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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 $$\oint_{|\omega|=x}d\omega\:(1-\omega)^{-3/4}(x-\omega)^{-3/4}\omega^{-3/4}$$ enter image description here

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...?

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The general idea is to ensure your contour doesn't cross the cut. Have your contour start and finish at the cut. For instance, if you're taking a circular contour and the cut is the negative real axis, $-r\exp(it)$ would be the contour to use along with the interval $(0,2\pi)$. –  J. M. Nov 4 '11 at 17:19
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1 Answer

up vote 3 down vote accepted

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:

enter image description here

Notice that the purported answer you gave in your post can not be correct, as it is not purely imaginary.

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Sorry for the late response. Thanks a lot, I really like your solution and It makes very good sense to me. But I am very confused now, the solution I gave should be the correct one (as it is used in several papers), but your solution and numerical calculations disagree. Is it possible that there is an ambiguity in how the contour should be interpreted? And the correct contour is slightly different from what you have used...? –  Heidar Nov 7 '11 at 16:07
@Heidar I do not see how. If the result was also mentioned in different article, can you isolate the common source, i.e. where the purported result was first obtained ? By the way, $I(x)$ in my post has not expression in terms of Gauss's hypergeometric function, rather in terms of Appell's $F_1$, $I(x) = \frac{4 i \sqrt{2}}{\sqrt{x}} \cdot F_1\left(\frac{1}{4};\frac{3}{4},\frac{3}{4};\frac{5}{4};-1,-x\right)$. –  Sasha Nov 7 '11 at 16:45
Yeah, I need to look into it. There are just many integrals of the same type in this field and all of them gives hypergeometric functions (this can we shown to be the correct solution by other independent methods). I think the conclusion is that I'm misunderstanding some details in what these papers do. I am now pretty convinced that the way I understood the integrals to begin with, cannot be correct. Thanks for your nice answer, it has been very helpful. –  Heidar Nov 7 '11 at 23:52
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