Contour Integration and Branch Cuts I have a embarrassingly simple question. For some reason I haven't really studied complex before and I'm suffering under this now. I need to evaluate contour integrals of multi-valued functions and I'm confused about a few details (the standard examples in the textbooks avoid the difficulties I encounter with my integrals). Let me take some examples from the paper arXiv:1008.5194 (see pages 53-54).
There they want to evaluate the integral
$$\oint_{|\omega|=x}d\omega\:(1-\omega)^{-3/4}(x-\omega)^{-3/4}\omega^{-3/4}$$

where $x\in]0,1[$ and they get
$$(-1+e^{-i\frac{3\pi}2})\int_0^xd\omega\:(1-\omega)^{-3/4}(x-\omega)^{-3/4}\omega^{-3/4}.$$
(Which is essentially the integral representation of hypergeometric functions). Since the integrand is multi-valued and has branch points at $\omega = 0, x, 1$ we should probably make a branch cut on the positive real axis. If I naively deform the contour into a line from $x$ to $0$ above the real axis and a line from $0$ to $x$ below the real axis, I seem to get the right result. But that can't be correct; due to the branch cut the circle is not closed and we need to cross the cut several times to get back to the original Riemann sheet and close the contour. But when I do this, taking into account the phases accumulated while going around the Riemann surface, I don't get the correct result. How is this done right?
Another example is 
$$\oint_{|\omega|=1}d\omega\:(1-\omega)^{-3/4}(x-\omega)^{-3/4}\omega^{-3/4} = (1-e^{-i\frac{3\pi}2})\int_1^{\infty}d\omega\:(1-\omega)^{-3/4}(x-\omega)^{-3/4}\omega^{-3/4}.$$
Here I don't get the result, even with the "naive way" and I don't understand why the integration should be over the interval $]1,\infty[$.

UPDATE: Sasha gave a very good answer and his solution seems to be correct, it agrees with numerical calculation. But it's different from what the paper finds and I think the paper is correct, so is there another non-equivalent way of interpreting the contour? Or might there be a problem in this (and lots of other) papers...?
 A: Consider  $f(\omega) = \omega^{-3/4} (x-\omega)^{-3/4} (1-\omega)^{-3/4}$  and $ I(x) = \oint_{\vert \omega \vert = x} f(\omega) \mathrm{d} \omega$.
The function $f(\omega)$ is discontinuous at $\omega = -x$ along $\vert \omega \vert = x$ and has integrable singularity at $\omega = x$. Making a change of variables, $\omega = x z$:
$$  
  I(x) = \frac{1}{\sqrt{x}} \oint_{\vert z \vert = 1} z^{-3/4} (1-z)^{-3/4} (1-x z)^{-3/4} \mathrm{d} z = \frac{1}{\sqrt{x}} \oint_{\vert z \vert = 1} h\left(z\right) \, \mathrm{d} z
$$
Let $\mathcal{C}$ denote the circle $\vert z \vert = 1$. Let $C_{-1, \delta}$ denote segment of $\mathcal{C}$ which crosses the negative axis and of length $2 \pi \delta$, with $\pi \delta$ above and $\pi \delta$ below the negative axis.
It is clear that integral along $\mathcal{C}_{-1,\delta}$ is vanishing as $\delta  \to 0$:
$$ \begin{eqnarray}
   \left\vert \int_{\mathcal{C}_{-1,\delta}} z^{-3/4} (1-z)^{-3/4} (1-x z)^{-3/4} \mathrm{d} z \right\vert &\le& (2(1+x))^{-3/4} \left\vert \int_{\mathcal{C}_{-1,\delta}} z^{-3/4} \mathrm{d} z \right\vert \\ 
 &=& \left(2 (1+x) \right)^{-3/4} \frac{8\sqrt{2}}{7} \sin\left(\frac{7 \delta}{8}\right) \left( \cos\left(\frac{7 \delta}{8}\right) - \sin\left(\frac{7 \delta}{8}\right)  \right)
\\ &\le& \sqrt{2} \delta \left(2 (1+x) \right)^{-3/4}
 \end{eqnarray}
$$
Let's complete $\mathcal{C} \backslash \mathcal{C}_{-1,\delta}$ with integration along $(-1,0)$ above the axis and then along $(0,-1)$ below the axis so as to complete the contour, and call the completed contour $\mathcal{L}$. Then
$$
\begin{eqnarray}
   \oint_{\mathcal{C}} h\left(z\right) \, \mathrm{d} z &=& \oint_{\mathcal{L}} h\left(z\right) \, \mathrm{d} z + \int_0^1 \left( h\left(-y - i \epsilon \right) - h\left(-y + i \epsilon \right) \right)\, \mathrm{d} y \\ 
  &=& \oint_{\mathcal{L}} h\left(z\right) \, \mathrm{d} z + \left( \mathrm{e}^{i \frac{3 \pi}{4}} - \mathrm{e}^{-i \frac{3 \pi}{4}} \right) \int_0^1 y^{-3/4}(1+y)^{-3/4}(1+x y)^{-3/4} \mathrm{d} y
\end{eqnarray}
$$
The claim is that the principal value of $\oint_\mathcal{L} h(z) \mathrm{d} z = 0$, so we get
$$
  I(x) = \frac{2 i \sin\left( 3/4 \pi \right)}{\sqrt{x}} \int_0^1 y^{-3/4}(1+y)^{-3/4}(1+x y)^{-3/4} \mathrm{d} y
$$
Now, let's check this with quadratures:

Notice that the purported answer you gave in your post can not be correct, as it is not purely imaginary.
A: 
I think the authors get the answer just by contour deformation from the circle to the green line. Suppose above the branch cut $(0,x)$, the integrated function is
$$(1-\omega)^{-3/4}(x-\omega)^{-3/4}\omega^{-3/4}$$
When the contour rotates around $\omega=0$, we have
$$\omega\rightarrow \omega e^{2\pi i}$$
and the integrated function becomes
$$e^{-3\pi i/2}(1-\omega)^{-3/4}(x-\omega)^{-3/4}\omega^{-3/4}$$
