Does improper integral sin(x)/x converge?

**Pn0yS0ld13r** · December 28th 2009, 06:46 PM

Does $\int^\infty_0 \frac{\sin x}{x} \, dx$ converge?

I'm having trouble understanding my professor's solution:

$\lim_{t \to 0^{+}} \int^\pi_t \frac{\sin x}{x} \,dx$ exists because $\frac{\sin x}{x}$ extends to a continuous function at x = 0.

Now this is where I'm having trouble. How are these equations derived?

Let $\lfloor x \rfloor$ be the floor function or greatest integer function. Then

$\lim_{t \to \infty} \int^{t}_\pi \frac{\sin x}{x} \, dx = \lim_{t \to \infty} \left( \sum_{n=2}^{\lfloor t/ \pi \rfloor} \int^{n \pi}_{(n-1)\pi} \frac{\sin x}{x} \, dx + \int^t_{\lfloor t/ \pi \rfloor \pi} \frac{\sin x}{x} \, dx \right)$

$= \lim_{t \to \infty} \left( \sum_{n=2}^{\lfloor t/ \pi \rfloor} \int_0^\pi (-1)^{n-1} \frac{\sin x}{x + (n-1)\pi} \,dx + \int_{\lfloor t/ \pi \rfloor \pi}^t \frac{\sin x}{x} \,dx \right)$ .

Since $\frac{\sin x}{x + (n-1)\pi} \le \frac{\sin x}{x + n \pi}$ for $0 \le x \le \pi$ , the first term has a finite limit as $t \to \infty$ , by the Alternating Series Test. The second term is bounded in absolute value by

$\int_0^\pi \frac{\sin x}{x+ \lfloor t/\pi \rfloor \pi} \,dx \le \frac{1}{\lfloor t/\pi \rfloor}$ ,

so it converges to 0 by the squeeze theorem. So the limit exists, and therefore the integral converges.

**Krizalid** · December 28th 2009, 08:16 PM

Simplest solution:

Spoiler:

Thread: Does improper integral sin(x)/x converge?

LinkBack

Thread Tools

Display

Does improper integral sin(x)/x converge?

Similar Math Help Forum Discussions

Improper Integral/ Converge-Diverge.

Improper integral converge/diverge

Can someone please check my work here? (Improper integrals, converge/diverge)

Converge and diverge of an improper integral

Integral (Converge?)

Search Tags