## Tuesday, March 20, 2007

### This Integral Not-Diverges. Topic: Calculus/S&S.

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.

1. Weak bound:

from $k \pi + \pi/4$ 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.

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

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

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

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

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

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

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

diverges?

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

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