Tuesday, 27 September 2016

calculus - Closed-form of intlimits10left(fracleft(x2+1right)arcsin(x)sqrt1x2+2xlnleft(x2+1right)right)fraclnxx3+x,dx



I've conjectured the following closed-form:
I=10((x2+1)arcsin(x)1x2+2xln(x2+1))lnxx3+xdx=2Gln2,
where G is Catalan's constant.

Numerically
I1.2697979381877088371491554851603117320986537271546606092465
How to prove it?


Answer



Apply the obvious substitution xsinx to the first integral
I=π20xln(sinx)sinx dx+102lnxln(1+x2)1+x2 dx
The latter integral has been addressed here and is equivalent to
102lnxln(1+x2)1+x2 dx=4Li3(1i)2Gln2+3π316+π4ln22

As for the first integral,
π20xln(sinx)sinx dx=210arctanxln(2x1+x2)x dx=2Ti2(1)ln2+210arctanxlnxx dx210arctanxln(1+x2)x dx=2Gln22n=0(1)n(2n+1)3+210lnxln(1+x2)1+x2 dx+410xarctanxlnx1+x2 dx
and the integral
410xarctanxlnx1+x2 dx=8Li3(1i)5π316π2ln22
has also been established in the link above. (Both integrals were covered in the evaluation of J2.) Hence the closed form is indeed
I=2Gln2


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