Author: John Quintanilla

I'm a Professor of Mathematics and a University Distinguished Teaching Professor at the University of North Texas. For eight years, I was co-director of Teach North Texas, UNT's program for preparing secondary teachers of mathematics and science.

How I Impressed My Wife: Part 5h

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.
green lineSince 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}

The four roots of the denominator satisfy

u^2 = \displaystyle 1 - 2b^2 \pm 2|b| \sqrt{b^2 - 1}

So far, I’ve handled the cases |b| = 1 and |b| > 1. In today’s post, I’ll start considering the case |b| < 1.

Factoring the denominator is a bit more complicated if |b| < 1. Using the quadratic equation, we obtain

u^2 = \displaystyle 1 - 2b^2 \pm 2|b| i \sqrt{1-b^2}

However, unlike the cases |b| \ge 1, the right-hand side is now a complex number. So, To solve for u, I’ll use DeMoivre’s Theorem and some surprisingly convenient trig identities. Notice that

(1-2b^2)^2 + (2|b| \sqrt{1-b^2})^2 = 1 - 4b^2 + b^4 + 4b^2 (1 - b^2) = 1 - 4b^2 + b^4 + 4b^2 - b^4 = 1.

Therefore, the four complex roots of the denominator satisfy |u^2| = 1, or |u| = 1. This means that all four roots can be written in trigonometric form so that

u^2 = \cos 2\phi + i \sin 2\phi,

where 2\phi is some angle. (I chose the angle to be 2\phi instead of \phi for reasons that will become clear shortly.)

I’ll begin with solving

u^2 = \displaystyle 1 - 2b^2 + 2|b| i \sqrt{1-b^2}.

Matching the real and imaginary parts, we see that

\cos 2\phi = 1-2b^2,

\sin 2\phi = 2|b| \sqrt{1-b^2}

This completely matches the form of the double-angle trig identities

\cos 2\phi = 1 - 2\sin^2 \phi,

\sin 2 \phi = 2 \sin\phi \cos \phi,

and so the problem reduces to solving

u^2 = \cos 2\phi + i \sin 2\phi,

where $\sin \phi = |b|$ and $\cos \phi = \sqrt{1-b^2}$. By De Moivre’s Theorem, I can conclude that the two solutions of this equation are

u = \pm(\cos \phi + i \sin \phi),

or

u = \pm( \sqrt{1-b^2} + i |b|).

I could re-run this argument to solve u^2 = \displaystyle 1 - 2b^2 - 2|b| i \sqrt{1-b^2} and get the other two complex roots. However, by the Conjugate Root Theorem, I know that the four complex roots of the denominator u^4 + (4 b^2 - 2) u^2 + 1 must come in conjugate pairs. Therefore, the four complex roots are

u = \pm \sqrt{1-b^2} \pm i |b|.

Therefore, I can factor the denominator as follows:

u^4 + (4 b^2 - 2) u^2 + 1 = (u - [\sqrt{1-b^2} + i|b|])(u - [\sqrt{1-b^2} - i|b|])

\qquad \times (u - [-\sqrt{1-b^2} + i|b|])(u - [-\sqrt{1-b^2} + i|b|])

= (u - \sqrt{1-b^2} - i|b|)(u - \sqrt{1-b^2} + i|b])

\qquad \times (u +\sqrt{1-b^2} + i|b|)(u +\sqrt{1-b^2} + i|b|)

= ([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)

To double-check my work, I can directly multiply this product:

([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)

= (u^2 - 2u \sqrt{1-b^2} + 1 - b^2 + b^2) (u^2 + 2u \sqrt{1-b^2} + 1 - b^2 + b^2)

= ([u^2 +1] - 2u\sqrt{1-b^2})([u^2+1] + 2u\sqrt{1-b^2})

= [u^2+1]^2 - [2u\sqrt{1-b^2}]^2

= u^4 + 2u^2 + 1 - 4u^2 (1-b^2)

= u^4 + u^2 (2 - 4[1-b^2]) + 1

= u^4 + u^2 (4b^2 - 2) + 1.

So, at last, I can rewrite the integral Q as

Q = \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)}

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I’ll continue with this fourth evaluation of the integral, continuing the case |b| < 1, in tomorrow’s post.

How I Impressed My Wife: Part 5g

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.
green lineSince 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} \left[ \displaystyle \left( \frac{2 - 2k_1^2}{k_2^2 - k_1^2} \right) \frac{1}{u^2 + k_1^2} + \displaystyle \left( \frac{2 k_2^2 - 2}{k_2^2 - k_1^2} \right) \frac{1}{u^2 + k_2^2} \right] du,

where k_1 and k_2 are the positive numbers so that

k_1^2 = 2b^2 + 2|b| \sqrt{b^2-1} - 1,

k_2^2 = 2b^2 - 2|b| \sqrt{b^2-1} - 1,

so that

u^4 + (4 b^2 - 2) u^2 + 1 = (u^2 + k_1^2) (u^2 + k_2^2)

