Previously in this series, I showed that

So far, I have shown that

,

where , , and is a certain angle (that will soon become irrelevant).

I now write as a new sum by dividing the region of integration:

,

.

For , I employ the substitution , so that and . Also, the interval of integration changes from to , so that

Next, I employ the trigonometric identity :

,

where I have changed the dummy variable from back to .

Therefore, becomes

.

Next, I employ the substitution , so that and the interval of integration changes from to :

.

Almost by magic, the mysterious angle has completely disappeared, making the integral that much easier to compute.

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

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