How I Impressed My Wife: Part 5j

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.

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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:

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

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

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

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

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

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

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

if $|b| < 1$ . (The cases $|b| = 1$ and $|b| > 1$ have already been handled earlier in this series.)

To complete the calculation, I employ the now-familiar antiderivative

$\displaystyle \int \frac{dx}{x^2 + b^2} = \displaystyle \frac{1}{|b|} \tan^{-1} \left( \frac{x}{b} \right)$ .

Using this antiderivative and a simple substitution, I see that

$Q = \displaystyle \left[ \frac{1}{|b|} \tan^{-1} \left( \frac{u - \sqrt{1-b^2}}{b} \right) + \frac{1}{|b|} \tan^{-1} \left( \frac{u - \sqrt{1-b^2}}{b} \right) \right]^{\infty}_{-\infty}$

$= \displaystyle \frac{1}{|b|} \left[ \left( \frac{\pi}{2} + \frac{\pi}{2} \right) - \left( \frac{-\pi}{2} + \frac{-\pi}{2} \right) \right]$

$= \displaystyle \frac{2\pi}{|b|}$ .

This completes the fourth method of evaluating the integral $Q$ , using partial fractions.

There’s at least one more way that the integral $Q$ can be calculated, which I’ll begin with tomorrow’s post.

How I Impressed My Wife: Part 5i

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.

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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:

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

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

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

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

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

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

if $|b| < 1$ . (The cases $|b| = 1$ and $|b| > 1$ have already been handled earlier in this series.)

I will evaluate this integral using partial fractions. The denominator factors as the product of two irreducible quadratics, and so I must solve

$\displaystyle \frac{ 2(1+u^2)}{ ([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)} = \displaystyle \frac{Au + B}{[u - \sqrt{1-b^2}]^2 +b^2} + \frac{Cu + D}{[u + \sqrt{1-b^2}]^2 +b^2}$ ,

or

$\displaystyle \frac{ 2(1+u^2)}{ ([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)} = \displaystyle \frac{Au + B}{u^2 - 2u \sqrt{1-b^2} +1} + \frac{Cu + D}{u^2 + 2u\sqrt{1-b^2} + 1}$ .

I now clear out the denominators:

$2(1+u^2) = (Au+B)(u^2 + 2u\sqrt{1-b^2} + 1) + (Cu+D)(u^2 - 2u\sqrt{1-b^2} + 1)$

Now I multiply out the right-hand side:

$Au^3 + 2Au^2 \sqrt{1-b^2} + Au + Bu^2 + 2Bu\sqrt{1-b^2} + B$

$+ Cu^3 - 2Cu^2 \sqrt{1-b^2} + Cu + Du^2 - 2Du\sqrt{1-b^2} + D$

Equating this with $2u^2 + 2$ and matching coefficients yields the following system of four equations in four unknowns:

$A + C = 0$

$2A\sqrt{1-b^2} + B - 2C \sqrt{1-b^2} + D = 2$

$Au + 2Bu \sqrt{1-b^2} + Cu - 2Du \sqrt{1-b^2} = 0$

$B + D = 2$

Ordinarily, four-by-four systems of equations are rather painful to solve, but this system isn’t so bad.

From the first equation, I see that $C = -A$ .

From the third equation, I see that

$Au + 2Bu \sqrt{1-b^2} + Cu - 2Du \sqrt{1-b^2} = 0$

$Au + 2Bu \sqrt{1-b^2} -Au - 2Du \sqrt{1-b^2} = 0$

$2Bu \sqrt{1-b^2} - 2Du \sqrt{1-b^2} = 0$

$B - D = 0$

$B = D$ .

From the fourth equation, I see that

$B + D = 2$

$B + B = 2$

$2B = 2$

$B = 1$ ,

so that $D =1$ as well. Finally, from the second equation, I see that

$2A\sqrt{1-b^2} + B - 2C \sqrt{1-b^2} + D = 2$

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

$4A\sqrt{1-b^2} = 0$

$A= 0$ ,

so that $C = 0$ as well. This yields the partial fractions decomposition

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

This can be confirmed by directly adding the fractions on the right-hand side:

$\displaystyle \frac{1}{[u - \sqrt{1-b^2}]^2 +b^2} + \frac{1}{[u + \sqrt{1-b^2}]^2 +b^2}$

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

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

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

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 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.

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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

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

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

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

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.

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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:

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

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

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

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

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

$= \displaystyle \int_{-\infty}^{\infty} \left[ \displaystyle \left( \frac{2 - 2k_1^2}{k_2^2 - k_1^2} \right) \frac{1}{u^2 + k_1^2} + \displaystyle \left( \frac{2 k_2^2 - 2}{k_2^2 - k_1^2} \right) \frac{1}{u^2 + k_2^2} \right] du$ ,

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.

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.

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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.

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 line

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:

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

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

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

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

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

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 $|b| = 1$ .

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.

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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:

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

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

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

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

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.

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 line

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:

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

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

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

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

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

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

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.