Very neat trick! My first reaction was that this must be related to to the Laplace transform, and looking through the Wikipedia article seems that this is basic

Very neat trick! My first reaction was that this must be related to to the Laplace transform, and looking through the Wikipedia article seems that this is basically the same trick: https://en.wikipedia.org/wiki/Laplace_transform#Evaluating_integrals_over_the_positive_real_axis https://en.wikipedia.org/wiki/Laplace_transform#Evaluating_i...: \int f(x) g(x) \dx = \int L[f] L^-1[g] \dt In their sine wave example we have f(x)=sin(x) and g(x)=1/x, and luckily for us the inverse Laplace of 1/x is just 1 (at least from 0 to \infty). I've learnt that finding the inverse Laplace is practically impossible (no good algorithm), but the regular Laplace can be often be found by "just" integrating. So I'm guessing this technique is mostly useful when we have a term with 1/x^n since the inverse is trivial. I had no idea Laplace transforms could do this so this was a nice discovery!