Question Number 115193 by mnjuly1970 last updated on 24/Sep/20

...advanced mathematics... :: digamma limit :: if k>0 then prove that lim_(x→0) (1/x)(ψ(((k+x)/(2x))) − ψ((k/(2x)))) =(1/k) ✓ m.n.july.1970...

Commented byTawa11 last updated on 06/Sep/21

great

Answered by mathdave last updated on 24/Sep/20

solution let I=lim_(x→0) ((1/x)(ψ((m/(2x))+(1/2))−ψ((m/(2x)))) we known ∫_0 ^1 (t^(n−1) /(1+t))dt=(1/2)(ψ((k/2)+(1/2))−ψ((k/2))) let k=(m/x) I=(2/x)∫_0 ^1 (t^((m/x)−1) /(1+t))dt z=(m/x) I=(2/m)∫_0 ^1 ((zt^(z−1) )/(1+t))dt ( let ∫dv=∫zt^(z−1) dz,v=t^z and u=(1/(1+t)),du=−(1/((1+t)^2 ))) using IBP I=(2/m)((t^z /(1+t)))_0 ^1 +(2/m)∫_0 ^1 (t^z /((1+t)^2 ))dt=(1/m)+(2/m)lim_(z→∞) ∫_0 ^1 (t^z /((1+t)^2 ))dt let y=(1/t),dy=−(1/t^2 ) I=(1/m)+(2/m)lim_(z→∞) ∫_∞ ^1 (y^(−z) /((1+(1/y))^2 ))×−(1/y^2 )dy I=(1/m)+(2/m)lim_(z→∞) ∫_0 ^∞ (y^2 /(y^z (1+y)^2 ))×(dy/y^2 )=((1/m)+(2/m)lim_(z→∞) ∫_1 ^∞ (dy/(y^z (1+y)^2 )))=(1/m) ∵lim_(x→0) (1/x)(ψ((m/(2x))+(1/2))−ψ((m/(2x))))=(1/m) Q.E.D by mathdave(24/09/2020)

Commented bymnjuly1970 last updated on 24/Sep/20

good work