Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

2 Answers 2

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 $|\zeta| = 1$.

share|improve this answer
    
What do you mean by $(P,Q)=1$? –  Ben Crowell Feb 1 '12 at 16:26
3  
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
1  
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
1  
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: $- x \ {}_2F_1\left(\frac{1}{n},1; 1+\frac{1}{n}; x^n\right)$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.