3 ms·
One of the points link to a second article that reads: (The fact that φ is zero outside a finite interval mean the “uv” term from integration by parts is zero.
by esturk 8y ago
One of the points link to a second article that reads:
(The fact that φ is zero outside a finite interval mean the “uv” term from integration by parts is zero.)
Can anyone elaborate on this? I'm not too sure how this trick works.
- pedrosorio 8y agoThis is the article you're referring to: https://www.johndcook.com/blog/2009/10/25/how-to-differentiate-a-non-differentiable-function/ https://www.johndcook.com/blog/2009/10/25/how-to-differentia... This is the wikipedia page on integration by parts: https://en.wikipedia.org/wiki/Integration_by_parts https://en.wikipedia.org/wiki/Integration_by_parts He's setting u = phi(x), v = f(x). Since he's integrating on the real line, a = -infinity, b = +infinity, so on the right hand side for "uv" you'd have u(+inf)v(+inf) - u(-inf)v(-inf) - and since u (which is phi) is 0 outside of a finite interval, you know these terms are 0. [You obviously can't evaluate functions at +-inf, and you have to take limits to evaluate u(x)v(x) for an improper integral, but you can see the result is the same if u(x) is zero outside of a finite interval]
- esturk 8y agoOh I see. So in the evaluation of u(+inf), it becomes 0, hence uv is 0. I have never thought of it that way. Thanks!