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Regard the function $f(x) = x^{-1/2}$ on the non-negative real line.

The point $z=0$ is 'distinguished' because it is the boundary of the domain of $f$, and because $f$ has a pole there. It seems natural to ask whether there exists a expansion of $f$ near $0$. What is the natural candidate for this? AFAIK most series' (like Laurent) are only developed in the interior of a domain.

The motivation is compute an integral that involves $f$ (addendum: inside a more complicated expression) by computing the terms of a series, where the terms are assumed to be simpler (rational functions at most) and the method converges. -

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It has a very simple expression as a Puiseux series, see en.wikipedia.org/wiki/Puiseux_series –  Gerry Myerson Feb 1 '13 at 11:33

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There is no expansion of a function about a branch point singularity. Terms such as $x^{\alpha}$ for noninteger $\alpha$ and $\log{x}$ are examples of such singularities in which the argument of the term is undefined. For example, functions resulting from differential equations having a regular singular point at $x=0$ may be expressed, by the Froebenius method, in a series involving powers of $x$ and terms in $x^{\alpha}$ and $\log{x}$. (See, for example, many of the solutions of the second kind of the orthogonal basis functions, e.g., Bessel, Legendre, etc., at regular singular points of their generating differential equations.)

Away from zero, you may expand about some point that is not a branch point as you would normally do in a Taylor or Laurent series.

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