The antiderivative of 1/(x^4+1): Part 2

This antiderivative has arguable the highest ratio of “really hard to compute” to “really easy to write”:

$\displaystyle \int \frac{1}{x^4 + 1} dx$

To compute this integral, I will use the technique of partial fractions. This requires factoring the denominator over the real numbers, which can be accomplished by finding the roots of the denominator. In other words, I need to solve

$x^4 + 1 = 0$ ,

$z^4 = -1$ .

I switched to the letter $z$ since the roots will be complex. The four roots of this quartic equation can be found with De Moivre’s Theorem by writing

$z = r (\cos \theta + i \sin \theta)$ ,

where $r$ is a real number, and

$-1 + 0i = 1(\cos \pi + \i \sin \pi)$

By De Moivre’s Theorem, I obtain

$r^4 (\cos 4\theta + i \sin 4 \theta) = 1 (\cos \pi + i \sin \pi)$ .

Matching terms, I obtain the two equations

$r^4 = 1$ and $4\theta = \pi + 2\pi n$

$r = 1$ and $\theta = \displaystyle \frac{\pi}{4} + \displaystyle \frac{\pi n}{2}$

$r = 1$ and $\theta = \displaystyle \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$ .

This yields the four solutions

$z = 1 \left[ \cos \displaystyle \frac{\pi}{4} + i \sin \frac{\pi}{4} \right] = \displaystyle \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}$

$z = 1 \left[ \cos \displaystyle \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right] = -\displaystyle \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}$

$z = 1 \left[ \cos \displaystyle \frac{5\pi}{4} + i \sin \frac{5\pi}{4} \right] = -\displaystyle \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}$

$z = 1 \left[ \cos \displaystyle \frac{7\pi}{4} + i \sin \frac{7\pi}{4} \right] = \displaystyle \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}$

Therefore, the denominator $x^4 + 1$ can be written as the following product of linear factors over the complex plane:

$\displaystyle \left(x - \left[ \displaystyle \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right] \right)\left(x - \left[ \displaystyle \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \right] \right) \left(x - \left[ -\displaystyle \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right] \right) \left(x - \left[ - \displaystyle \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \right] \right)$

$\displaystyle \left(\left[x - \displaystyle \frac{\sqrt{2}}{2} \right] - i \frac{\sqrt{2}}{2} \right)\left( \left[ x - \displaystyle \frac{\sqrt{2}}{2} \right] + i \frac{\sqrt{2}}{2} \right) \left( \left[ x + \displaystyle \frac{\sqrt{2}}{2} \right] - i \frac{\sqrt{2}}{2} \right) \left( \left[ x + \displaystyle \frac{\sqrt{2}}{2} \right] + i \frac{\sqrt{2}}{2} \right)$

$\displaystyle \left(\left[x - \displaystyle \frac{\sqrt{2}}{2} \right]^2 - \left[ i \frac{\sqrt{2}}{2} \right]^2 \right) \left( \left[ x + \displaystyle \frac{\sqrt{2}}{2} \right]^2 - \left[i \frac{\sqrt{2}}{2} \right]^2 \right)$

$\displaystyle \left(x^2 - x \sqrt{2} + \displaystyle \frac{1}{2} + \displaystyle \frac{1}{2}\right) \left(x^2 + x \sqrt{2} + \displaystyle \frac{1}{2} + \displaystyle \frac{1}{2}\right)$

$\displaystyle \left(x^2 - x \sqrt{2} + 1 \right) \left(x^2 + x \sqrt{2} + 1\right)$ .

We have thus factored the denominator over the real numbers:

$\displaystyle \int \frac{dx}{x^4 + 1} = \displaystyle \int \frac{dx}{\left(x^2 - x \sqrt{2} + 1 \right) \left(x^2 + x \sqrt{2} + 1\right)}$ ,

and the technique of partial fractions can be applied.

I’ll continue the calculation of this integral with tomorrow’s post.

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The antiderivative of 1/(x^4+1): Part 2

Published by John Quintanilla

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