At this point in the calculation, we can employ the now familiar antiderivative

\displaystyle \int \frac{du}{u^2 +k^2} = \displaystyle \frac{1}{k} \tan^{-1} \left( \frac{u}{k} \right) + C,

so that

Q = \left[ \displaystyle \left( \frac{2 - 2k_1^2}{k_2^2 - k_1^2} \right) \frac{1}{k_1}\tan^{-1} \left( \frac{u}{k_1} \right) + \displaystyle \left( \frac{2 k_2^2 - 2}{k_2^2 - k_1^2} \right) \displaystyle \frac{1}{k_2} \tan^{-1} \left( \frac{u}{k_2} \right) \right]^{\infty}_{-\infty}

= \displaystyle \left( \frac{2 - 2k_1^2}{k_2^2 - k_1^2} \right) \frac{1}{k_1} \left( \frac{\pi}{2} - \frac{-\pi}{2} \right) + \displaystyle \left( \frac{2 k_2^2 - 2}{k_2^2 - k_1^2} \right) \displaystyle \frac{1}{k_2} \tan^{-1} \left(\frac{\pi}{2} - \frac{-\pi}{2} \right)

= \displaystyle\frac{\pi}{k_2^2-k_1^2} \left( \frac{2 - 2k_1^2}{k_1} + \frac{2 .k_2^2 - 2}{k_2} \right)

= \displaystyle\frac{\pi}{k_2^2-k_1^2} \left( \frac{(2 - 2k_1^2)k_2 + (2 k_2^2 - 2)k_1 }{k_1 k_2} \right)

= \displaystyle\frac{\pi}{k_2^2-k_1^2} \left( \frac{2 k_2 - 2k_1^2 k_2 + 2 k_1 k_2^2 - 2 k_1 }{k_1 k_2} \right)

= \displaystyle\frac{2\pi}{k_2^2-k_1^2} \left( \frac{k_2 - k_1^2 k_2 + k_1 k_2^2 - k_1 }{k_1 k_2} \right)

= \displaystyle\frac{2\pi}{k_2^2-k_1^2} \left( \frac{k_2 - k_1 - k_1^2 k_2 + k_1 k_2^2 }{k_1 k_2} \right)

= \displaystyle\frac{2\pi}{(k_2-k_1)(k_2+k_1)} \left( \frac{(k_2 - k_1) + k_1 k_2 (k_2 - k_1)}{k_1 k_2} \right)

= \displaystyle\frac{2\pi}{(k_2-k_1)(k_2+k_1)} \left( \frac{(k_2 - k_1)(1+ k_1 k_2)}{k_1 k_2} \right)

= \displaystyle 2\pi \left( \frac{1+ k_1 k_2}{(k_2+k_1) k_1 k_2} \right).

Now it remains to simplify this fraction. To do this, we note that

u^4 + (4 b^2 - 2) u^2 + 1 = (u^2 + k_1^2) (u^2 + k_2^2),

so that

u^4 + (4b^2 - 2)u^2 + 1 = u^4 + (k_1^2 + k_2^2)u^2 + k_1^2 k_2^2

Matching coefficients, we see that

k_1^2 + k_2^2 = 4b^2 -2,

k_1^2 k_2^2 = 1

Since both k_1 and k_2 are positive, we see that k_1 k_2 = 1 from the last equation. Therefore,

k_1^2 + 2 k_1 k_2 + k_2^2 = 4b^2 -2 + 2

(k_1+k_2)^2 = 4b^2

k_1 + k_2 = 2|b|

Plugging these in, we finally conclude that

Q = \displaystyle 2\pi \left( \frac{1+ 1}{2 |b| \cdot 1} \right) = \displaystyle \frac{2\pi}{|b|},

again matching our earlier result.

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Using the fourth method, I’ve shown that Q = \displaystyle \frac{2\pi}{|b|} for the cases |b| = 1 and |b| > 1. With tomorrow’s post, I’ll consider the remaining case of |b| < 1.

How I Impressed My Wife: Part 5f

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.
green lineSince 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}

The four roots of the denominator satisfy

u^2 = \displaystyle 1 - 2b^2 \pm 2|b| \sqrt{b^2 - 1}

In yesterday’s post, I handled the case |b| = 1. In today’s post, I’ll consider the case |b| > 1, so that u^2 is a real number for the four roots of the denominator.

For the sake of simplicity, let me define the positive numbers k_1 and k_2 so that

k_1^2 = 2b^2 + 2|b| \sqrt{b^2-1} - 1,

k_2^2 = 2b^2 - 2|b| \sqrt{b^2-1} - 1.

Clearly 2b^2 + 2|b| \sqrt{b^2-1} - 1 > 0 if |b| > 1, and so we can choose k_1 to be positive. For k_2, notice that

(2b^2 - 1)^2 = 4b^4 - 4b^2 + 1,

while

\left[ 2|b| \sqrt{b^2-1} \right]^2 = 4b^2 (b^2 - 1) = 4b^4 - 4b^2.

Therefore,

(2b^2 - 1)^2 > \left[ 2|b| \sqrt{b^2-1} \right]^2

2b^2 - 1 > 2|b| \sqrt{b^2-1}

2b^2 - 2|b| \sqrt{b^2-1} - 1 > 0

So k_2 can also be chosen to be a positive number.

