Previously in this series, I showed that

So far, I have shown that

where and (and is a certain angle that is now irrelevant at this point in the calculation).

We now employ the substitution

,

so that

.

Before going much further, let’s take a closer look at and to make sure that is positive (so that the square root is defined).

First, both and $R$ are clearly positive, and so .

Next, notice that

So as long as . Therefore, since :

So, since and , we have , and so the above substitution is well-defined.

We now employ the above substitution. The endpoints of integration remain unchanged, and so

In the above calculation, I used the fact that , which was derived above. Also, I was careful to avoid a common algebraic mistake.

I’ll complete this different method of evaluating this integral in tomorrow’s post.

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