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:

if . (The cases and have already been handled earlier in this series.)

I will evaluate this integral using partial fractions. The denominator factors as the product of two irreducible quadratics, and so I must solve

,

or

.

I now clear out the denominators:

Now I multiply out the right-hand side:

Equating this with and matching coefficients yields the following system of four equations in four unknowns:

Ordinarily, four-by-four systems of equations are rather painful to solve, but this system isn’t so bad.

From the first equation, I see that .

From the third equation, I see that

.

From the fourth equation, I see that

,

so that as well. Finally, from the second equation, I see that

,

so that as well. This yields the partial fractions decomposition

.

This can be confirmed by directly adding the fractions on the right-hand side:

.

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

I'm a Professor of Mathematics and a University Distinguished Teaching Professor at the University of North Texas. For eight years, I was co-director of Teach North Texas, UNT's program for preparing secondary teachers of mathematics and science.
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