Re: " A way to solve Int{u=ln(x) to ln(x +2*Pi) , g(u)du } = tan(x) "

Dear friends ,

Pretty thanks for your quick replies .
I am very interested by alternative simpler methods.
We've got with G(x) antiderivative of g(x) a clear
functional equation :
G(ln(x + 2*Pi) = G(ln(x) + tan(x) ,
rewritten as
G(ln(x + 2*Pi) )/ tan(x +2Pi )= G(ln(x)) / tan(x) + 1
this form represents an Abel's equation, the occurences of
(x +2*Pi) being counted by G(ln(x))/ tan(x) ; but it is also
known (x +2*Pi) may be counted by (x + c )/(2*Pi) ,
c any constant .
So G(ln(x)) / tan(x) = (x +c) /(2*pi) works , or
G(x) =tan(exp(x))*(exp(x) +c) / (2*pi) (1) ,
derivation gives g(x) .

All this may be generalized , instead of c , insert in (1)
an invariant term for ln(x) => ln(x +2*Pi) for instance
p(sin(exp(x)) , p a derivable function ,

Alain

.