Calculate: ∫0∞ln2(x)(1−x2)2dx

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Calculate:  ∫0∞ln2(x)(1−x2)2dx

Stella Calderon · Accepted Answer

You&#039;re definetly on the right track with that substitution of x=1tBasically we have:I=∫0∞ln2(x)(1−x2)2dx=∫0∞x2ln2x(1−x2)2dxNow what if we add them up?2I=∫0∞ln2x1+x2(1−x2)2dxIf you don&#039;t know how to deal easily with the integral∫1+x2(1−x2)2dx=x1−x2+CAnyway we have, integrating by parts:2I=x1−x2ln2x∣0∞⏟=0+2∫0∞ln⁡xx2−1dx⏟=π24⇒2I=2⋅π24⇒I=π24

Fasaniu · Accepted Answer

This is a variant of TheSimpliFires

Vasquez · Accepted Answer

Here is yet another slight variation on a theme. Let I=∫0∞ln2⁡x(1−x2)2dx then I=∫01ln2⁡x(1−x2)2dx+∫1∞ln2⁡x(1−x2)2dx =∫01(1+x2)ln2⁡x(1−x2)2dx (1) after a substitution of ↦1/xZ has been enforced in the second of the integrals. As 11−x2=∑n=0∞x2n, |x|&amp;lt;1 differentiating with respect to x gives 1(1−x2)2=∑n=1∞nx2n−2 On substituting the above series expansion in (1), after interchanging the order of the summation with the integration we have I=∑n=1∞n∫01(x2n−2+x2n)ln2⁡xdx Integrating be parts twice, we are left with I=∑n=1∞(2n(2n−1)3+2n(2n+1)3) As the series is absolutely convergent, terms can be rearranged without changing its sum. Doing so we have I=∑n=1∞[1(2n−1)2+1(2n−1)3+1(2n+1)2−1(2n+1)3]=∑n=1∞[1(2n−1)2+1(2n−1)3]+∑n=1∞[1(2n+1)2−1(2n+1)3]=∑n=1∞[1(2n+1)2+1(2n+1)3]+∑n=1∞[1(2n+1)2−1(2n+1)3]=2+2∑n=1∞1(2n+1)2=2∑n=0∞1(2n+1)2=2[∑n=1∞1n2−∑n=1∞1(2n)2]=2(1−14)∑n=1∞1n2=32∑n=1∞1n2=32⋅π26=π24 as expected.

Calculate: \int_0^\infty\frac{\ln^2(x)}{(1-x^2)^2}dx

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