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solving for simple Integration symbol

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A C
A C on 10 Apr 2020
Edited: David Goodmanson on 10 Apr 2020
When integrate for rho in this equation b/(1-b*rho), I get -log(b*rho-1) which is wrong. It should come out to be -log(1-b*rho).
My code:
syms A B Z a b R T rho
q1 = b/(1-b*rho)
I1 = int(q1,rho)
I1 = - log(b*rho - 1)

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Accepted Answer

David Goodmanson
David Goodmanson on 10 Apr 2020
Edited: David Goodmanson on 10 Apr 2020
HI AC
d/drho (-)*log(b*rho-1) = (-)*1/(b*rho-1)*b = b/(1-b*rho) = q1
so it is correct. But your result is correct as well.
Ignoring the (-) in front for the moment, your result is
log(1-b*rho) = log((-1)*(b*rho-1)) = log(b*rho-1) + log(-1)
= log(b*rho-1) +i*pi
which differs from the 'int' result by a constant of integration i*pi. Whichever result you want to use would usually be decided by keeping the argument of the log function to be positive.

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