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)}

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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