How can one evaluate the improper Riemann integral of (sin x)/x on [0, infinity)?

Question

The value of the integral is pi/2.

This integral cannot be evaluated using the Fundamental Theorem of Calculus--at least, not in the way one might expect--because (sin x) / x does not have a nice-looking antiderivative.

I found one proof online using the Laplace transform, but I am wondering if there is another way (pehaps using a series, or the Fourier transform, or a contour in the complex plane, or inequalities, or some other technique.)

Answer 1

It can be done by using a contour integral over the top half-plane with the function e^(ix)/x. The interal of sin(x)/x is then half of the imaginary part. To do the contour integral, notice that the integral over a small half-circle close to the origin is i*pi and the integral over a large half circle goes to 0 as the radius goes to infinity.

Answer 2

The indefinite integral, or antiderivative, is in fact well behaved, beginning from 0 at x = 0 and thereafter oscillating to the asymptotic value of π/2. Sin(x)/x is a well known fourier transform inverse of the "hat", or rectangle, function.

Answer 3

Um, those are some big words and stuff.