How I Impressed My Wife: Part 2f

Some husbands try to impress their wives by lifting extremely heavy objects or other extraordinary feats of physical prowess.

That will never happen in the Quintanilla household in a million years.

But she was impressed that I broke an impasse in her research and resolved a discrepancy between Mathematica 4 and Mathematica 8 by finding the following integral by hand in less than an hour:

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

In this series, I’ll explore different ways of evaluating this integral.green lineSo far in this series, I’ve shown 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 \frac{2}{|b|} \int_{-\infty}^{\infty} \frac{ dw }{w^2 +1}

This last integral can be evaluated using a standard trick. Let \theta = \tan^{-1} w, so that w = \tan \theta. We differentiate this last equation with respect to w:

\displaystyle \frac{dw}{dw} = \sec^2 \theta \cdot \displaystyle \frac{d\theta}{dw}

Employing a Pythagorean identity, we have

1 = (1+ \tan^2 \theta) \cdot \displaystyle \frac{d\theta}{dw}

Since w = \tan \theta, we may rewrite this as

1 = (1+ w^2) \cdot \displaystyle \frac{d\theta}{dw}

\displaystyle \frac{1}{1+w^2} = \displaystyle \frac{d\theta}{dw}

\displaystyle \frac{1}{1+w^2} = \displaystyle \frac{d}{dw} \tan^{-1} w

Integrating both sides with respect to w, we obtain the antiderivative

\displaystyle \int \frac{1}{1+w^2} = \tan^{-1} w + C

We now employ this antiderivative to evaluate Q:

Q = \displaystyle \frac{2}{|b|} \int_{-\infty}^{\infty} \frac{ dw }{w^2 +1}

= \displaystyle \frac{2}{|b|} \displaystyle \left[ \tan^{-1} w \right]^{\infty}_{-\infty}

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

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

And so, at long last, we have arrived at the solution for the integral Q. Surprisingly, the answer is independent of the parameter a.

green line

These last few posts illustrated the technique that I used to compute this integral for my wife in support of her recent paper in Physical Review A. However, I had more than a few false starts along the way… or, at the time, I thought they were false starts. It turns out that there are multiple ways of evaluating this integral, and I’ll explore another method of attack beginning with tomorrow’s post.

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1 Comment

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

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