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Question Number 108283 by Ar Brandon last updated on 16/Aug/20
GiventhefunctionΓdefinedbyΓ(x)=∫0+∞tx−1e−tdt1.WhatisthedomainofdefinitionofΓ?2.Showthat∀x∈DΓ,xΓ(x)=Γ(x+1)anddeducethevalueofΓ(n),n∈N∗3.Assuming∫0+∞e−u2=π2,calculateΓ(12)anddeducethatΓ(n+12)=(2n)!π22nn!
Answered by mathmax by abdo last updated on 16/Aug/20
1)Γ(x)=∫0∞tx−1e−tdt=∫01tx−1e−tdt+∫1+∞tx−1e−tdtatv(0)tx−1e−t∼tx−1and∫01tx−1dt=∫01dtt1−xconv⇔1−x<1⇒x>0at+∞wehavelimt→+∞t2tx−1e−t=limt→0tx+1e−t=0⇒∫1∞tx−1e−tdtconvergeforx>0⇒DΓ=]0,+∞[2)Γ(x+1)=∫0∞txe−tdt=bypsrts[−txe−t]t=0∞+∫0∞xtx−1e−tdt=x∫0∞tx−1e−tdt=xΓ(x)⇒fornnaturalΓ(n+1)=nΓ(n−1)=n(n−1)Γ(n−1)=...=n!Γ(1)=n!generalyΓ(x+n)=(x+n−1)(x+n−2)....(x+1)xΓ(x)3)Γ(12)=∫0∞t−12e−tdt=∫0∞e−ttdt=t=u∫0∞e−u2u(2u)du=2∫0∞e−u2du=2.π2=π⇒Γ(12)=π4)Γ(12+n)=(12+n−1)(12+n−2).....(12+1)12Γ(12)=2n−12.2n−32.......32.12π=1.3.5.....(2n−3)(2n−1)2nπ=1.2.3.4.5.....(2n−1)(2n)2n(2.4.6......(2n))π=(2n)!22nn!π
Commented by Ar Brandon last updated on 16/Aug/20
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