I'm trying to compute the following integral
integrand = (-2 (Log[2] - Log[1/t + t])^2)/(-1 + t^2)
Numerically integrating it gives the expected result
1/2*NIntegrate[integrand, {t, 0, 1}]
=> 1.0518
N[7/8*Zeta[3]]
=> 1.0518
However when I try to integrate it symbolically I get
1/2*Integrate[integrand, {t, 0, 1}]
=> 1/16 (\[Pi] (I \[Pi] + Log[4]) (\[Pi] + I Log[64]) + 14 Zeta[3])
N[%]
=> -0.658472 + 3.06993 I
It is interesting that Mathematica returns the correct result 14/16 Zeta[3]
as part of its solution, but in addition it returns some strange imaginary terms, which I couldn't get rid of. What can I do so that Mathematica will compute this integral correctly?
Update: My version of Mathematica:
11.0.1 for Mac OS X x86 (64-bit) (September 21, 2016)
11.0.1 for Mac OS X x86 (64-bit) (September 21, 2016)
using either1/2*Integrate[integrand, {t, 0, 1}, PrincipalValue -> True]
or1/2*Integrate[integrand, {t, 0, 1}, Assumptions -> 0 < t < 1]
they evaluate to-((7*Zeta[3])/8)
. Note your "correction" is not the negative of the originalintegrand
$\endgroup$(7 Zeta[3])/8
directly. Can you tryIntegrate[Log[(2 t)/(1 + t^2)]^2/(1 - t^2), {t, 0, 1}]
? $\endgroup$7/8*Zeta[3]
. $\endgroup$