[SOLVED] Bounding the sine integral
Let Si(x) = integral( sin(t)/t dt, 0, x)
I'm studying for a preliminary exam and I need to show the following:
1) limit as x--> infinity of Si(x) is finite
2) integral( sin(x)/x dx, 1, infinity) < 2/pi
integral( sin(x)/x dx, 0, infinity) = integral( (sin(x)/x)^2, 0, infinity)
I've done a Taylor series expansion of Si(x), and then for the first question above, I've broken the integral into two parts, from 0 to 1, and from 1 to infinity. I can use the Taylor series expansion to evaluate the first integral, and then for the second integral, I can integrate by parts to get:
integral( sin(t)/t dt) = -cos(t)/t - integral( cos(t)/t^2 dt)
evaluating this at the limits of 1 and infinity is fine for the first term, and for the second term, I know instinctively that it is bounded but I can't figure out a way to determine the bound so that I can show part 2) of above. As for part 3, I have no idea how to tackle it. Everywhere I look online seems to use contour integration for this, but I feel there must be another easier way.
Any help would be greatly appreciated!!!