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Question Number 33915 by prof Abdo imad last updated on 27/Apr/18

let Γ(x)=∫_0 ^∞  t^(x−1)  e^(−t)  dt with x>0  1) find Γ^((n)) (x) with n∈ N^★   2) calculate Γ(n +(3/2)) for n integr.

letΓ(x)=0tx1etdtwithx>01)findΓ(n)(x)withnN2)calculateΓ(n+32)fornintegr.

Commented by prof Abdo imad last updated on 29/Apr/18

1) we have Γ(x)= ∫_0 ^∞  e^((x−1)lnt)  e^(−t) dt ⇒  Γ^′ (x) = ∫_0 ^∞  ln(t)e^((x−1)ln(t)) e^(−t)  dt  Γ^(′′) (x) = ∫_0 ^∞ (lnt)^2  e^((x−1)ln(t))  e^(−t) dt and its easy   to prove by recurrence that  Γ^((n)) (x) = ∫_0 ^∞ (lnt)^n  t^(x−1)  e^(−t)  dt  ∀ n∈ N  (Γ^((0)) =Γ)

1)wehaveΓ(x)=0e(x1)lntetdtΓ(x)=0ln(t)e(x1)ln(t)etdtΓ(x)=0(lnt)2e(x1)ln(t)etdtanditseasytoprovebyrecurrencethatΓ(n)(x)=0(lnt)ntx1etdtnN(Γ(0)=Γ)

Commented by prof Abdo imad last updated on 29/Apr/18

we knew that Γ(x+1)=xΓ(x) ⇒  Γ(n+(3/2)) =Γ( (n+(1/2)) +1)=(n+(1/2))Γ(n+(1/2))  =(n+(1/2))Γ(n−(1/2)+1)=(n+(1/2))(n−(1/2))Γ(n−(1/2))  =(n+(1/2))(n−(1/2))(n−(3/2))Γ(n −(3/2))  =(n+(1/2))(n−(1/2))(n−(3/2))....(n−((2n−1)/2))Γ(n−((2n−1)/2))  =(n+(1/2))(n−(1/2))(n −(3/2))...(1/2)Γ((1/2))   but   Γ((1/2))=∫_0 ^∞  t^(−(1/2))  e^(−t) dt = ∫_0 ^∞    (e^(−t) /(√t))dt  =_((√t)=x)  ∫_0 ^∞    (e^(−x^2 ) /x) 2xdc =2 ∫_0 ^∞   e^(−x^2 ) dx= 2((√π)/2) =(√π)  Γ(n+(3/2)) =(√π)(n+(1/2))(n−(1/2))(n−(3/2))....(3/2)(1/2)  and this quantity can be given by factoriels.

weknewthatΓ(x+1)=xΓ(x)Γ(n+32)=Γ((n+12)+1)=(n+12)Γ(n+12)=(n+12)Γ(n12+1)=(n+12)(n12)Γ(n12)=(n+12)(n12)(n32)Γ(n32)=(n+12)(n12)(n32)....(n2n12)Γ(n2n12)=(n+12)(n12)(n32)...12Γ(12)butΓ(12)=0t12etdt=0ettdt=t=x0ex2x2xdc=20ex2dx=2π2=πΓ(n+32)=π(n+12)(n12)(n32)....3212andthisquantitycanbegivenbyfactoriels.

Commented by prof Abdo imad last updated on 29/Apr/18

Γ(n+(3/2)) =((√π)/2) Π_(k=1) ^n  ((2k+1)/2)  =((√π)/2) (1/2^n ) (3.5.7.....(2n+1))  =((√π)/2^(n+1) ) (2.3.4.5.6.....(2n)(2n+1))(2.4.6....(2n))^(−1)   =((√π)/2^(n+1) ) (((2n+1)!)/(2^n n!)) = ((√π)/2^(2n+1) ) (((2n+1)!)/(n!)) .

Γ(n+32)=π2k=1n2k+12=π212n(3.5.7.....(2n+1))=π2n+1(2.3.4.5.6.....(2n)(2n+1))(2.4.6....(2n))1=π2n+1(2n+1)!2nn!=π22n+1(2n+1)!n!.

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Apr/18

2.⌈(n+3/2)     =⌈(n+1+1/2)     =(n+1/2)⌈(n+1/2)      =(n+1/2)⌈(n−1/2 +1)      =(n+1/2)(n−1/2)⌈(n−1/2)       =(n+1/2)(n−1/2)(n−3/2)⌈(n−3/2)       =(n+1/2)(n−1/2)(n−3/2)(n−5/2)⌈(n−5/2  the expression can not be in factorial form  since extreme right factor can not be 1

2.(n+3/2)=(n+1+1/2)=(n+1/2)(n+1/2)=(n+1/2)(n1/2+1)=(n+1/2)(n1/2)(n1/2)=(n+1/2)(n1/2)(n3/2)(n3/2)=(n+1/2)(n1/2)(n3/2)(n5/2)(n5/2theexpressioncannotbeinfactorialformsinceextremerightfactorcannotbe1

Answered by MJS last updated on 27/Apr/18

1.     Γ^((n)) (x)=∫_0 ^∞ t^(x−1) ln(t)^n e^(−t) dt  2.     ((√π)/2)×Π_(i=1) ^n ((2i+1)/2)

1.Γ(n)(x)=0tx1ln(t)netdt2.π2×ni=12i+12

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