Contour integral of $\frac{e^{iz}}{\sqrt{z}}$ I attempt to integrate $f(z) = \frac{e^{iz}}{\sqrt{z}}$ over the "keyhole contour"
where the branch cut for the square root deletes the positive real axis, and the argument $\theta$ is restricted to $(0, 2\pi)$.

The integral over the large circle vanishes by Jordan's lemma. EDIT: This is not true, as Daniel Fischer points out. Thank you!
Then integral
over the smaller circle also vanishes, since we have the estimate
$|\int f(z) \; dz| \leq c\sqrt{\epsilon} \rightarrow 0$.
The integral over the lower ray $\gamma_L$ that tends to the real axis with reversed orientation is
$$\int_\gamma \frac{e^{iz}}{\sqrt{z}} \; dz = \int_\infty^0 \frac{e^{ix}}{-\sqrt{x}} \; dx = \int_0^\infty \frac{e^{ix}}{\sqrt{x}} \; dx$$
The integral over the upper ray $\gamma_U$ also tends towards the same integral.
Since there is no singularity inside the contour, the contour integral is zero. So we have
$$2 \int_0^\infty \frac{e^{ix}}{\sqrt{x}} \; dx = 0$$
which is not true.
 A: You are right that there are no poles inside the contour, but the big circle does not vanish. We have
\begin{equation}
\int_{C_R} + \int_{C_\gamma} + \int_{\gamma_U} + \int_{\gamma_L} = 0,
\end{equation}
where $C_R$ is the big circle and $C_\gamma$ is the small circle around the branch point whose contribution vanishes in the limit $\gamma\rightarrow0$. So you obtain
\begin{equation}
\int_{C_R} dz \frac{e^{iz}}{\sqrt{z}} = -\left(\int_{\gamma_U} + \int_{\gamma_L} \right) dz \frac{e^{iz}}{\sqrt{z}} = -2 \int_0^R dx \frac{e^{ix}}{\sqrt{x}}
\end{equation}
As we let $R\rightarrow \infty$, we find
\begin{equation}
2 \int_0^\infty dx \frac{e^{ix}}{\sqrt{x}} = \int_{-\infty}^\infty dt e^{it^2} = e^{i\frac{\pi}{4}} \int_{-\infty}^\infty ds e^{-s^2} = e^{i\frac{\pi}{4}} \sqrt{\pi}.
\end{equation}
The substitutions are $x = t^2$ and $t = e^{i\frac{\pi}{4}}s$. In the second step we also rotated the contour back to the real axis which is allowed since the integrand is holomorphic and vanishes at real infinity (the contribution between $R$ and $Re^{i\pi/4}$ vanishes for $R\rightarrow\infty$).
