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Question Number 108692 by 150505R last updated on 18/Aug/20

Answered by Dwaipayan Shikari last updated on 18/Aug/20

∫_0 ^(π/2) ((2log(tanθ))/(tan^2 θ+1))sec^2 θdθ   (x=tanθ, 1=sec^2 θ(dθ/dx))  2∫_0 ^(π/2) log(sinθ)−log(sinθ) dθ=2∫_0 ^(π/2) log(cosθ)−log(sinθ)=I  2I=0  I=0

0π22log(tanθ)tan2θ+1sec2θdθ(x=tanθ,1=sec2θdθdx)20π2log(sinθ)log(sinθ)dθ=20π2log(cosθ)log(sinθ)=I2I=0I=0

Commented by 150505R last updated on 18/Aug/20

thanks

thanks

Answered by mathmax by abdo last updated on 18/Aug/20

i think the question is ∫_0 ^∞  (((lnx)^2 )/(1+x^2 ))dx

ithinkthequestionis0(lnx)21+x2dx

Commented by 150505R last updated on 18/Aug/20

yes i has sent it again.

yesihassentitagain.

Commented by mnjuly1970 last updated on 18/Aug/20

Answer is = (π^3 /8)   hint   put  f(a)=∫_0 ^(  ∞) (x^a /(1+x^(2 ) )) dx=^(x^2 =t) (1/2)∫_0 ^( ∞) (t^((a/2)−(1/2)) /(1+t))=  =(1/2)β(((a+1)/2) , ((1−a)/2) )=(1/2)Γ(((a+1)/2) )Γ(((1−a)/2))  =(1/2) (π/(sin((((1+a)/2))π))) =(1/2) (π/(cos(((πa)/2))))  f^( ′′) (a) =(1/2)π(d/da)[(((π/2)sin(((πa)/2)))/(cos^2 (((πa)/2))=g(a)))]=(π^2 /4) (((π/2)cos(((πa)/2))g(a)−g^′ (a)sin(((πa)/2)))/(g^2 (a)))∣_(a=0)   on the other hand  :: f^( ′′) (a)=∫_0 ^( ∞) ((x^a ln^2 (x))/(1+x^2 )) dx   goal:=f^( ′′) (0) =???  f^( ′′) (0)=(π^3 /(8  ))= goal.....

Answeris=π38hintputf(a)=0xa1+x2dx=x2=t120ta2121+t==12β(a+12,1a2)=12Γ(a+12)Γ(1a2)=12πsin((1+a2)π)=12πcos(πa2)f(a)=12πdda[π2sin(πa2)cos2(πa2)=g(a)]=π24π2cos(πa2)g(a)g(a)sin(πa2)g2(a)a=0ontheotherhand::f(a)=0xaln2(x)1+x2dxgoal:=f(0)=???f(0)=π38=goal.....

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