Maxima takes integral
factor(integrate(log(1+cos(x)),x,0,2*%pi));
correctly:
-2%pi(log(4)-log(2))
But for almost the same integral:
factor(integrate(log(1+sin(x)),x,0,2*%pi));
it gives:
-%pi(2log(2)-3i %pi)
Obviuously, the answer can not be complex number.
Maxima 5.34.0 using Lisp SBCL 1.2.2
Function log(sin(x)+1) has singularity at
x=3*%pi/2
.Integral
'integrate(log(sin(x)+1),x,0,2*%pi)
is equal.best
Aleksas D
Last edit: Robert Dodier 2023-05-03
I just tried these examples, both in the original report and in the comment dated 2014-09-15, and I get the same result in each case.
The bug appears to be that Maxima is not looking for singularities on the interval of integration.