Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We know that $$\int_0^\infty \left(\frac{\sin x}{x}\right)^2 dx=\int_0^\infty \frac{\sin x}{x} dx=\frac{\pi}{2}.$$

How do I show that $$\int_0^\infty \left\vert\frac{\sin x}{x}\right\vert dx$$ converges?

share|cite|improve this question
up vote 15 down vote accepted

It doesn't. Using the convexity of $1/x$,

$$\int_0^\infty \left\vert\frac{\sin x}{x}\right\vert \,\mathrm{d}x=\sum_{k=0}^\infty\int_{k\pi}^{(k+1)\pi}\left\vert\frac{\sin x}{x}\right\vert \,\mathrm{d}x>\sum_{k=0}^\infty\int_{k\pi}^{(k+1)\pi}\frac{\left\vert\sin x\right\vert}{(k+1/2)\pi} \,\mathrm{d}x=\frac{2}{\pi}\sum_{k=0}^\infty\frac{1}{k+1/2}\;,$$

which diverges since the harmonic series diverges.

share|cite|improve this answer
thanks for the neat explanation. – user7815 Mar 7 '11 at 20:06
@user7815: You're welcome! – joriki Mar 7 '11 at 20:08
I don't see how you do it using the convexity of $1/x$. You can take in the denominator $k\pi$ since $1/x$ is decreasing. – Theta33 Mar 7 '11 at 20:09
@Mielu: You're right, that would have been easier :-) I'm using the convexity by replacing the argument of $1/x$ in the sum of the contributions at $(k+1/2)\pi-\Delta$ and $(k+1/2)\pi+\Delta$ (which have the same value of $\lvert\sin x\rvert$) by the average value $(k+1/2)\pi$. – joriki Mar 7 '11 at 20:13
+1 That was quite nice. – Felix Marin Jan 27 '14 at 22:25

It doesn't. This function is a sequence of bumps of decreasing size. The $n^{\text{th}}$ "bump" bounds area on the order of $\frac{1}{n}$ (there are a number of ways to show this), but $\sum_{n=1}^\infty \frac{1}{n} = \infty$.

share|cite|improve this answer

$\int_0^\infty \frac{\sin x}{x} dx$ is in fact my most favorite example of a function which is Riemann integrable but not Lesbegue integrable. (just to tease people which think that Lesbegue is so much better than Riemann; in fact this integral might appear more frequent in applications than any integral which is Lesbegue but not Riemann integrable)

share|cite|improve this answer
In defense of your teasing victims, if you allow improper Riemann integrals, then why not improper Lebesgue integrals? The limit of Lebesgue integrals $\lim_{b\to \infty}\int_0^b\frac{\sin x}{x}$ is just the same as that for the Riemann integrals, which after all is what this "improper" integral means. But I do agree that it is a good example, not just for teasing people, but also to clarify the differences in the definitions of improper Riemann integrals versus ordinary Lebesgue integrals on unbounded domains. – Jonas Meyer Mar 7 '11 at 20:01
I do hope that your teasing victims have no delusions that application to functions in the real world is the reason that Lebesgue integration is so much better than Riemann integration. Anyway, +1. – Jonas Meyer Mar 7 '11 at 20:05
@Meyer: I agree and I don't want to give the impression that I don't see the beauty of Lesbegues (e.g., the dominated convergence theorem), but Lesbegues doesn't solve all problems and the sinc integral stays an improper integral... – Fabian Mar 7 '11 at 21:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.