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

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

,

while

.

Therefore,

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,

Therefore, the integrand has the partial fractions decomposition

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

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