How I Impressed My Wife: Part 5c

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 $Q$ . For example, let me substitute $a =0$ :

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

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

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

So that I can employ the magic substitution $u = \tan x/2$ , I’ll divide the interval of integration into two pieces and then perform the substitution $x = t + 2\pi$ on the second piece:

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

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

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

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

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

I’ll continue with this fourth evaluation of the integral in tomorrow’s post.

How I Impressed My Wife: Part 5b

Amazingly, the integral below has a simple solution:

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

Even more amazingly, the integral $Q$ ultimately does not depend on the parameter $a$ . For several hours, I tried to figure out a way to demonstrate that $Q$ is independent of $a$ , but I couldn’t figure out a way to do this without substantially simplifying the integral, but I’ve been unable to do so (at least so far).

So here’s what I have been able to develop to prove that $Q$ is independent of $a$ without directly computing the integral $Q$ .

Earlier in this series, I showed that

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

$= 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{1 + 2 a \tan x + (a^2 + b^2) \tan^2 x}$

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

$= \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{dv}{v^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }$

$= \displaystyle 2 \int_{-\infty}^{\infty} \frac{(a^2+ b^2) dv}{(a^2 + b^2) v^2 + b^2 }$

Yesterday, I showed used the substitution $w = (a^2 + b^2) v$ to show that $Q$ was independent of $a$ . Today, I’ll use a different method to establish the same result. Let

$Q(a) = \displaystyle 2 \int_{-\infty}^{\infty} \frac{(a^2+b^2) dv}{(a^2+b^2)^2 v^2 + b^2 }$ .

Notice that I’ve written this integral as a function of the parameter $a$ . I will demonstrate that $Q'(a) = 0$ , so that $Q(c)$ is a constant with respect to $a$ . In other words, $Q(a)$ does not depend on $a$ .

To do this, I differentiate under the integral sign with respect to $a$ (as opposed to $x$ ) using the Quotient Rule:

$Q'(a) = \displaystyle 2 \int_{-\infty}^{\infty} \frac{ 2a \left[ (a^2+b^2)^2 v^2 + b^2\right] - 2 (a^2+b^2) \cdot (a^2+b^2) v^2 \cdot 2a }{\left[ (a^2+b^2)^2 v^2 + b^2 \right]^2} dv$

$Q'(a) = \displaystyle 4a \int_{-\infty}^{\infty} \frac{(a^2+b^2)^2 v^2 + b^2- 2 (a^2+b^2)^2 v^2}{\left[ (a^2+b^2)^2 v^2 + b^2 \right]^2} dv$

$Q'(a) = \displaystyle 4a \int_{-\infty}^{\infty} \frac{b^2-(a^2+b^2)^2 v^2}{\left[ (a^2+b^2)^2 v^2 + b^2 \right]^2} dv$

I now apply the trigonometric substitution $v = \displaystyle \frac{b}{a^2+b^2} \tan \theta$ , so that

$(a^2+b^2)^2 v^2 = (a^2+b^2)^2 \displaystyle \left[ \frac{b}{a^2+b^2} \tan \theta \right]^2 = b^2 \tan^2 \theta$

and

$dv = \displaystyle \frac{b}{a^2+b^2} \sec^2 \theta \, d\theta$

The endpoints of integration change from $-\infty < v < \infty$ to $-\pi/2 < \theta < \pi/2$ , and so

$Q'(a) = \displaystyle 4a \int_{-\pi/2}^{\pi/2} \frac{b^2- b^2 \tan^2 \theta}{\left[ b^2 \tan^2 \theta + b^2 \right]^2} \frac{b}{a^2+b^2} \sec^2 \theta \, d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1- \tan^2 \theta] \sec^2 \theta}{\left[ \tan^2 \theta +1 \right]^2} d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1-\tan^2 \theta] \sec^2 \theta}{\left[ \sec^2 \theta \right]^2} d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1-\tan^2 \theta] \sec^2 \theta}{\sec^4 \theta} d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1- \tan^2 \theta]}{\sec^2 \theta} d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} [1- \tan^2 \theta] \cos^2 \theta \, d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} [\cos^2 \theta -\sin^2 \theta] d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \cos 2\theta \, d\theta$

