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 .

Yesterday, I showed used the substitution to show that was independent of . Today, I’ll use a different method to establish the same result. Let

.

Notice that I’ve written this integral as a function of the parameter . I will demonstrate that , so that is a constant with respect to . In other words, does not depend on .

To do this, I differentiate under the integral sign with respect to (as opposed to ) using the Quotient Rule:

I now apply the trigonometric substitution , so that

and

The endpoints of integration change from to , and so

.

I’m not completely thrilled with this demonstration that is independent of , mostly because I had to do so much simplification of the integral to get this result. As I mentioned in yesterday’s post, I’d love to figure out a way to directly start with

and demonstrate that is independent of , perhaps by differentiating with respect to and demonstrating that the resulting integral must be equal to 0. However, despite several hours of trying, I’ve not been able to establish this result without simplifying first.

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