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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.
$${let}\:\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:{t}^{{x}−\mathrm{1}} \:{e}^{−{t}} \:{dt}\:{with}\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{find}\:\Gamma^{\left({n}\right)} \left({x}\right)\:{with}\:{n}\in\:{N}^{\bigstar} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\Gamma\left({n}\:+\frac{\mathrm{3}}{\mathrm{2}}\right)\:{for}\:{n}\:{integr}. \\ $$
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)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:\Gamma\left({x}\right)=\:\int_{\mathrm{0}} ^{\infty} \:{e}^{\left({x}−\mathrm{1}\right){lnt}} \:{e}^{−{t}} {dt}\:\Rightarrow \\ $$$$\Gamma^{'} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:{ln}\left({t}\right){e}^{\left({x}−\mathrm{1}\right){ln}\left({t}\right)} {e}^{−{t}} \:{dt} \\ $$$$\Gamma^{''} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \left({lnt}\right)^{\mathrm{2}} \:{e}^{\left({x}−\mathrm{1}\right){ln}\left({t}\right)} \:{e}^{−{t}} {dt}\:{and}\:{its}\:{easy}\: \\ $$$${to}\:{prove}\:{by}\:{recurrence}\:{that} \\ $$$$\Gamma^{\left({n}\right)} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \left({lnt}\right)^{{n}} \:{t}^{{x}−\mathrm{1}} \:{e}^{−{t}} \:{dt}\:\:\forall\:{n}\in\:{N}\:\:\left(\Gamma^{\left(\mathrm{0}\right)} =\Gamma\right) \\ $$
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.
$${we}\:{knew}\:{that}\:\Gamma\left({x}+\mathrm{1}\right)={x}\Gamma\left({x}\right)\:\Rightarrow \\ $$$$\Gamma\left({n}+\frac{\mathrm{3}}{\mathrm{2}}\right)\:=\Gamma\left(\:\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\:+\mathrm{1}\right)=\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({n}−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)=\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({n}−\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left({n}\:−\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$=\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({n}−\frac{\mathrm{3}}{\mathrm{2}}\right)….\left({n}−\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({n}−\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({n}\:−\frac{\mathrm{3}}{\mathrm{2}}\right)…\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:{but}\: \\ $$$$\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\int_{\mathrm{0}} ^{\infty} \:{t}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:{e}^{−{t}} {dt}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−{t}} }{\:\sqrt{{t}}}{dt} \\ $$$$=_{\sqrt{{t}}={x}} \:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−{x}^{\mathrm{2}} } }{{x}}\:\mathrm{2}{xdc}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{x}^{\mathrm{2}} } {dx}=\:\mathrm{2}\frac{\sqrt{\pi}}{\mathrm{2}}\:=\sqrt{\pi} \\ $$$$\Gamma\left({n}+\frac{\mathrm{3}}{\mathrm{2}}\right)\:=\sqrt{\pi}\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({n}−\frac{\mathrm{3}}{\mathrm{2}}\right)….\frac{\mathrm{3}}{\mathrm{2}}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${and}\:{this}\:{quantity}\:{can}\:{be}\:{given}\:{by}\:{factoriels}. \\ $$
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!)) .
$$\Gamma\left({n}+\frac{\mathrm{3}}{\mathrm{2}}\right)\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:\prod_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}} \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}}\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\left(\mathrm{3}.\mathrm{5}.\mathrm{7}…..\left(\mathrm{2}{n}+\mathrm{1}\right)\right) \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}^{{n}+\mathrm{1}} }\:\left(\mathrm{2}.\mathrm{3}.\mathrm{4}.\mathrm{5}.\mathrm{6}…..\left(\mathrm{2}{n}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)\right)\left(\mathrm{2}.\mathrm{4}.\mathrm{6}….\left(\mathrm{2}{n}\right)\right)^{−\mathrm{1}} \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}^{{n}+\mathrm{1}} }\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\mathrm{2}^{{n}} {n}!}\:=\:\frac{\sqrt{\pi}}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{{n}!}\:. \\ $$
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
$$\mathrm{2}.\lceil\left({n}+\mathrm{3}/\mathrm{2}\right) \\ $$$$\:\:\:=\lceil\left({n}+\mathrm{1}+\mathrm{1}/\mathrm{2}\right) \\ $$$$\:\:\:=\left({n}+\mathrm{1}/\mathrm{2}\right)\lceil\left({n}+\mathrm{1}/\mathrm{2}\right) \\ $$$$\:\:\:\:=\left({n}+\mathrm{1}/\mathrm{2}\right)\lceil\left({n}−\mathrm{1}/\mathrm{2}\:+\mathrm{1}\right) \\ $$$$\:\:\:\:=\left({n}+\mathrm{1}/\mathrm{2}\right)\left({n}−\mathrm{1}/\mathrm{2}\right)\lceil\left({n}−\mathrm{1}/\mathrm{2}\right) \\ $$$$\:\:\:\:\:=\left({n}+\mathrm{1}/\mathrm{2}\right)\left({n}−\mathrm{1}/\mathrm{2}\right)\left({n}−\mathrm{3}/\mathrm{2}\right)\lceil\left({n}−\mathrm{3}/\mathrm{2}\right) \\ $$$$\:\:\:\:\:=\left({n}+\mathrm{1}/\mathrm{2}\right)\left({n}−\mathrm{1}/\mathrm{2}\right)\left({n}−\mathrm{3}/\mathrm{2}\right)\left({n}−\mathrm{5}/\mathrm{2}\right)\lceil\left({n}−\mathrm{5}/\mathrm{2}\right. \\ $$$${the}\:{expression}\:{can}\:{not}\:{be}\:{in}\:{factorial}\:{form} \\ $$$${since}\:{extreme}\:{right}\:{factor}\:{can}\:{not}\:{be}\:\mathrm{1} \\ $$$$ \\ $$
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)
$$\mathrm{1}.\:\:\:\:\:\Gamma^{\left({n}\right)} \left({x}\right)=\underset{\mathrm{0}} {\overset{\infty} {\int}}{t}^{{x}−\mathrm{1}} \mathrm{ln}\left({t}\right)^{{n}} \mathrm{e}^{−{t}} {dt} \\ $$$$\mathrm{2}.\:\:\:\:\:\frac{\sqrt{\pi}}{\mathrm{2}}×\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{\mathrm{2}{i}+\mathrm{1}}{\mathrm{2}} \\ $$

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