$= \displaystyle \left[ \frac{2ab^3}{a^2+b^2} \sin 2\theta \right]^{\pi/2}_{-\pi/2}$

$= \displaystyle \frac{2ab^3}{a^2+b^2} \left[ \sin \pi - \sin (-\pi) \right]$

$= \displaystyle \frac{2ab^3}{a^2+b^2} \left[ 0- 0 \right]$

$= 0$ .

I’m not completely thrilled with this demonstration that $Q$ is independent of $a$ , mostly because I had to do so much simplification of the integral $Q$ to get this result. As I mentioned in yesterday’s post, I’d love to figure out a way to directly start with

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

and demonstrate that $Q$ is independent of $a$ , perhaps by differentiating $Q$ with respect to $a$ and demonstrating that the resulting integral must be equal to 0. However, despite several hours of trying, I’ve not been able to establish this result without simplifying $Q$ first.

How I Impressed My Wife: Part 5a

Amazingly, the integral below has a simple solution:

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

So here’s what I have been able to develop to prove that $Q$ is independent of $a$ without directly computing the integral $Q$ .

Earlier in this series, I showed that

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

$= 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{1 + 2 a \tan x + (a^2 + b^2) \tan^2 x}$

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

$= \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{dv}{v^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }$

I now multiply the top and bottom of this last integral by $a^2 + b^2$ :

$Q = \displaystyle \frac{2}{(a^2+b^2)^2} \int_{-\infty}^{\infty} \frac{(a^2+ b^2) dv}{v^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }$

$= \displaystyle 2 \int_{-\infty}^{\infty} \frac{(a^2+ b^2) dv}{(a^2 + b^2) v^2 + b^2 }$

I now employ the substitution $w = (a^2 + b^2) v$ , so that $dw = (a^2 + b^2) v$ . Since $a^2 + b^2 > 0$ , the endpoints of integration do not change, and so

$Q = \displaystyle 2 \int_{-\infty}^{\infty} \frac{dw}{w^2 + b^2 }$ .

This final integral is independent of $a$ .

Since $Q$ is independent of $a$ , I can substitute any convenient value of $a$ that I wish. For example, I can let $a = 0$ without altering the value of $Q$ :

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x} = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

This provides a considerable simplification of the integral $Q$ which also opens up additional methods of evaluation.

How I Impressed My Wife: Part 4h

So far in this series, I have used three different techniques to show that

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

For the third technique, a key step in the calculation was showing that the residue of the function

$f(z) = \displaystyle \frac{1}{z^2 + 2\frac{S}{R}z + 1} = \displaystyle \frac{1}{(z-r_1)(z-r_2)}$

at the point

$r_1 = \displaystyle \frac{-S + \sqrt{S^2 -R^2}}{R}$

was equal to

$\displaystyle \frac{R}{ 2 \sqrt{S^2-R^2} }$ .

Initially, I did this by explicitly computing the Laurent series expansion about $z = r_1$ and identifying the coefficient for the term $(z-r_1)^{-1}$ .

In this post, I’d like to discuss another way that this residue could have been obtained.

Notice that the function $f(z)$ has the form $\displaystyle \frac{g(z)}{(z-r) h(z)}$ , where $g$ and $h$ are differentiable functions so that $g(r) \ne 0$ and $h(r) \ne 0$ . Therefore, we may rewrite this function using the Taylor series expansion of $\displaystyle \frac{g(z)}{h(z)}$ about $z = r$ :

$f(z) = \displaystyle \frac{1}{z-r} \left[ \frac{g(z)}{h(z)} \right]$

$f(z) = \displaystyle \frac{1}{z-r} \left[ a_0 + a_1 (z-r) + a_2 (z-r)^2 + a_3 (z-r)^3 + \dots \right]$

