Definition (or formula from a theorem) of contour integral of complex function $f(z)$ over a contour $C$: let $\gamma: [a, b] \rightarrow \mathbb{C}$ be a parameterization of $C$. Then $\int_C f(z) dz = \int _a^b f(\gamma(t)) \gamma'(t) dt$.

Your expression for the integral seems to be missing the factor $ \gamma'(t)$, which for $\gamma(t) = 0.5 e^{i t}$ should be $0.5 i e^{i t}$.

I do not disagree with the result, but rather meant to suggest that some kind of rigorous justification is really needed for such things, as in the math.stackexchange reference I cited.

Counting the residue at the pole on the contour itself seems to require a bit of justification, as in answers to: https://math.stackexchange.com/questions/1075962/pole-on-the-contour-using-the-residu-theorem-what-is-this-formula-of-plemelj

Effectively I'm letting $z(t)=.5e^{it}$, and computing $\int_0^{2\pi}\frac{\exp(z(t))\cos(\pi z(t))}{z(t)^2+z(t)}dt$. The parameter of integration and its bounds are specified by {t,0,2\[Pi]}, just notation. Mathematica can typeset it nicely too, though with

With[{z=Exp[I t]}, Inactive@Integrate[(E^z Cos[\[Pi] z])/(z^2 + z),{t,0,2\[Pi]}]]

On the subject of contour integrals, I suppose Mathematica can't handle complex regions--most of their Region related functions are explicitly $\mathbb R^n$. That being said, it handles arbitrary real regions as integration bounds no problem, and even some basic complex bounds (ha) like piecewise linear or circular, consider (from the docs)

Integrate[1/(z + 1/2), {z, 1, E^((I \[Pi])/3), E^((2 I \[Pi])/3), -1, E^(-((2 I \[Pi])/3)), E^(-((I \[Pi])/3)), 1}]

I did do the usual parameterization to evaluate the integral, but using Mathematica and found it was divergent.

Hey! thanks for trying to help me. I will clarify the function: exp(z)*cos(pi * z)/(z^2+2) around abs(z), How would you insert that in the Wolfram!Alpha?