ABSTRACT
The problem, of course, is that the logarithm function has a branch cut. Hence, the two logarithms in (14.2) may not be on the same lliemann sheet.
The correct way to evaluate (14.1) is to separate the region of integration into two sub-intervals, with the division point being the value where ax + b may vanish. An easier way, for this integral, is to first write the integral as
I(a,b) = ~11 ~/ =~log (x+ ~) 11• a 0 x+ a a a 0 No matter what the sign is oflm(b/a), the argument of the logarithm never crosses the cut (since x is real). Thus, the answer is
Note that since 1 + bfa and bfa have the same imaginary part, we may combine them to obtain our final answer
1 a+b I(a,b) =~log-b-.
