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find-value-of-0-1-ln-1-ix-2-dx-and-0-1-ln-1-ix-2-dx-with-i-1-




Question Number 79758 by mathmax by abdo last updated on 27/Jan/20
find value of ∫_0 ^1 ln(1+ix^2 )dx and ∫_0 ^1 ln(1−ix^2 )dx with i=(√(−1))
findvalueof01ln(1+ix2)dxand01ln(1ix2)dxwithi=1
Commented by mathmax by abdo last updated on 29/Jan/20
let z from C and f(z)=∫_0 ^1 ln(1+zx^2 )dx ⇒  f^′ (z)=∫_0 ^1  (x^2 /(1+zx^2 ))dx =(1/z) ∫_0 ^1 ((zx^2 +1−1)/(1+zx^2 ))dx  =(1/z) −(1/z) ∫_0 ^1  (dx/(1+zx^2 )) and   ∫_0 ^1  (dx/(1+zx^2 )) =_(x(√z)=t)   ∫_0 ^(√z)    (dt/( (√z)(1+t^2 )))  =(1/( (√z))) arctan((√z)) ⇒f^′ (z)=(1/z)−((arctan((√z)))/(z(√z))) ⇒  f(z)=lnz  +∫_1 ^z  ((arctan((√u)))/(u(√u))) du +c    (u=t^2 )  =lnz +∫_1 ^(√z)   ((arctan(t))/t^3 )(2t)dt +c =ln(z)+2∫_1 ^(√z)   ((arctan(t))/t^2 )dt +c  f(1) =∫_0 ^1 ln(1+x^2 )dx =c ⇒  f(z) =lnz +2∫_1 ^(√z)   ((arctan(t))/t^2 )dt +∫_0 ^1  ln(1+x^2 )dx  we have by parts  ∫_1 ^(√z)  ((arctan(t))/t^2 )dt =[−((arctant)/t)]_1 ^(√z) +∫_1 ^(√z)  (1/(t(1+t^2 )))dt  =(π/4)−((arctan((√z)))/( (√z))) +∫_1 ^(√z) ((1/t)−(t/(t^2 +1)))dt  =(π/4)−((arctan((√z)))/( (√z))) +ln((√z))−[(1/2)ln(t^2  +1)]_1 ^(√z)   =(π/4)−((arctan((√z)))/( (√z))) +(1/2)ln(z)−(1/2){ln(z+1)−ln(2)} ⇒  f(z)=ln(z)+(π/2)−((2arctan((√z)))/( (√z))) +ln(z)−ln(z+1)+ln(2)+∫_0 ^1 ln(1+x^2 )dx  f(z)=2ln(z)+(π/2) −2((arctan((√z)))/( (√z))) −ln(z+1)+ln(2)+∫_0 ^1 ln(1+x^2 )dx  by parts ∫_0 ^1 ln(1+x^2 )dx =[xln(1+x^2 )]_0 ^1 −∫_0 ^1 ((2x^2 )/(1+x^2 ))dx  =ln(2)−2 ∫_0 ^1  ((1+x^2 −1)/(1+x^2 ))dx =ln(2)−2 +2 ×(π/4) =ln(2)−2+(π/2) ⇒  f(z) =2ln(z) +2ln(2) +π −2 −2((arctan((√z)))/( (√z))) −ln(z+1)  =∫_0 ^1 ln(1+zx^2 )dx ⇒  ∫_0 ^1 ln(1+ix^2 )dx =f(i) =2ln(i)+2ln(2)+π−2−2 ×((arctan((√i)))/( (√i)))  −ln(i+1)  =2×((iπ)/2) +2ln(2)+π−2 −2 ×((arctan(e^((iπ)/4) ))/e^((iπ)/4) )−ln((√2)e^((iπ)/4) )  =iπ +2ln(2)+π−2 −2e^(−((iπ)/4))  arctan(e^((iπ)/4) )−(1/2)ln(2)−((iπ)/4)
letzfromCandf(z)=01ln(1+zx2)dxf(z)=01x21+zx2dx=1z01zx2+111+zx2dx=1z1z01dx1+zx2and01dx1+zx2=xz=t0zdtz(1+t2)=1zarctan(z)f(z)=1zarctan(z)zzf(z)=lnz+1zarctan(u)uudu+c(u=t2)=lnz+1zarctan(t)t3(2t)dt+c=ln(z)+21zarctan(t)t2dt+cf(1)=01ln(1+x2)dx=cf(z)=lnz+21zarctan(t)t2dt+01ln(1+x2)dxwehavebyparts1zarctan(t)t2dt=[arctantt]1z+1z1t(1+t2)dt=π4arctan(z)z+1z(1ttt2+1)dt=π4arctan(z)z+ln(z)[12ln(t2+1)]1z=π4arctan(z)z+12ln(z)12{ln(z+1)ln(2)}f(z)=ln(z)+π22arctan(z)z+ln(z)ln(z+1)+ln(2)+01ln(1+x2)dxf(z)=2ln(z)+π22arctan(z)zln(z+1)+ln(2)+01ln(1+x2)dxbyparts01ln(1+x2)dx=[xln(1+x2)]01012x21+x2dx=ln(2)2011+x211+x2dx=ln(2)2+2×π4=ln(2)2+π2f(z)=2ln(z)+2ln(2)+π22arctan(z)zln(z+1)=01ln(1+zx2)dx01ln(1+ix2)dx=f(i)=2ln(i)+2ln(2)+π22×arctan(i)iln(i+1)=2×iπ2+2ln(2)+π22×arctan(eiπ4)eiπ4ln(2eiπ4)=iπ+2ln(2)+π22eiπ4arctan(eiπ4)12ln(2)iπ4

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