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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 by123456 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 by112358 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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