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

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
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