$f(z) = \displaystyle \frac{a_0}{z-r} + a_1 + a_2 (z-r) + a_3 (z-r)^2 + \dots$

Clearly,

$\displaystyle \lim_{z \to r} (z-r) f(z) = \displaystyle \lim_{z \to r} \left[ a_0 + a_1 (z-r) + a_2 (z-r)^2 + a_3 (z-r)^3 + \dots \right] = a_0$

Therefore, the residue at $z = r$ can be found by evaluating the limit $\displaystyle \lim_{z \to r} (z-r) f(z)$ . Notice that

$\displaystyle \lim_{z \to r} (z-r) f(z) = \displaystyle \lim_{z \to r} \frac{(z-r) g(z)}{(z-r) h(z)}$

$= \displaystyle \lim_{z \to r} \frac{(z-r) g(z)}{H(z)}$ ,

where $H(z) = (z-r) h(z)$ is the original denominator of $f(z)$ . By L’Hopital’s rule,

$a_0 = \displaystyle \lim_{z \to r} \frac{(z-r) g(z)}{H(z)} = \displaystyle \lim_{z \to r} \frac{g(z) + (z-r) g'(z)}{H'(z)} = \displaystyle \frac{g(r)}{H'(r)}$ .

For the function at hand, $g(z) \equiv 1$ and $H(z) = z^2 + 2\frac{S}{R}z + 1$ , so that $H'(z) = 2z + 2\frac{S}{R}$ . Therefore, the residue at $z = r_1$ is equal to

$\displaystyle \frac{1}{2r_1+2 \frac{S}{R}} = \displaystyle \frac{1}{2 \displaystyle \frac{-S + \sqrt{S^2 -R^2}}{R} + 2 \frac{S}{R}}$

$= \displaystyle \frac{1}{ ~ 2 \displaystyle \frac{\sqrt{S^2 -R^2}}{R} ~ }$

$= \displaystyle \frac{R}{2 \sqrt{S^2-R^2}}$ ,

matching the result found earlier.

How I Impressed My Wife: Part 4g

So far in this series, I have used three different techniques to show that

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

For the third technique, a key step in the calculation was showing that the residue of the function

$f(z) = \displaystyle \frac{1}{z^2 + 2\frac{S}{R}z + 1} = \displaystyle \frac{1}{(z-r_1)(z-r_2)}$

at the point

$r_1 = \displaystyle \frac{-S + \sqrt{S^2 -R^2}}{R}$

was equal to

$\displaystyle \frac{R}{ 2 \sqrt{S^2-R^2} }$ .

Initially, I did this by explicitly computing the Laurent series expansion about $z = r_1$ and identifying the coefficient for the term $(z-r_1)^{-1}$ .

In this post and the next post, I’d like to discuss alternate ways that this residue could have been obtained.

$f(z) = \displaystyle \frac{1}{z-r} \left[ \frac{g(z)}{h(z)} \right]$

$f(z) = \displaystyle \frac{1}{z-r} \left[ a_0 + a_1 (z-r) + a_2 (z-r)^2 + a_3 (z-r)^3 + \dots \right]$

$f(z) = \displaystyle \frac{a_0}{z-r} + a_1 + a_2 (z-r) + a_3 (z-r)^2 + \dots$

Therefore, the residue at $z = r$ is equal to $a_0$ , or the constant term in the Taylor expansion of $\displaystyle \frac{g(z)}{h(z)}$ about $z = r$ . Therefore,

$a_0 = \displaystyle \frac{g(r)}{h(r)}$

For the function at hand $g(z) \equiv 1$ and $h(z) = z-r_2$ . Therefore, the residue at $z = r_1$ is equal to $\displaystyle \frac{1}{r_1 - r_2}$ , matching the result found earlier.

How I Impressed My Wife: Part 4f

Previously in this series, I have used two different techniques to show that

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

Originally, my wife had asked me to compute this integral by hand because Mathematica 4 and Mathematica 8 gave different answers. At the time, I eventually obtained the solution by multiplying the top and bottom of the integrand by $\sec^2 x$ and then employing the substitution $u = \tan x$ (after using trig identities to adjust the limits of integration).

