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Question Number 108283 by Ar Brandon last updated on 16/Aug/20

Given the function Γ defined by Γ(x)=∫_0 ^(+∞) t^(x−1) e^(−t) dt  1.  What is the domain of definition of Γ ?  2.  Show that ∀x∈ DΓ, xΓ(x)=Γ(x+1) and deduce the value of Γ(n), n∈N^∗   3.  Assuming ∫_0 ^(+∞) e^(−u^2 ) =((√π)/2), calculate Γ((1/2)) and deduce that       Γ(n+(1/2))=(((2n)!(√π))/(2^2^n  n!))

GiventhefunctionΓdefinedbyΓ(x)=0+tx1etdt1.WhatisthedomainofdefinitionofΓ?2.ShowthatxDΓ,xΓ(x)=Γ(x+1)anddeducethevalueofΓ(n),nN3.Assuming0+eu2=π2,calculateΓ(12)anddeducethatΓ(n+12)=(2n)!π22nn!

Answered by mathmax by abdo last updated on 16/Aug/20

1)Γ(x)=∫_0 ^∞ t^(x−1)  e^(−t) dt =∫_0 ^1  t^(x−1)  e^(−t)  dt +∫_1 ^(+∞)  t^(x−1)  e^(−t)  dt  at v(0)  t^(x−1)  e^(−t)  ∼t^(x−1)  and ∫_0 ^1  t^(x−1) dt =∫_0 ^1  (dt/t^(1−x) ) conv ⇔1−x<1 ⇒x>0  at  +∞  we have  lim_(t→+∞) t^2  t^(x−1 )  e^(−t)  =lim_(t→0)   t^(x+1)  e^(−t)  =0 ⇒  ∫_1 ^∞  t^(x−1)  e^(−t)  dt converge for x>0 ⇒D_Γ =]0,+∞[  2)Γ(x+1) =∫_0 ^∞  t^x  e^(−t)  dt =_(bypsrts)     [−t^x  e^(−t) ]_(t=0) ^∞ +∫_0 ^∞  xt^(x−1)  e^(−t) dt  =x ∫_0 ^∞  t^(x−1)  e^(−t)  dt =xΓ(x) ⇒for n natural  Γ(n+1) =nΓ(n−1)=n(n−1)Γ(n−1)=...=n!Γ(1)=n!  generaly Γ(x+n) =(x+n−1)(x+n−2)....(x+1)xΓ(x)  3)Γ((1/2)) =∫_0 ^∞  t^(−(1/2))  e^(−t)  dt =∫_0 ^∞   (e^(−t) /(√t))dt =_((√t)=u)   ∫_0 ^∞  (e^(−u^2 ) /u)(2u)du  =2∫_0 ^∞  e^(−u^2 ) du =2.((√π)/2) =(√π) ⇒Γ((1/2))=(√π)  4) Γ((1/2)+n) =((1/2)+n−1)((1/2)+n−2).....((1/2)+1)(1/2)Γ((1/2))  =((2n−1)/2).((2n−3)/2) .......(3/2).(1/2)(√π)  =((1.3.5.....(2n−3)(2n−1))/2^n )(√π)  =((1.2.3.4.5.....(2n−1)(2n))/(2^n (2.4.6......(2n))))(√π) =(((2n)!)/(2^(2n) n!))(√π)

1)Γ(x)=0tx1etdt=01tx1etdt+1+tx1etdtatv(0)tx1ettx1and01tx1dt=01dtt1xconv1x<1x>0at+wehavelimt+t2tx1et=limt0tx+1et=01tx1etdtconvergeforx>0DΓ=]0,+[2)Γ(x+1)=0txetdt=bypsrts[txet]t=0+0xtx1etdt=x0tx1etdt=xΓ(x)fornnaturalΓ(n+1)=nΓ(n1)=n(n1)Γ(n1)=...=n!Γ(1)=n!generalyΓ(x+n)=(x+n1)(x+n2)....(x+1)xΓ(x)3)Γ(12)=0t12etdt=0ettdt=t=u0eu2u(2u)du=20eu2du=2.π2=πΓ(12)=π4)Γ(12+n)=(12+n1)(12+n2).....(12+1)12Γ(12)=2n12.2n32.......32.12π=1.3.5.....(2n3)(2n1)2nπ=1.2.3.4.5.....(2n1)(2n)2n(2.4.6......(2n))π=(2n)!22nn!π

Commented by Ar Brandon last updated on 16/Aug/20

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