Is there a more elegant way to solve this integral?

Question

I was considering proposing the integral $$ \int \left(\frac{1}{1-\varepsilon \cos{\theta}}\right)^{2} \, d\theta $$ to an advanced high school physics class to show the derivation of Kepler's laws. As of now, my solution uses partial fraction decomposition and is not particularly simple. Is there a better way to evaluate this integral, even if it uses advanced math?

Answer 1

Use the substitution $$(1-\epsilon \cos\theta) (1+\epsilon\cos\phi) =1-\epsilon^2$$ so that $$\sin\theta =\frac{\sqrt{1-\epsilon ^2}\sin\phi}{1+\epsilon \cos\phi}$$ and note that $$\frac{d\theta} {d\phi} =\frac{\sqrt{1-\epsilon^2}}{1+\epsilon\cos\phi}$$ and thus the integral reduces to $$(1-\epsilon ^2)^{-3/2}\int(1+\epsilon \cos\phi)\, d\phi$$ which is integrated easily. For back substitution note that $$\cos\phi=\frac{\cos\theta-\epsilon }{1-\epsilon \cos\theta}, \sin\phi=\frac{\sqrt{1-\epsilon ^2}\sin\theta} {1-\epsilon\cos\theta} $$ and thus the final answer is $$\frac{1}{(1-\epsilon ^2)^{3/2}}\arccos\left(\frac{\cos\theta-\epsilon}{1-\epsilon \cos\theta} \right)+\frac{\epsilon} {1-\epsilon ^2}\cdot\frac{\sin\theta}{1-\epsilon \cos\theta} +C$$

Answer 2

Assuming this is meant to be the integral over a full period, I worked out four versions of it in this AoPS thread (posts #6 and #8).

First, there was the elementary method, applying the Weierstrass substitution $t=\tan\frac{\theta}{2}$ and later another trig substitution to find an antiderivative. It's about ten lines of routine but messy calculus.

Second, complex analysis. We convert it to an integral of a rational function on the circle and apply the residue theorem using partial fractions. Fundamentally routine, but complicated algebraically; it takes even more space than the real version finding the antiderivative.

Third, Fourier series. We use the series expansion $\frac1{(1-t)^2}=1+2t+3t^2+\cdots$ to expand the integrand into a trigonometric series, convert to complex form and expand out each $\cos^n$ term. Then, extract the coefficient of $e^{0i\theta}$ as a series, which is the only one that contributes to the integral. We recognize this series as a case of a known power series, and there's the answer. It's the shortest of these methods, but it requires a lot of cleverness.

Fourth, there's the area of an ellipse. There's a lot of setup to start with, deriving the equation $r=\frac{D^2-f^2}{2D}\cdot\frac{1}{1-\epsilon\cos\theta}$ for an ellipse with focus at the origin in polar coordinates ($f$ is the focal length, $D$ is the major axis, $\sqrt{D^2-f^2}$ is the minor axis). But then, if you're going to do Kepler's laws, you need this setup anyway. Once that's done, we use the polar area formula $\frac12\int_a^b r^2\,d\theta$ and compare to the rectangular area formula for the ellipse: \begin{align*}\frac12\int_0^{2\pi}\left(\frac{D^2-f^2}{2D}\cdot\frac{1}{1-\varepsilon\cos\theta}\right)^2\,d\theta &= \frac{\pi}{4}D\sqrt{D^2-f^2}\\ \frac{(D^2-f^2)^2}{8D^2} I(\varepsilon) &= \frac{\pi}{4}D\sqrt{D^2-f^2}\\ I(\varepsilon) &= \frac{\pi}{4} D\sqrt{D^2-f^2}\cdot\frac{8D^2}{(D^2-f^2)^2}\\ I(\varepsilon) &= 2\pi \frac{D^3}{(D^2-f^2)^{3/2}}\\ I(\varepsilon) &= 2\pi\frac{D^3}{\left(D^2(1-\varepsilon^2)\right)^{3/2}} = \frac{2\pi}{(1-\varepsilon^2)^{3/2}}\end{align*} I followed this last one with my own derivation of Kepler's laws. It's definitely my choice for the purpose indicated in the question.

Answer 3

Differentiate $$ \left(\frac{\varepsilon\sin\theta}{1-\varepsilon \cos{\theta}}\right)’ = \frac{1-\varepsilon^2}{(1-\varepsilon\cos\theta)^2} -\frac1{1-\varepsilon\cos\theta} $$ and integrate both sides to simplify the integral $$ \int\frac{1-\varepsilon^2}{(1-\varepsilon\cos\theta)^2}d\theta=\frac{\varepsilon\sin\theta}{1-\varepsilon \cos{\theta}} +\int \frac1{1-\varepsilon\cos\theta}d\theta $$

Answer 4

The tangent half-angle substitution, $$\theta = 2 \arctan t, \qquad d\theta = \frac{2 \,dt}{1 + t^2}$$ transforms the integral to $$2 \int \frac{1 + t^2}{[(1 + \epsilon) t^2 + (1 - \epsilon)]} \,dt .$$

• In the elliptic case ($\epsilon < 1$) the form of the denominator suggests the substitution $$u = \sqrt\frac{1 + \epsilon}{1 - \epsilon} \,t,$$ which transforms the integral to \begin{align*} \frac{2}{(1 - \epsilon^2)^{3 / 2}} \int \frac{(1 - \epsilon) u^2 + (1 + \epsilon)}{(1 + u^2)^2} \,dv &= \frac{2}{(1 - \epsilon^2)^{3 / 2}} \int \left[\frac{2 \epsilon}{(1 + u^2)^2} + \frac{1 - \epsilon}{1 + u^2}\right] \,du \\ &= \frac{2}{(1 - \epsilon^2)^{3 / 2}} \left(\frac{\epsilon u}{1 + u^2} + \arctan u\right) + C. \end{align*}
• In the parabolic case ($\epsilon = 1$), the integrand in $t$ is $$\frac{1}{2} \left(t^{-2} + t^{-4}\right).$$ Alternatively, we can rewrite the original integrand in $\theta$ as $$\csc^2 \theta \cot^2 \theta + 2 \csc^3 \theta \cot \theta + \csc^4 \theta .$$
• The hyperbolic case ($\epsilon > 1)$ is similar to the elliptic case, proceeding instead with the substitution $v = \sqrt\frac{\epsilon + 1}{\epsilon - 1}$, which leads to an inverse hyperbolic tangent term in place of the arctangent one.