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

Lennie Davis

Lennie Davis

Answered question

2022-01-03

How can I prove that?
01ln(x)x21dx=π28

Answer & Explanation

Janet Young

Janet Young

Beginner2022-01-04Added 32 answers

01ln(x)x21dx=12(01ln(u)u21du+01ln(v)v21dv)
=12(01ln(u)u21du+1ln(v)v21dv)
=12(01ln(u)u21du+1ln(u)u21du01ln(u)u21du)
=14011u2[ln(1+v1+uv)]v=0v=du
=14011u2(0(11+vu21+u2v)dv)du
=14001(1+v)(1+u2v)dvdu=140(11+v011+u2vdu)dv
Fasaniu

Fasaniu

Beginner2022-01-05Added 46 answers

We have
01ln(x)x21dx=01ln(1x)(1x)21dx=01ln(1x)x(x2)dx
We will generalize by introducing parameter α such that
I(α)=01ln(1αx}{x(x2)}dx
And we have I(0)=0 Then
I(α)=011(1αx)(x2)dx=12α1[ln(x21αx)]01=ln(22α)12α
And we have
I(α)=ln(22α)12α
I(α)=12Li2(2α1)+c
I(0)=12Li2(1)+c=0c=12Li2(1)=π224
I(α)=12Li2(2α1)+π224
I(1)=12Li2(2α1)+π224
=π212+π224
=π28
I(1)=π28
karton

karton

Expert2022-01-11Added 613 answers

Write the integrand as a sum of fractions and use the polylogarithm function Li2:
f(x):=ln(x)x21dx=12(ln(x)x1ln(x)x+1)dx
=12ln(x)dxx112ln(x)dxx+1=12Li2(1x)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)x21dx=π224+π212=π28

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