I'm wondering if there is another way to calculate the contour integral of $\int(\tan(z/2)/(z-1))$ in the square w/ sides $Re(z)=+/-2$, $Im(z)=+/- 2$ other than using the residue theorem. The cauchy integral formula only works if the function, in this case $tan(z/2)$ is analytic inside the contour, which it isn't.
Cheers
You are wrong on at least two points:
1) $\tan{(z/2)}$ is indeed analytic inside the square. Its smallest poles are at $z=\pm \pi$.
2) The residue theorem/Cauchy's integral theorem is designed to work with contour integrals with integrands that have poles inside the contour. As a reminder, with $f$ analytic inside a contour $C$ and $a$ inside $C$:
$$\oint_C dz \frac{f(z)}{z-a} = i 2 \pi f(a) $$
So, that's the easy way of evaluating the integral. You are of course welcome to do it the hard way if that's what you need to do.