But this wasn’t the only method I tried. Indeed, I tried two or three different methods before deciding they were too messy and trying something different. So, for the rest of this series, I’d like to explore different ways that the above integral can be computed.

Previously in this series, I have used two different techniques to show that

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

Here’s my progress so far:

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

$= \displaystyle \int_0^{2\pi} \frac{2 \, dx}{1+\cos 2x + 2 a \sin 2x + (a^2 + b^2)(1-\cos 2x)}$

$= 2 \displaystyle \int_0^{2\pi} \frac{d\theta}{(1+a^2+b^2) + 2 a \sin \theta + (1 - a^2 - b^2) \cos \theta}$

$= 2 \displaystyle \int_{0}^{2\pi} \frac{d\theta}{S + R \cos (\theta - \alpha)}$

$= 2 \displaystyle \int_{0}^{2\pi} \frac{d\phi}{S + R \cos \phi}$

$= \displaystyle -\frac{4i}{R} \oint_C \frac{dz}{z^2 + 2\frac{S}{R}z + 1}$ ,

where this last integral is taken over the complex plane on the unit circle, a closed contour oriented counterclockwise. In these formulas, $R = \sqrt{(2a)^2 + (1-a^2-b^2)^2}$ and $S = 1 + a^2 + b^2$ . (Also, $\alpha$ is a certain angle that is now irrelevant at this point in the calculation).

This contour integral looks complicated; however, it’s an amazing fact that integrals over closed contours can be easily evaluated by only looking at the poles of the integrand. In recent posts, I established that there was only one pole inside the contour, and the residue at this pole was equal to $\displaystyle \frac{R}{ 2 \sqrt{S^2-R^2} }$ .

This residue can be used to evaluate the contour integral. Ordinarily, integrals are computed by subtracting the values of the antiderivative at the endpoints. However, there is an alternate way of computing a contour integral using residues. It turns out that the value of the contour integral is $2\pi i$ times the sum of the residues within the contour; see Wikipedia and Mathworld for more information.

Therefore,

$Q = \displaystyle -\frac{4i}{R} \oint_C \frac{dz}{(z - r_1)(z- r_2)}$

$= \displaystyle -\frac{4i}{R} \cdot 2\pi i \cdot \frac{R}{ 2 \sqrt{S^2-R^2} }$

$= \displaystyle \frac{4\pi}{\sqrt{S^2-R^2}}$

Next, I use some algebra to simplify the denominator:

$S^2 - R^2 = (1+a^2+b^2)^2 - (1-a^2-b^2)^2 - (2a)^2$

$S^2 - R^2 = [(1 + a^2 + b^2) + (1-a^2-b^2)][(1 + a^2 + b^2) - (1 - a^2 -b^2)] - 4a^2$

$S^2 - R^2 = 2[2 a^2 + 2b^2] - 4a^2$

$S^2 - R^2 = 4b^2$

Therefore,

$Q = \displaystyle \frac{4\pi}{\sqrt{4b^2}} = \displaystyle \frac{4\pi}{2|b|} = \frac{2\pi}{|b|}$

Once again, this matches the solution found with the previous methods… and I was careful to avoid a common algebraic mistake.

In tomorrow’s post, I’ll discuss an alternative way of computing the residue.

