Let $f(z)=\pi e^{\pi\bar z}$ be a function on complex numbers. If $\gamma$ is the square with vertices $0,1,1+i \ \text{and} \ i$ in the counterclockwise sense. I would like to compute:
$$\int_{\gamma} f(z)dz$$
My guess is using Cauchy theorems the value of this integrations is zero. The problem is when I tried to calculate this integral by brute force I didn't get zero as a value to this integral.
On
By "Cauchy theorems" I'm assuming you're mainly referring to this:
If $f(z)$ is analytic on and inside a simple closed contour $C$ and $f'(z)$ is continuous on and inside $C$ then $\displaystyle \int f(z) \, dz = 0.$
But this (and pretty much any related theorem) doesn't apply here because $f(z)$ is not analytic. A necessary condition for analyticity is $df/d\overline z =0$, but this is clearly not satisfied by the given $f$.
Direct calculation is the way to go here.
$f$ is not holomorphic and not even meromorphic because $z \mapsto \bar z$ is not differentiable anywhere:
$$\frac{\bar z - \bar 0}{z - 0} = \frac{(x-iy)^2}{x^2 + y^2} = \begin{cases} -i, \text{ on the path $y=x$} \\ i, \text{ on the path $y = -x$} \end{cases}$$
the same can be done for any other point.
So you can't apply Cauchy's theorem or the Residue theorem. You have to calculate it directly.