Where is the error in this complex integral?

Question

I got myself confused over the following problem: Compute $$\int_\gamma\frac{1}{\sqrt{z}}dz$$ where $\gamma$ is the lower unit circle arc from $-1$ to $1$.

Isn't it correct that I can either choose $$\gamma:[-\pi,0]\to\mathbb C,t\mapsto e^{it}$$ or $$\gamma:[\pi,2\pi]\to\mathbb C, t\mapsto e^{it}$$

It probably must be wrong because in the first case we have $$\int_{-\pi}^0ie^{it/2}dt=2(1+i)$$ where in the second case we have $$\int_\pi^{2\pi}ie^{it/2}dt=-2(1+i)$$ but for whatever reason I can't figure out what's wrong.

Answer 1

$\sqrt z$ is two valued, one being the negative of the other. So the answer depends on how you define $\sqrt z$. With appropriate choice of this function both answers can be considered correct!.

Answer 2

Let $u = t -2\pi$ and $du = dt$

At $t = \pi \ , u = -\pi$

and at $t = 2\pi \ , u = 0$

So, $\int^{2\pi}_{\pi} ie^{it/2}dt = -(2+i)$

$\implies \int^{0}_{-\pi} ie^{i(u+2\pi)/2}du = -(2+i)$

$\implies e^{i\pi}\int^{0}_{-\pi} ie^{iu}du = -(2+i)$

As, $e^{i\pi} = -1$

$-\int^{0}_{-\pi} ie^{iu}du = -(2+i)$

or

$$\int^{0}_{-\pi} ie^{iu}du = (2+i)$$

which is identical as the first one ($t$ replaced by $u$ )