3 ms·
Break the integral from 0 to infinity as a sum over the intervals demarcated by zeros of (sin(x))/x. Since the integrand is always positive or always negative o
by spekcular 6y ago
Break the integral from 0 to infinity as a sum over the intervals demarcated by zeros of (sin(x))/x. Since the integrand is always positive or always negative on such an integral, and the area under the curve is decreasing in absolute value, you can use the alternating series test. (This is a bit clearer if you graph it with e.g. Wolfram Alpha.)
Note that it does not converge absolutely, however.
- kgwgk 6y ago“The story is told of G. H. Hardy (and of other people) that during a lecture he said ‘It is obvious… Is it obvious?’ left the room, and returned fifteen minutes later, saying ‘Yes, it's obvious.’ I was present once when Rogosinski asked Hardy whether the story were true. Hardy would admit only that he might have said ‘It is obvious… Is it obvious?’ (brief pause) ‘Yes, it's obvious.’”