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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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For |x|<1 you can use power series to get a very good approximation. – Potato Feb 1 '12 at 7:37
You know of others that software would fail at, like something a first year Calc student could understand? – yiyi Nov 9 '12 at 4:57
up vote 10 down vote accepted

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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What do you mean by $(P,Q)=1$? – Ben Crowell Feb 1 '12 at 16:26
It means $P$ and $Q$ are relatively prime (as polynomials over the complex numbers). Equivalently (by the Fundamental Theorem of Algebra) $P$ and $Q$ have no roots in common. – Robert Israel Feb 1 '12 at 22:55
To expand a bit on zulon's last paragraph: if $\alpha$ is a non-real root and the partial fraction decomposition includes $c/(z-\alpha)$ then it also includes $\overline{c}/(z - \overline{\alpha})$, and an antiderivative of $$\frac{c}{z-\alpha} + \frac{\overline{c}}{z - \overline{\alpha}} = \frac{2 \text{Re}(c)(z-\text{Re}(\alpha)) - 2 \text{Im}(c)\text{Im}(\alpha)}{(z-\alpha)(z-\overline{\alpha})} $$ is $\text{Re}(c) \ln((z - \text{Re}(\alpha))^2 + \text{Im}(\alpha)^2) - 2 \text{Im}(c) \arctan\left(\frac{z-\text{Re}(\alpha)}{\text{Im}(\alpha)}\right)$ – Robert Israel Feb 1 '12 at 23:14
An advantage of this over the form $c \ln(z-\alpha) + \overline{c} \ln(z - \overline{\alpha})$ is that (once you have the real and imaginary parts of $c$ and $\alpha$) it doesn't explicitly involve complex quantities when $z$ is real. From the point of view of complex analysis, both forms are equally valid, but (if you use the principal branches) their branch cuts are in different places. – Robert Israel Feb 1 '12 at 23:24
Two great answers! I'm marking this one as accepted because it taught me something new about partial fractions. – Ben Crowell Feb 2 '12 at 0:14

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? – Ben Crowell Feb 2 '12 at 0:16

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