Amazingly, the integral below has a simple solution:
Even more amazingly, the integral ultimately does not depend on the parameter
. For several hours, I tried to figure out a way to demonstrate that
is independent of
, but I couldn’t figure out a way to do this without substantially simplifying the integral, but I’ve been unable to do so (at least so far).
So here’s what I have been able to develop to prove that is independent of
without directly computing the integral
.
Earlier in this series, I showed that
I now multiply the top and bottom of this last integral by :
I now employ the substitution , so that
. Since
, the endpoints of integration do not change, and so
.
This final integral is independent of .
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