\Gamma\left({b}+\mathrm{1}\right)+\Gamma\left({a}+\mathrm{1}\right)\Gamma\left({b}+\mathrm{2}\right)}{\Gamma\left({a}+{b}+\mathrm{3}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\Gamma\left({a}+\mathrm{1}+\mathrm{1}\right)\Gamma\left({b}+\mathrm{1}\right)+\Gamma\left({a}+\mathrm{1}\right)\Gamma\left({b}+\mathrm{1}+\mathrm{1}\right)}{\Gamma\left({a}+{b}+\mathrm{2}+\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({a}+\mathrm{1}\right)\Gamma\left({a}+\mathrm{1}\right)\Gamma\left({b}+\mathrm{1}\right)+\left({b}+\mathrm{1}\right)\Gamma\left({a}+\mathrm{1}\right)\Gamma\left({b}+\mathrm{1}\right)}{\left({a}+{b}+\mathrm{2}\right)\Gamma\left({a}+{b}+\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({a}+{b}+\mathrm{2}\right)\Gamma\left({a}+\mathrm{1}\right)\Gamma\left({b}+\mathrm{1}\right)}{\left({a}+{b}+\mathrm{2}\right)\Gamma\left({a}+{b}+\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\Gamma\left({a}+\mathrm{1}\right)\Gamma\left({b}+\mathrm{1}\right)}{\Gamma\left({a}+{b}+\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{I}\left({a},{b}\right) \\ $$$$\frac{{b}}{{a}+\mathrm{1}}\mathrm{I}\left({a}+\mathrm{1},{b}−\mathrm{1}\right)=\frac{{b}}{{a}+\mathrm{1}}\centerdot\frac{\Gamma\left({a}+\mathrm{2}\right)\Gamma\left({b}\right)}{\Gamma\left({a}+{b}+\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\Gamma\left({a}+\mathrm{1}+\mathrm{1}\right){b}\Gamma\left({b}\right)}{\left({a}+\mathrm{1}\right)\Gamma\left({a}+{b}+\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({a}+\mathrm{1}\right)\Gamma\left({a}+\mathrm{1}\right)\Gamma\left({b}+\mathrm{1}\right)}{\left({a}+\mathrm{1}\right)\Gamma\left({a}+{b}+\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\Gamma\left({a}+\mathrm{1}\right)\Gamma\left({b}+\mathrm{1}\right)}{\Gamma\left({a}+{b}+\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{I}\left({a},{b}\right) \\ $$
Commented by 123456 last updated on 29/Aug/15
B(x,y)=∫_0 ^1 t^(x−1) (1−t)^(y−1) dt this is the beta function Γ(x)=∫_0 ^(+∞) t^(x−1) e^(−t) dt this is the gamma function there a relation betwen these two function B(x,y)=((Γ(x)Γ(y))/(Γ(x+y))) Γ(x+1)=xΓ(x) also there a relation betwen gamma function and factorial Γ(x+1)=x!
$$\mathrm{B}\left({x},{y}\right)=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{t}^{{x}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{{y}−\mathrm{1}} {dt} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{beta}\:\mathrm{function} \\ $$$$\Gamma\left({x}\right)=\underset{\mathrm{0}} {\overset{+\infty} {\int}}{t}^{{x}−\mathrm{1}} {e}^{−{t}} {dt} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{gamma}\:\mathrm{function} \\ $$$$\mathrm{there}\:\mathrm{a}\:\mathrm{relation}\:\mathrm{betwen}\:\mathrm{these}\:\mathrm{two}\:\mathrm{function} \\ $$$$\mathrm{B}\left({x},{y}\right)=\frac{\Gamma\left({x}\right)\Gamma\left({y}\right)}{\Gamma\left({x}+{y}\right)} \\ $$$$\Gamma\left({x}+\mathrm{1}\right)={x}\Gamma\left({x}\right) \\ $$$$\mathrm{also}\:\mathrm{there}\:\mathrm{a}\:\mathrm{relation}\:\mathrm{betwen}\:\mathrm{gamma}\:\mathrm{function}\:\mathrm{and}\:\mathrm{factorial} \\ $$$$\Gamma\left({x}+\mathrm{1}\right)={x}! \\ $$
Commented by 112358 last updated on 29/Aug/15
That′s cool stuff. Thanks for the information.
$${That}'{s}\:{cool}\:{stuff}.\:{Thanks}\:{for}\:{the} \\ $$$${information}.\: \\ $$

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