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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