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Is there any way to evaluate this indefinite integral using pencil and paper? A closed-form solution exists, because $1/(x^{10000}-1)$ can be expressed as a partial fraction decomposition of the form $\sum c_m/(x-a_m)$, where the $a_m$ are the 10,000-th roots of unity. But brute-force computation of the $c_m$ is the kind of fool's errand that a human would never embark on, and that software stupidly attempts and fails to accomplish. (Maxima, Yacas, and Wolfram Alpha all try and fail.)

This is not homework.

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    You know of others that software would fail at, like something a first year Calc student could understand?2012-11-09

2 Answers 2

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You can use the fact that, in a partial fraction decomposition, for a simple root $\alpha$ of the denominator (say $F = {P \over Q}$ where $(P,Q) = 1$) then the coefficient of ${1 \over X-\alpha}$ is {P(\alpha) \over Q'(\alpha)}. Since $X^{1000} - 1$ only has simple roots (the 1000th powers of unity), which you can express easily as $\omega^k, k \in \{0,\dots,999\}$ where $\omega = e^{2i\pi \over 1000}$. Then it's just a matter of computing a sum, since the integral of ${1 \over x-\alpha}$ is easy enough to compute.

Beware though, $\alpha$ is complex here, so the antiderivative is not just $\log(x-\alpha)$…. But another trick you can use is that you can naturally pair the roots of unity, as $\bar{\zeta} = \zeta^{-1}$ for $|\zeta| = 1$.

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    Two great answers! I'm marking this one as accepted because it taught me something new about partial fractions.2012-02-02
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For $|x|<1$ we have $1/(x^n - 1) = - \sum_{k=0}^\infty x^{nk}$, so an antiderivative of this is $ - \sum_{k=0}^\infty \frac{x^{nk+1}}{nk+1}$. This can be written as a hypergeometric function: $- x \ {}_2F_1\left(\frac{1}{n},1; 1+\frac{1}{n}; x^n\right)$.

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    Cool! I'm curious about your thought process. Did you write out the series and then look it up in a table to identify it as a hypergeometric?2012-02-02