Earlier in this series, I gave three different methods of showing that
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
In yesterday’s post, I handled the case . In today’s post, I’ll consider the case , so that is a real number for the four roots of the denominator.
For the sake of simplicity, let me define the positive numbers and so that
Clearly if , and so we can choose to be positive. For , notice that
So can also be chosen to be a positive number.
Using and , I can write
and so the integrand must have the partial fractions decomposition
Notice that ordinarily, when the denominator contains an irreducible quadratic, the numerator of the partial fractions decomposition has the form and not . However, there are no and terms in the denominator, I can treat as the variable for the purposes of the decomposition. Since the right-hand side has linear terms in , it suffices to use a constant for finding the decomposition.
To solve for the constants and , I clear out the denominator:
Matching coefficients, this yields the system of equations
Substituting into the second equation, I get
Therefore, the integrand has the partial fractions decomposition
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