Question Number 2818 by prakash jain last updated on 27/Nov/15

show that Γ′(1)=−γ Γ gamma function γ=lim_(n→∞) [H_n −ln n]

Answered by 123456 last updated on 30/Nov/15

Γ(x)=∫_0 ^(+∞) t^(x−1) e^(−t) dt (x>0) Γ′(x)=∫_0 ^(+∞) t^(x−1) ln t e^(−t) dt Γ′(1)=∫_0 ^(+∞) ln t e^(−t) dt (Q2435) e^(−t) =lim_(n→+∞) (1−(t/n))^n Γ(x)=lim_(n→+∞) ∫_0 ^n t^(x−1) (1−(t/n))^n dt u=t/n,du=dt/n t=0,u=0 t=n,u=1 Γ(x)=lim_(n→∞) ∫_0 ^1 (un)^(x−1) (1−u)^n (du∙n) =lim_(n→∞) n^x ∫_0 ^1 u^(x−1) (1−u)^n du integral by parts a=(1−u)^n ,da=−n(1−u)^(n−1) du db=u^(x−1) du,b=u^x /x =lim_(n→∞) n^x (n/x)∫_0 ^1 u^x (1−u)^(n−1) du ⋮ =lim_(n→∞) ((n^x n...1)/(x(x+1)...(x+n−1)))∫_0 ^1 u^(x+n−1) du =lim_(n→∞) ((n^x n!)/(x(x+1)...(x+n))) (Q2845) =lim_(n→∞) ((n^x n!)/(Π_(l=0) ^n x+l)) Γ(z+1)=zΓ(z) ln Γ(z+1)=ln z+ln Γ(z) =lim_(n→+∞) ln (n!n^z )−Σ_(l=1) ^n ln (z+k) ψ(z+1)=lim_(n→∞) ln n−Σ_(l=1) ^n (1/(z+l)) (1/(z+l))=(l/(l(z+l))) =(1/l)∙(l/(z+l)) =(1/l)∙((z+l−z)/(z+l)) =(1/l)(1−(z/(z+l))) =(1/l)−(z/(l(z+l))) so ψ(z+1)=lim_(n→∞) ln n−Σ_(l=1) ^n (1/l)+Σ_(l=1) ^n (z/(l(z+l))) =−lim_(n→∞) (Σ_(l=1) ^n (1/l)−ln n)+Σ_(l=1) ^(+∞) (z/(l(z+l))) =−γ+Σ_(l=1) ^(+∞) (z/(l(z+l))) take z=0 ψ(1)=−γ ψ(x)=(d/dx)ln Γ(x) ψ(x)=((Γ′(x))/(Γ(x))) Γ′(x)=ψ(x)Γ(x) Γ′(1)=ψ(1)Γ(1)=ψ(1)=−γ □