How I Impressed My Wife: Part 4e

Previously in this series, I have used two different techniques to show that

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

Here’s my progress so far:

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

$= \displaystyle \int_0^{2\pi} \frac{2 \, dx}{1+\cos 2x + 2 a \sin 2x + (a^2 + b^2)(1-\cos 2x)}$

$= 2 \displaystyle \int_0^{2\pi} \frac{d\theta}{(1+a^2+b^2) + 2 a \sin \theta + (1 - a^2 - b^2) \cos \theta}$

$= 2 \displaystyle \int_{0}^{2\pi} \frac{d\theta}{S + R \cos (\theta - \alpha)}$

$= 2 \displaystyle \int_{0}^{2\pi} \frac{d\phi}{S + R \cos \phi}$

$= \displaystyle -\frac{4i}{R} \oint_C \frac{dz}{z^2 + 2\frac{S}{R}z + 1}$

$= \displaystyle -\frac{4i}{R} \oint_C \frac{dz}{(z - r_1)(z- r_2)}$

where this last integral is taken over the complex plane on the unit circle, a closed contour oriented counterclockwise. Also,

$r_1 = \displaystyle \frac{-S + \sqrt{S^2 -R^2}}{R}$

and

$r_2 = \displaystyle \frac{-S - \sqrt{S^2 -R^2}}{R}$ ,

are the two distinct roots of the denominator (as long as $b \ne 0$ ). In these formulas, $R = \sqrt{(2a)^2 + (1-a^2-b^2)^2}$ and $S = 1 + a^2 + b^2$ . (Also, $\alpha$ is a certain angle that is now irrelevant at this point in the calculation).

This contour integral looks complicated; however, it’s an amazing fact that integrals over closed contours can be easily evaluated by only looking at the poles of the integrand. In yesterday’s post, I established that $r_1$ lies inside the contour, but $r_2$ lies outside of the contour.

The next step of the calculation is finding the residue at $r_1$ ; see Wikipedia and Mathworld for more information. This means rewriting the rational function

$\displaystyle \frac{1}{(z - r_1)(z - r_2)}$

as a power series (technically, a Laurent series) about the point $z = r_1$ . This can be done by using the formula for an infinite geometric series (see here, here, and here):

$\displaystyle \frac{1}{(z - r_1)(z - r_2)} = \displaystyle \frac{1}{z-r_1} \times \frac{1}{z-r_2}$

$= \displaystyle \frac{-1}{z-r_1} \times \frac{1}{r_2-z}$

$= \displaystyle \frac{-1}{z-r_1} \times \frac{1}{(r_2-r_1) - (z-r_1)}$

$= \displaystyle \frac{-1}{z-r_1} \times \frac{1}{r_2-r_1} \times \frac{ 1}{ 1 - \displaystyle \frac{z-r_1}{r_2-r_1} }$

$= \displaystyle \frac{-1}{z-r_1} \times \frac{1}{r_2-r_1} \left[ 1 + \left( \displaystyle \frac{z-r_1}{r_2-r_1} \right) + \left( \displaystyle \frac{z-r_1}{r_2-r_1} \right)^2 + \left( \displaystyle \frac{z-r_1}{r_2-r_1} \right)^3 + \dots \right]$

$= \displaystyle \frac{-1}{z-r_1} \times \frac{1}{r_2-r_1} - \frac{1}{(r_2-r_1)^2} - \frac{z-r_1}{(r_2-r_1)^3} - \frac{(z-r_1)^2}{(r_2-r_1)^4} \dots$

The residue of the function at $z = r_1$ is defined to be the constant multiplying the $\displaystyle \frac{1}{z-r_1}$ term in the above series. Therefore,

The residue at $x = r_1$ is $\displaystyle \frac{-1}{r_2-r_1} = \displaystyle \frac{1}{r_1-r_2}$

From the definitions of $r_1$ and $r_2$ above,

$\displaystyle \frac{1}{r_1-r_2} = \displaystyle \frac{1}{\displaystyle \frac{-S + \sqrt{S^2 -R^2}}{R} - \frac{-S - \sqrt{S^2 -R^2}}{R}}$

$= \displaystyle \frac{1}{ ~ 2 \displaystyle \frac{\sqrt{S^2-R^2}}{R} ~ }$

$= \displaystyle \frac{R}{ 2 \sqrt{S^2-R^2} }$

Now that I’ve identified the residue of the only root that lies inside of the contour, we are in position to evaluate the contour integral above. I’ll discuss this in tomorrow’s post.

How I Impressed My Wife: Part 4d