The contour integrals I am familiar with end up on the same branch of the function as they started, perhaps circulating a pole. It is not clear whether integrating √z just once around the origin makes sense. Lets try.

For practice we integrate z around the unit circle. I won’t indulge in mathematical notation to depict the path but assume that we integrate about the unit circle, beginning and ending at 1. This is the limit of summing z dz but when z is near one then dz is in the direction of i! The indefinite integral, p, from 1 to x, as x moves along the unit circle begins to move up from the origin as x moves away from 1. When x reaches i the motion of p is towards −i. By the time x gets to −1 p has moved completely around a circle thru the origin and centered at −½. When x gets back to 1 p has circulated its circle twice. (If we had integrated a constant then p’s direction would have gone thru one circle and p would have ended at 0.) Bizzare! The integral is 0 as we have been taught to believe. If we had integrated 1/z then p would have moved up at a constant velocity and ended at 2πi, also as we have been led to believe.

Now we go for √z around the same path.
Again p is the indefinite integral.
p begins moving as it did before but its direction of motion turns slower.
In both cases its speed is a constant 1.
When z gets to i, p is moving in the direction e^{5πi/4}.
When z gets back to 1 the direction of p has changed thru 3π and we have traversed half a circle thru the origin.
As z returns to 1 the velocity of p is −i and p is not 0!
Remember √z is on another branch with the opposite sign.
If we integrate twice around, p returns to 0.

This does not bode well for the dilogarithm for the integrand does not repeat as z circulates, it continues to drift in the general direction i as we spiral the infinite set of branches of the logarithm.

This program numerically computes the integral around the branch point 1t 1 of −log(1−z)/z in order to evaluate the dilogarithm on different branches.
The routine `ln` there is the complex logarithm but remembers which branch you are on and demands that you move slowly.
We integrate counterclockwise around the unit circle with center = 1, beginning and ending at 0.

Oops, debugging this program brought me to realize that there are poles(!) on the other branches of the integrand. Despite appearances −log(1−z)/z behaves nicely at 0, but only as long as we are on the branch of the log that goes to 0 there. The only such branch is the ‘main branch’. The other branches of dilogarithm have a pole at 0.