Improper integral sin(x) on 0 to infinity. Area under sin(x) to infinity: area integral diverges.

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We compute the improper integral of sin(x) on zero to infinity.

The integral of sin(x) on 0 to infinity must diverge, because the total area never settles down to a single number. The area over an integer number of periods is zero, but integrating over the next half period gives a value of 2. Thus, the integral keeps oscillating between 0 and 2, so the limit of the area under sin(x) to infinity does not exist.

More formally, we approach the infinite sin(x) integral by replacing the infinite limit with a parameter "t", then taking the limit as t goes to infinity. We guess the antiderivative of sin(x), evaluate across the limits of integration, then take the limit as t goes to infinity. We then discover that the area integral diverges, because the limit of cos(t) as t goes to infinity does not exist (it keeps oscillating between -1 and 1, never settling down to a single number).

[neopalm]

A lot of sources that explain fourier transforms deal with the "improper sin integral" over all reals (being 0) and it always annoyed me. It's especially annoying that the results they get often turn out to not even be functions, but distributions. Distributions on which the inner product fails, removing any semblance of a Hilbert space there once was. I suppose the problem lies in fourier transforms being explained in terms of sine waves when sine wave functions have no place being in the Hilbert space (with their infinite norm).

[edgardofigueroa]

Thanks, I wasn't sure if this diverged, you saved me a few points on homework.