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

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on the computation of

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

Part 1: Introduction.

Part 2: Factoring the denominator using De Moivre’s Theorem.

Part 3: Factoring the denominator using the difference of two squares.

Part 4: The partial fractions decomposition of the integrand.

Part 5: Partial evaluation of the resulting integrals.

Part 6: Evaluation of the remaining integrals.

Part 7: An apparent simplification using a trigonometric identity.

Part 8: Discussion of the angles for which the identity holds.

Part 9: Proof of the angles for which the identity holds.

Part 10: Implications for using this identity when computing definite integrals.

How I Impressed My Wife: Index

Some husbands try to impress their wives by lifting extremely heavy objects or other extraordinary feats of physical prowess.

That will never happen in the Quintanilla household in a million years.

But she was impressed that I broke an impasse in her research and resolved a discrepancy between Mathematica 4 and Mathematica 8 by finding the following integral by hand in less than an hour:

$\displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

Yes, I married well indeed.

In this post, I collect the posts that I wrote last summer regarding various ways of computing this integral.

Part 1: Introduction

Part 2a, 2b, 2c, 2d, 2e, 2f: Changing the endpoints of integration, multiplying top and bottom by $\sec^2 x$ , and the substitution $u = \tan x$ .

Part 3a, 3b, 3c, 3d, 3e, 3f, 3g, 3h, 3i: Double-angle trig identity, combination into a single trig function, changing the endpoints of integration, and the magic substitution $u = \tan \theta/2$ .

Part 4a, 4b, 4c, 4d, 4e, 4f, 4g, 4h: Double-angle trig identity, combination into a single trig function, changing the endpoints of integration, and contour integration using the unit circle

Part 5a, 5b, 5c, 5d, 5e, 5f, 5g, 5h, 5i, 5j: Independence of the parameter $a$ , the magic substitution $u = \tan \theta/2$ , and partial fractions.

Part 6a, 6b, 6c, 6d, 6e, 6f, 6g:Independence of the parameter $a$ , the magic substitution $u = \tan \theta/2$ , and contour integration using the real line and an expanding semicircle.

Part 7: Concluding thoughts… and ways that should work that I haven’t completely figured out yet.

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

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

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

So far, I’ve shown that the denominator can be factored over the real numbers:

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

so that the technique of partial fractions can be applied. Since both quadratics in the denominator are irreducible (and the degree of the numerator is less than the degree of the denominator), the partial fractions decomposition has the form

$\displaystyle \frac{1}{\left(x^2 - x \sqrt{2} + 1 \right) \left(x^2 + x \sqrt{2} + 1\right)} = \displaystyle \frac{Ax + B}{\left(x^2 - x \sqrt{2} + 1 \right)} + \displaystyle \frac{Cx + D}{ \left(x^2 + x \sqrt{2} + 1\right)}$

Clearing out the denominators, I get

$1 = (Ax + B) \left(x^2 + x \sqrt{2} + 1\right) + (Cx + D) \left(x^2 - x \sqrt{2} + 1\right)$

$1 = Ax^3 + Bx^2 + Ax^2 \sqrt{2} + Bx\sqrt{2} + Ax + B + Cx^3 + Dx^2 - Cx^2 \sqrt{2} - Dx\sqrt{2} + Cx + D$

$0x^3 + 0x^2 + 0x + 1 = (A + C)x^3 + (A \sqrt{2} + B - C \sqrt{2} + D)x^2 + (A + B\sqrt{2} + C - D \sqrt{2} ) x + (B+D)$

Matching coefficients yields the following system of four equations in four unknowns:

$A + C = 0$

$A\sqrt{2} + B - C\sqrt{2} + D = 0$

$A + B \sqrt{2} + C - D\sqrt{2} = 0$

$B + D = 1$

Ordinarily, four-by-four systems of linear equations are somewhat painful to solve, but this system isn’t too bad. Since $A + C = 0$ from the first equation, the third equation becomes

$0 + B \sqrt{2} - D \sqrt{2} = 0$ , or $B = D$ .

From the fourth equation, I can conclude that $B = 1/2$ and $D = 1/2$ . The second and third equations then become

$A\sqrt{2} + \displaystyle \frac{1}{2} - C\sqrt{2} + \frac{1}{2} = 0$

$A + \displaystyle \frac{\sqrt{2}}{2} + C - \frac{\sqrt{2}}{2} = 0$ ,

$A - C = \displaystyle -\frac{\sqrt{2}}{2}$ ,

$A + C = 0$ .

Adding the two equations yields $2A = -\displaystyle \frac{\sqrt{2}}{4}$ , so that $A = -\displaystyle \frac{\sqrt{2}}{4}$ and $C = \displaystyle \frac{\sqrt{2}}{4}$ .

Therefore, the integral can be rewritten as

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

I’ll start evaluating this integral in tomorrow’s post.

How I Impressed My Wife: Part 5j

Earlier in this series, I gave three different methods of showing that

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.$

Using the fact that $Q$ is independent of $a$ , I’ll now give a fourth method.

Since $Q$ is independent of $a$ , I can substitute any convenient value of $a$ that I want without changing the value of $Q$ . As shown in previous posts, substituting $a =0$ yields the following simplification:

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\pi}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + (4 b^2 - 2) u^2 + 1}$

$= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{ ([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2}$

$\displaystyle \int_{-\infty}^{\infty} \left[ \frac{1}{[u - \sqrt{1-b^2}]^2 +b^2} + \frac{1}{[u + \sqrt{1-b^2}]^2 +b^2} \right] du$

if $|b| < 1$ . (The cases $|b| = 1$ and $|b| > 1$ have already been handled earlier in this series.)

To complete the calculation, I employ the now-familiar antiderivative

$\displaystyle \int \frac{dx}{x^2 + b^2} = \displaystyle \frac{1}{|b|} \tan^{-1} \left( \frac{x}{b} \right)$ .

Using this antiderivative and a simple substitution, I see that

$Q = \displaystyle \left[ \frac{1}{|b|} \tan^{-1} \left( \frac{u - \sqrt{1-b^2}}{b} \right) + \frac{1}{|b|} \tan^{-1} \left( \frac{u - \sqrt{1-b^2}}{b} \right) \right]^{\infty}_{-\infty}$

$= \displaystyle \frac{1}{|b|} \left[ \left( \frac{\pi}{2} + \frac{\pi}{2} \right) - \left( \frac{-\pi}{2} + \frac{-\pi}{2} \right) \right]$

$= \displaystyle \frac{2\pi}{|b|}$ .

This completes the fourth method of evaluating the integral $Q$ , using partial fractions.

There’s at least one more way that the integral $Q$ can be calculated, which I’ll begin with tomorrow’s post.

How I Impressed My Wife: Part 5i