How can I prove that? ∫01ln⁡(x)x2−1dx=π28

Question

How can I prove that?  ∫01ln⁡(x)x2−1dx=π28

Janet Young · Accepted Answer

∫01ln⁡(x)x2−1dx=12(∫01ln⁡(u)u2−1du+∫01ln⁡(v)v2−1dv)  =12(∫01ln⁡(u)u2−1du+∫1∞ln⁡(v)v2−1dv)  =12(∫01ln⁡(u)u2−1du+∫1∞ln⁡(u)u2−1du−∫01ln⁡(u)u2−1du)  =14∫0∞11−u2[ln⁡(1+v1+uv)]v=0v=∞du  =14∫0∞11−u2(∫0∞(11+v−u21+u2v)dv)du  =14∫0∞∫0∞1(1+v)(1+u2v)dvdu=14∫0∞(11+v∫0∞11+u2vdu)dv

Fasaniu · Accepted Answer

We have∫01ln⁡(x)x2−1dx=∫01ln⁡(1−x)(1−x)2−1dx=∫01ln⁡(1−x)x(x−2)dxWe will generalize by introducing parameter α such thatI(α)=∫01ln⁡(1−αx}{x(x−2)}dxAnd we have I(0)=0 ThenI′(α)=−∫011(1−αx)(x−2)dx=12α−1[ln⁡(x−21−αx)]01=ln⁡(2−2α)1−2αAnd we haveI′(α)=ln⁡(2−2α)1−2αI(α)=12Li2(2α−1)+cI(0)=12Li2(−1)+c=0⇒c=−12Li2(−1)=π224I(α)=12Li2(2α−1)+π224I(1)=12Li2(2α−1)+π224=π212+π224=π28I(1)=π28

karton · Accepted Answer

Write the integrand as a sum of fractions and use the polylogarithm function Li2: f(x):=∫ln⁡(x)x2−1dx=∫12(ln⁡(x)x−1−ln⁡(x)x+1)dx =12∫ln⁡(x)dxx−1−12∫ln⁡(x)dxx+1=−12Li2(1−x)−12(Li2(−x)+ln⁡(x)ln⁡(x+1)) Since Li2(0)=0 and ln⁡(x)ln⁡(x+1) vanishes at x=0 and x=1, we have f(0)=−12Li2(1)=−π212 and f(1)=−12Li2(−1)=π224 and the value of the integral is ∫01ln⁡(x)x2−1dx=π224+π212=π28

How can I prove that? \int_0^1\frac{\ln(x)}{x^2-1}dx=\frac{\pi^2}{8}

Answered question

Answer & Explanation

New Questions in Calculus 1