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}

green line

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

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1 Comment

  1. How I Impressed My Wife: Index | Mean Green Math

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