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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 $(0, 2\pi)$.

Keyhole contour

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.

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"The integral over the large circle vanishes by Jordan's lemma." No, it doesn't. Jordan's lemma is about semicircles in either the upper or lower half plane. Your almost-circle lies in both half planes, and $\lvert e^{iz}\rvert$ becomes large in the lower half plane. – Daniel Fischer Aug 30 '13 at 19:56
Wolfram products have the branch cut for $\sqrt{z}$ as going from $(-\infty,0)$, not the way you've shown it. See here. – rajb245 Aug 30 '13 at 19:58
Non-computers may choose branch cuts that help to solve a given problem. They may even change the branch cut from one problem to another. Strange but true! – GEdgar Aug 30 '13 at 21:48

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 $R\rightarrow\infty$).

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