Andrei Zorine <zoav1@uic.nnov.ru> writes:
> It doesn't help to ASSUME(0<z,z<1), because call to
> integrate(2/(1-z^2),z) returns log(z+1)-log(z-1), which is wrong
> when z<1!
I forgot to mention LOGABS, sorry. LOGNEGINT may also be of interest.
Look at `Definitions for Logarithms' in the manual.
(C1) assume(0<z,z<1);
(D1) [z > 0, z < 1]
(C2) logabs:true;
(D2) TRUE
(C3) integrate(2/(1-z^2),z);
LOG(z + 1) LOG(1 - z)
(D3) 2 (---------- - ----------)
2 2
Wolfgang