Earlier in this series, I gave three different methods of showing that

Using the fact that

is independent of

, I’ll now give a fourth method.

Since

is independent of

, I can substitute any convenient value of

that I want without changing the value of

. As shown in previous posts, substituting

yields the following simplification:

The four roots of the denominator satisfy

So far, I’ve handled the cases and . In today’s post, I’ll start considering the case .

Factoring the denominator is a bit more complicated if . Using the quadratic equation, we obtain

However, unlike the cases , the right-hand side is now a complex number. So, To solve for , I’ll use DeMoivre’s Theorem and some surprisingly convenient trig identities. Notice that

.

Therefore, the four complex roots of the denominator satisfy , or . This means that all four roots can be written in trigonometric form so that

,

where is some angle. (I chose the angle to be instead of for reasons that will become clear shortly.)

I’ll begin with solving

.

Matching the real and imaginary parts, we see that

,

This completely matches the form of the double-angle trig identities

,

,

and so the problem reduces to solving

,

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

,

or

.

I could re-run this argument to solve and get the other two complex roots. However, by the Conjugate Root Theorem, I know that the four complex roots of the denominator must come in conjugate pairs. Therefore, the four complex roots are

.

Therefore, I can factor the denominator as follows:

To double-check my work, I can directly multiply this product:

.

So, at last, I can rewrite the integral as

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

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