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