I would like to do the integral
$$I=\int_0^{2\pi}d\phi\frac{\ln(e^{i\phi}+e^{-i\phi}-\frac{5}{2})}{e^{i\phi}+e^{-i\phi}-\frac{5}{2}}.$$
Numerically, we readily find that it has a specific finite value:
fun = -(5/2) + E^(-I \[Phi]) + E^(I \[Phi]);
NIntegrate[ Log[fun]/fun, {\[Phi], 0, 2 \[Pi]}]
-0.493368 - 13.1595 I
Now, if we want to consider the integral analytically, we could substitute for instance
$$e^{i\phi}=z~~~,~~~d\phi=\frac{-i}{z}dz$$
which leads to
$$I=-i\oint_{|z|=1}\frac{\ln\left[\frac{1}{z}(z - \frac{1}{2}) (z - 2)\right]}{(z - \frac{1}{2}) (z - 2)}$$
This looks like there is a pole at z=1/2
within the unit circle. So I tried to get the residue:
Residue[-(( I Log[((-2 + z) (-(1/2) + z))/z])/((-2 + z) (-(1/2) + z))), {z, 1/2}]
Residue[-(( I Log[((-2 + z) (-(1/2) + z))/z])/((-2 + z) (-(1/2) + z))), {z, 1/2}]
which just gave back the input. Also, the 1/z
term inside the logarithm seems to blow up inside the unit circle as well. This integral is confusing and does not seem to be accessible via straightforward analytical methods. Is there a way to evaluate it exactly using Mathematica?
Series
atz=1/2
doesn't have the term(z-1/2)^-1
. $\endgroup$