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lim-z-0-z-z-2-




Question Number 120593 by snipers237 last updated on 01/Nov/20
   lim_(z→0)   ((Γ(z)+Γ(−z))/2) =^?  −γ
limz0Γ(z)+Γ(z)2=?γ
Answered by mnjuly1970 last updated on 02/Nov/20
solution:      note:=  Γ(z+1)=zΓ(z)        note:=ψ(z)=(d/dz)(ln(Γ(z))       lim_(z→0) (((Γ(z)+Γ(−z))/2))        =lim_(z→0) ((1/z)(((zΓ(z)+zΓ(−z))/2)))  =lim_(z→0)  (1/2)(((Γ(z+1)−Γ(1−z))/z))     =^(hopital rule) lim_(z→0) (1/2)(((Γ′(z+1)+Γ′(1−z))/1))      =(1/2)(Γ′(1)+Γ^′ (1))=Γ′(1)       =Γ(1)ψ(1)=−γ : euler constant.               ...♣m.n.july.1970♣...
solution:note:=Γ(z+1)=zΓ(z)note:=ψ(z)=ddz(ln(Γ(z))limz0(Γ(z)+Γ(z)2)=limz0(1z(zΓ(z)+zΓ(z)2))=limz012(Γ(z+1)Γ(1z)z)=hopitalrulelimz012(Γ(z+1)+Γ(1z)1)=12(Γ(1)+Γ(1))=Γ(1)=Γ(1)ψ(1)=γ:eulerconstant.m.n.july.1970
Commented by I want to learn more last updated on 02/Nov/20
Help sir.
Helpsir.
Commented by I want to learn more last updated on 02/Nov/20
Commented by Dwaipayan Shikari last updated on 03/Nov/20
∫(5x^4 +4x^3 −x^2 )^(−(1/2)) dx  =∫(1/( (√(5x^4 +4x^3 −x^2 ))))dx  =∫(1/(x(√(5x^2 +4x−1))))dx=∫(1/(x(√((5x−1)(x+1)))))dx  =∫((1/x^2 )/( (√((5−(1/x))(1+(1/x))))))dx    t=1+(1/x)  =−∫(dt/( (√(t(6−t)))))  =−∫(dt/( (√(9−(t−3)^2 ))))               t−3=3sin𝛉⇒1=3cos𝛉(d𝛉/dt)  =−∫((3cos𝛉d𝛉)/( (√(9−9sin^2 𝛉))))  =−∫d𝛉=−𝛉+C  =−sin^(−1) (((t−3)/3))+C  =−sin^(−1) (((1+(1/x)−3)/3))+c  =−sin^(−1) (((−2x+1)/(3x)))+c
(5x4+4x3x2)12dx=15x4+4x3x2dx=1x5x2+4x1dx=1x(5x1)(x+1)dx=1x2(51x)(1+1x)dxt=1+1x=dtt(6t)=dt9(t3)2t3=3sinθ1=3cosθdθdt=3cosθdθ99sin2θ=dθ=θ+C=sin1(t33)+C=sin1(1+1x33)+c=sin1(2x+13x)+c
Commented by I want to learn more last updated on 03/Nov/20
Wow, thanks sir, i appreciate.
Wow,thankssir,iappreciate.
Commented by Dwaipayan Shikari last updated on 03/Nov/20
With pleasure bro. I am not a sir!
Withpleasurebro.Iamnotasir!
Commented by I want to learn more last updated on 09/Nov/20
Ohh, ma
Ohh,ma

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