Well... d1 and d2 are lengths - in this particular case, their value is 4e8 and 5e4, respectively but having a symbolic expression will be useful later. When I try the series expansion of the integrand, I get: > MultiSeries[series](ss, t, 3); Error, (in MultiSeries:-multiseries) need to determine the sign of -4*Pi*Re(d1*N/L) Thanks again, Pedro

Well... d1 and d2 are lengths - in this particular case, their value is 4e8 and 5e4, respectively but having a symbolic expression will be useful later. When I try the series expansion of the integrand, I get: > MultiSeries[series](ss, t, 3); Error, (in MultiSeries:-multiseries) need to determine the sign of -4*Pi*Re(d1*N/L) Thanks again, Pedro

First of all, thanks for the reply. I'm sorry for the input in the first post. I hope this is ok: int(ln(1-8*Pi^2*ns*e^2*N/(m*(omega^2-e^2*B^2/m^2)*L*sin(t)*kappa*((kappa*tanh(2*Pi*N*d1/(L*sin(t)))+1)/(kappa+tanh(2*Pi*N*d1/(L*sin(t))))+(kappa*tanh(2*Pi*N*d2/(L*sin(t)))+1)/(kappa+tanh(2*Pi*N*d2/(L*sin(t))))))), t = 0 .. 1/2*Pi) All parameters are real and positive - except B that may be negative - and N is an integer. At first I thought the integral had a lot of discontinuites but Maple10 says the integrand is continuous in the interval considered. I already tried setting options like AllSolutions, CauchyPrincipalValue and Continuous with no luck. I'm greatly appreciated for your help, Pedro

First of all, thanks for the reply. I'm sorry for the input in the first post. I hope this is ok: int(ln(1-8*Pi^2*ns*e^2*N/(m*(omega^2-e^2*B^2/m^2)*L*sin(t)*kappa*((kappa*tanh(2*Pi*N*d1/(L*sin(t)))+1)/(kappa+tanh(2*Pi*N*d1/(L*sin(t))))+(kappa*tanh(2*Pi*N*d2/(L*sin(t)))+1)/(kappa+tanh(2*Pi*N*d2/(L*sin(t))))))), t = 0 .. 1/2*Pi) All parameters are real and positive - except B that may be negative - and N is an integer. At first I thought the integral had a lot of discontinuites but Maple10 says the integrand is continuous in the interval considered. I already tried setting options like AllSolutions, CauchyPrincipalValue and Continuous with no luck. I'm greatly appreciated for your help, Pedro