Using k_1 and k_2, I can write

u^4 + (4 b^2 - 2) u^2 + 1 = (u^2 + k_1^2)(u^2 + k_2^2),

and so the integrand must have the partial fractions decomposition

\displaystyle \frac{ 2(1+u^2)}{u^4 + (4 b^2 - 2) u^2 + 1} = \displaystyle \frac{A}{u^2 + k_1^2} + \displaystyle \frac{B}{u^2 + k_2^2},

Notice that ordinarily, when the denominator contains an irreducible quadratic, the numerator of the partial fractions decomposition has the form Au + B and not A. However, there are no u^3 and u terms in the denominator, I can treat u^2 as the variable for the purposes of the decomposition. Since the right-hand side has linear terms in u^2, it suffices to use a constant for finding the decomposition.

To solve for the constants A and B, I clear out the denominator:

2u^2 + 2 = A \left[ u^2 + k_2^2 \right] + B \left[ u^2 + k_1^2 \right]

Matching coefficients, this yields the system of equations

A + B = 2

A k_2^2 + B k_1^2 = 2

Substituting B = 2-A into the second equation, I get

A k_2^2 + (2-A) k_1^2 = 2

2 k_1^2 + A (k_2^2 - k_1^2) = 2

A (k_2^2 - k_1^2) = 2 - 2k_1^2

A = \displaystyle \frac{2 - 2k_1^2}{k_2^2 - k_1^2}

Therefore,

B = 2 - A = 2 - \displaystyle \frac{2 - 2k_1^2}{k_2^2 - k_1^2}

B = \displaystyle \frac{2(k_2^2 - k_1^2) - (2 - 2k_1^2)}{k_2^2 - k_1^2}

B = \displaystyle \frac{2 k_2^2 - 2}{k_2^2 - k_1^2}

Therefore, the integrand has the partial fractions decomposition

\displaystyle \frac{ 2(1+u^2)}{u^4 + (4 b^2 - 2) u^2 + 1} = \displaystyle \left( \frac{2 - 2k_1^2}{k_2^2 - k_1^2} \right) \frac{1}{u^2 + k_1^2} + \displaystyle \left( \frac{2 k_2^2 - 2}{k_2^2 - k_1^2} \right) \frac{1}{u^2 + k_2^2}

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I’ll continue with this fourth evaluation of the integral, continuing the case |b| > 1, in tomorrow’s post.

How I Impressed My Wife: Part 5e

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.
green lineSince 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}

To evaluate this integral, I need to find the four complex roots of the denominator:

u^4 + (4b^2 - 2) u^2 + 1 = 0

u^2 = \displaystyle \frac{ 2 - 4b^2 \pm \sqrt{(4b^2 - 2)^2 - 4}}{2}

u^2 = \displaystyle \frac{2 - 4b^2 \pm \sqrt{16b^4 - 16b^2 - 4 + 4}}{2}

u^2 = \displaystyle \frac{2 - 4b^2 \pm \sqrt{16b^4 - 16b^2}}{2}

u^2 = \displaystyle \frac{2 - 4b^2 \pm 4|b| \sqrt{b^2 - 1}}{2}

u^2 = \displaystyle 1 - 2b^2 \pm 2|b| \sqrt{b^2 - 1}

To solve for u, there are three separate cases that have to be considered: |b| = 1, |b| > 1, and |b| < 1. I’ll begin with the easiest case of |b| = 1. In this case, the integral Q is easy to evaluate:

= Q \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^4 + (4 \cdot [1^2] - 2) u^2 + 1}

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

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

=\displaystyle \int_{-\infty}^{\infty} \frac{ 2 du}{1+u^2}

=\displaystyle \left[ 2\tan^{-1} x \right]^{\infty}_{-\infty}

=\displaystyle \left[ 2 \frac{\pi}{2} - 2 \frac{-\pi}{2} \right]

= 2\pi

This matches the expected answer of Q = \displaystyle \frac{2\pi}{|b|} since I used the assumption that |b| = 1.

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I’ll continue with this fourth evaluation of the integral, examining the two remaining cases, in future posts.

How I Impressed My Wife: Part 5d

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.
green lineSince 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}

I now employ the magic substitution u = \tan x/2, so that

\cos x = \displaystyle \frac{1-u^2}{1+u^2},

\sin x = \displaystyle \frac{2u}{1+u^2},

dx = \displaystyle \frac{2 \, du}{1+u^2}.

The endpoints change from -\pi < x < \pi to -\infty < u < \infty, and so

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

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

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

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

I have transformed the integral Q into a new integral involving a fairly simple rational function that can be evaluated using standard (and non-standard) techniques.

Hypothetically, the magic substitution can be applied to the original integral. Unfortunately, I was unable to make any headway in finding the four complex roots of the resulting rational function. However, since I made the replacement a =0 at the start, this new rational function is much more tractable.

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

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.
green lineSince 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}

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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.

green lineEarlier 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.

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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|}.

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.

green lineEarlier 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.

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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.
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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.