Menu Close

calculate-0-ln-x-1-x-3-dx-




Question Number 79096 by mathmax by abdo last updated on 22/Jan/20
calculate ∫_0 ^∞  ((ln(x))/((1+x)^3 ))dx
calculate0ln(x)(1+x)3dx
Commented by mind is power last updated on 22/Jan/20
sam idee will worck  for  ∫_0 ^(+∞) ((ln(x)^p )/((x+1)^r ))dx  withe r>1,p∈N  ln(x)^p =(∂^p /∂t^p )x^t ∣_(t=0)    and sam idea like previous one  but we will have mor complicated function Ψ_p (x)....
samideewillworckfor0+ln(x)p(x+1)rdxwither>1,pNln(x)p=ptpxtt=0andsamidealikepreviousonebutwewillhavemorcomplicatedfunctionΨp(x).
Answered by mind is power last updated on 22/Jan/20
ln(x)=(d/dx)x^t ∣_(t=0)   f(t)=∫_0 ^(+∞) (x^t /((1+x)^3 ))dx,we want f′(0)  x=tg^2 (y)⇒  f(t)=∫_0 ^(π/2) ((tg^(2t) (y))/((1+tg^2 (y))^3 )).2(1+tg^2 (y))tg(y)dy  =2∫_0 ^(π/2) ((tg^(2t+1) (y))/((1+tg^2 (y))^2 ))dy  =2∫_0 ^(π/2) sin^(2t+1) (y)cos^(−2t+3) (y)dy  =β(t+1,−t+2)=((Γ(t+1)Γ(2−t))/(Γ(3)))=((Γ(t+1)Γ(2−t))/2)  f(t)=((Γ(t+1)Γ(2−t))/2)  f′(t)=(1/2){Γ′(t+1)Γ(2−t)−Γ′(2−t)Γ(t+1)}  =(1/2){Ψ(t+1)Γ(t+1)Γ(2−t)−Ψ(2−t)Γ(2−t)Γ(t+1)}  f′(0)=(1/2){Ψ(1)−Ψ(2)}=(1/2){−γ−(1−γ)}=−(1/2)=−0.5
ln(x)=ddxxtt=0f(t)=0+xt(1+x)3dx,wewantf(0)x=tg2(y)f(t)=0π2tg2t(y)(1+tg2(y))3.2(1+tg2(y))tg(y)dy=20π2tg2t+1(y)(1+tg2(y))2dy=20π2sin2t+1(y)cos2t+3(y)dy=β(t+1,t+2)=Γ(t+1)Γ(2t)Γ(3)=Γ(t+1)Γ(2t)2f(t)=Γ(t+1)Γ(2t)2f(t)=12{Γ(t+1)Γ(2t)Γ(2t)Γ(t+1)}=12{Ψ(t+1)Γ(t+1)Γ(2t)Ψ(2t)Γ(2t)Γ(t+1)}f(0)=12{Ψ(1)Ψ(2)}=12{γ(1γ)}=12=0.5
Commented by msup trace by abdo last updated on 23/Jan/20
thank you sir now try to solve it  complex analysis...
thankyousirnowtrytosolveitcomplexanalysis

Leave a Reply

Your email address will not be published. Required fields are marked *