Monday Math 23

For today, we have the solution to another definite integral:



Back in Monday Math 6, I demonstrated that for any real n, . A simple change of variables from this allows us to see that as well (the change is ; then rename the variable to x).

Now, looking at our present integral, one might note that
. Thus .
The key is to now replace x in this expression with n, and a with , and we have something we can obtain from the integrand of . Specifically:
.

The internal integral is the one we already established is , for any real value of y. Hence
.

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