desolve question



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! In this case, INTEGRATIONCONSTANT is necessary, it equals to 
log(-1), so that log(z-1)-log(-1) turns into log(1-z), which is correct 
when z<1. INTEGRATIONCONSTANT is inserted only in 
integrate(diff(f(z),z)=2/(1-z^2),z).... This behaviour is not stressed 
in the manuals, neither is any variable affecting the 
inclusion/not-inclusion of INTEGRATIONCONSTANT into indefinite 
integration results.

--
Andrei Zorine

> 
> No need for fancy ODE machinery.  Just iterate INTEGRATE three times,
> being careful with the integration constants.  It helps to ASSUME(0<z,
> z<1).  And check that it is possible to interchange summation over n
> and the limit for z -> 1-.
>