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.

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:

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

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

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

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

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

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

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

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

One thought on “How I Impressed My Wife: Part 5f”

How I Impressed My Wife: Part 5f

Published by John Quintanilla

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