**Problem**: Show that the integral $ \displaystyle \int_0^{\infty} \frac{\sin{x}}{x}dx $ converges.

**Solution**: Consider the intervals $ [2k \pi, (2k+2) \pi] $ for $ k = 0, 1, 2, \ldots $. We can rewrite the given integral as

$ \displaystyle \int_0^{\infty} \frac{\sin{x}}{x}dx = \sum_{k=1}^{\infty} \int_{2k \pi}^{(2k+2) \pi} \frac{\sin{x}}{x} dx+C $,

where $ C $ is some unimportant constant. So how can we go about bounding the integral

$ \displaystyle \int_{2k \pi}^{(2k+2) \pi} \frac{\sin{x}}{x} dx $?

Well, first note that $ \sin{x} = - \sin{(x+\pi)} $ so we can say

$ \displaystyle \int_{2k \pi}^{(2k+2) \pi} \frac{\sin{x}}{x} dx = \int_{2k \pi}^{(2k+1)\pi} \left( \frac{\sin{x}}{x}+\frac{\sin{(x+\pi)}}{x+\pi} \right) dx = \int_{2k \pi}^{(2k+1)\pi} \left( \frac{\sin{x}}{x}-\frac{\sin{x}}{x+\pi}\right) dx $.

Then, putting the last expression under a common denominator, we get

$ \displaystyle \int_{2k \pi}^{(2k+1)\pi} \left( \frac{\sin{x}}{x}-\frac{\sin{x}}{x+\pi}\right) dx = \int_{2k \pi}^{(2k+1)\pi} \frac{\pi \sin{x}}{x(x+\pi)} dx $,

which we can easily bound with $ \sin{x} \le 1 $ and $ x \ge 2k \pi $. This gives us

$ \displaystyle \int_{2k \pi}^{(2k+1)\pi} \frac{\pi \sin{x}}{x(x+\pi)} dx < [(2k+1)\pi-2k \pi] \cdot \frac{\pi}{(2k \pi)(2k \pi + \pi)} = \frac{1}{2k(2k+1)} $.

Hence we know that

$ \displaystyle \int_0^{\infty} \frac{\sin{x}}{x}dx = \sum_{k=1}^{\infty} \int_{2k \pi}^{(2k+2) \pi} \frac{\sin{x}}{x} dx+C < \sum_{k=1}^{\infty} \frac{1}{2k(2k+1)}+C $

and this converges by a $ p $-series test. QED.

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Comment: A pretty neat problem, though it is a standard convergence/divergence exercise. I'm sure there are many ways of doing this, but it's always nice to come up with a cool way of showing that a series converges or diverges. It's also interesting to note that the practice problem integral, which is only slightly different from this one, diverges.

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Practice Problem: Show that the integral $ \displaystyle \int_0^{\infty} \frac{|\sin{x}|}{x}dx $ diverges.

Weak bound:

ReplyDeletefrom $ k \pi + \pi/4 $ to $ k \pi + 3\pi/4 $ we can bound below by $ (\sqrt{2} / 2) / (k \pi + 3\pi / 4) $ which clearly diverges by p-series.

Oh, right, you enabled TeX. Okay, more interesting question then:

ReplyDeleteWhat about $ \displaystyle \int_{\pi}^{\infty} \frac{1 - |\sin x|}{x} \, dx $?

If we multiply the top by $ 1 \le 1+|\sin{x}| \le 2 $, it should not affect the convergence. The integral then becomes

ReplyDelete$ \displaystyle \int_{\pi}^{\infty} \frac{\cos^2{x}}{x} dx $,

which diverges by a similar argument that you had above (i.e. using intervals from $ k\pi-\frac{\pi}{4} $ to $ k\pi+\frac{\pi}{4} $).

Hmm. So: does there exist a constant $k$ such that

ReplyDelete$ \int_{\pi}^{\infty} \frac{k - | \sin x |}{x} \, dx $

diverges?

Wait, I think k = 1/2 trivially. Or something. Dang.

ReplyDeleteSo, what's the RANGE of k such that the integral converges? (Oops. Yeah, I meant converges.)

ReplyDelete