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Question Number 65690 by mathmax by abdo last updated on 02/Aug/19
calculate ∫_0 ^1 ((ln^2 (x))/(1+x^2 ))dx
calculate∫01ln2(x)1+x2dx
Commented by mathmax by abdo last updated on 02/Aug/19
let A =∫_0 ^1 ((ln^2 x)/(1+x^2 ))dx changement lnx =−t give x =e^(−t) ⇒ A = −∫_0 ^(+∞) (t^2 /(1+e^(−2t) ))(−e^(−t) )dt = ∫_0 ^∞ ((t^2 e^(−t) )/(1+e^(−2t) ))dt A =∫_0 ^∞ t^2 e^(−t) (Σ_(n=0) ^∞ (−1)^n e^(−2nt) )dt =Σ_(n=0) ^∞ (−1)^n ∫_0 ^∞ t^2 e^(−(2n+1)t) dt=Σ_(n=0) ^∞ (−1)^n A_n A_n =_((2n+1)t =u) ∫_0 ^∞ (u^2 /((2n+1)^2 )) e^(−u) (du/(2n+1)) =(1/((2n+1)^3 ))∫_0 ^∞ u^2 e^(−u) du by parts ∫_0 ^∞ u^2 e^(−u) du =[−u^2 e^(−u) ]_0 ^(+∞) −∫_0 ^∞ 2u(−e^(−u) )du =2 ∫_0 ^∞ u e^(−u) du =2{[−ue^(−u) ]_0 ^(+∞) −∫_0 ^∞ (−e^(−u) )du} =2∫_0 ^∞ e^(−u) du =2[−e^(−u) ]_0 ^(+∞) =2 ⇒A_n =(2/((2n+1)^3 )) ⇒ A =2Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^3 )) be continued....
letA=∫01ln2x1+x2dxchangementlnx=−tgivex=e−t⇒A=−∫0+∞t21+e−2t(−e−t)dt=∫0∞t2e−t1+e−2tdtA=∫0∞t2e−t(∑n=0∞(−1)ne−2nt)dt=∑n=0∞(−1)n∫0∞t2e−(2n+1)tdt=∑n=0∞(−1)nAnAn=(2n+1)t=u∫0∞u2(2n+1)2e−udu2n+1=1(2n+1)3∫0∞u2e−udubyparts∫0∞u2e−udu=[−u2e−u]0+∞−∫0∞2u(−e−u)du=2∫0∞ue−udu=2{[−ue−u]0+∞−∫0∞(−e−u)du}=2∫0∞e−udu=2[−e−u]0+∞=2⇒An=2(2n+1)3⇒A=2∑n=0∞(−1)n(2n+1)3becontinued….
Commented by ~ À ® @ 237 ~ last updated on 04/Aug/19
Let named I that integral.when changing u=(1/x) dx=((−du)/u^2 ) I=∫_1 ^∞ ((ln^2 (u))/(1+u^2 ))du and then ∫_0 ^∞ ((ln^2 (u))/(1+u^2 ))du=2I 1+u^2 =(u−i)(u+i) With the residus theorem 2I=2iπRes(f. i) with f(x)=((ln^2 (x))/(1+x^2 )) Res(f. i)=lim_(z−>i) ((ln^2 (z))/(z+i))=(([ln(e^(i(π/2)) )]^2 )/(2i))=(((−π^2 )/4)/(2i))=((−π^2 )/(8i)) Then 2I=((−π^3 )/4) so finally I= ((−π^3 )/8)
LetnamedIthatintegral.whenchangingu=1xdx=−duu2I=∫1∞ln2(u)1+u2duandthen∫0∞ln2(u)1+u2du=2I1+u2=(u−i)(u+i)Withtheresidustheorem2I=2iπRes(f.i)withf(x)=ln2(x)1+x2Res(f.i)=limz−>iln2(z)z+i=[ln(eiπ2)]22i=−π242i=−π28iThen2I=−π34sofinallyI=−π38

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