contour integrating 1/z over unit square equal zero? After I have studied contour integration and related theorems for some time, I encounter a quite confusing problem:

integrating $\frac{1}{z}$ about a unit square centered at the origin(positive orientation).

Let $F(z)$ be the antiderivative of $\frac{1}{z}$, seperating the contour into 4 straight lines, then 

$$\int_\Gamma \frac{1}{z}dz = (F(0.5+0.5i) - F(0.5-0.5i)) + (F(-0.5+0.5i) - F(0.5+0.5i)) +... = 0$$

I expect the answer to be $2\pi i$ however that results in zero. What mistake did I make?

p.s. please do not use residue theorem or any theorems to directly show that the integral equal $2\pi i$. I would like to see some sort of direct integration. Thanks.
Hint: $F(z)=log(z)$ has a branch cut. See the Wikipedia page on complex logarithm.
Observe the domain (i mean simply connected region) where the branch of logarithm fn is analytic.Now see if you take closed curve which is homotopic to a circle in that simply connected region.And then you do integral and you find it will be 2 *pi *i *n where n is a positive integer,it is the winding number,it depends no. of times of rotation on the circle or the closed curve.