This series was inspired by a question that my wife asked me: calculate
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:
,
where is the contour in the complex plane shown below (graphic courtesy of Mathworld).
Amazingly, contour integrals can be simply computed by evaluating the residues at every pole located inside of the contour. (See Wikipedia and Mathworld for more details.) I have already handled the case of
and
. Today, I begin the final case of
.
Earlier in this series, I showed that
if , and so the quadratic formula can be used to find the four poles of the integrand:
,
,
,
.
Of these, only two lie ( and
) within the contour for sufficiently large
(actually, for
since all four poles lie on the unit circle in the complex plane).
As shown earlier in this series, the residue at each pole is given by
I’ll now simplify this considerably by using the fact that at each pole:
.
Therefore, to evaluate the contour integral, I simply the sum of the residues within the contour and multiply the sum by :
.
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