Show that $$ \int_{1}^{\infty} \left| \frac{\sin x}{x} \right| \,\mathrm{d}x $$ diverges.

I know that there are many questions in which this problem is solved, but I want to know if my proof is correct.

My attempt.


$$ \int_{1}^{\pi} \left| \frac{\sin x}{x} \right| \,\mathrm{d}x $$ converges, it suffices to show that

$$ \int_{\pi}^{\infty} \left| \frac{\sin x}{x} \right| \,\mathrm{d}x $$

diverges. We can use

$$\int_{\pi}^{\infty} \left|\frac{\sin x}{x} \right| \,\mathrm{d}x = \left[\sum_{n=1}^{\infty} \int_{2n\pi}^{(2n+1)\pi} \frac{\sin t}{t} \, \mathrm{d}t \right] - \left[\sum_{n=0}^{\infty} \int_{(2n+1)\pi}^{2(n+1)\pi} \frac{\sin \omega}{\omega} \,\mathrm{d}\omega \right] $$

With the substitution $x = t - 2n\pi \implies \mathrm{d}t = \mathrm{d}x$, $$\int_{2n\pi}^{(2n+1)\pi} \frac{\sin t}{t} \,\mathrm{d}t = \int_{0}^{\pi} \frac{\sin x}{x + 2n\pi} \,\mathrm{d}x $$

and, with the substitution $x = \omega - (2n+1)\pi \implies \mathrm{d}\omega = \mathrm{d}x$,

$$ \int_{(2n+1)\pi}^{2(n+1)\pi} \frac{\sin \omega}{\omega}\,\mathrm{d}\omega = -\int_{0}^{\pi}\frac{\sin x}{x + (2n+1)\pi}\,\mathrm{d} x$$,


$$\int_{\pi}^{\infty} \left|\frac{\sin x}{x} \right| \,\mathrm{d}x = \left[\int_{0}^{\pi}\sin x\sum_{n=1}^{\infty}\frac{1}{x + 2n\pi} \,\,\mathrm{d}x\right] + \left[\int_{0}^{\pi} \sin x \sum_{n=0}^{\infty}\frac{1}{x + (2n+1)\pi} \,\, \mathrm{d}x\right]$$

Since $x \in [0,\pi]$, both $\displaystyle{\sum_{n=1}^{\infty}\frac{1}{x + 2n\pi}}$ and $\displaystyle{\sum_{n=0}^{\infty}\frac{1}{x + (2n+1)\pi}}$ diverge, then the integral $$\int_{\pi}^{\infty} \left|\frac{\sin x}{x} \right| \,\mathrm{d}x $$ diverges. Therefore, the integral $$ \int_{1}^{\infty} \left| \frac{\sin x}{x} \right| \,\mathrm{d}x $$ will do so. $\square$

Edit: 08-12-14 at 00:38

Does this make the proof better in any way?

$$\int_{\pi}^{\infty} \left|\frac{\sin x}{x} \right| \,\mathrm{d}x = \left[ \sum_{n=1}^{\infty}\int_{0}^{\pi}\frac{\sin x}{x+2n\pi}\,\mathrm{d}x \right] + \left[ \sum_{n=0}^{\infty}\int_{0}^{\pi}\frac{\sin x}{x+(2n+1)\pi}\,\mathrm{d}x \right]$$

Since $\forall\,x \in [0, \pi] \,\, \sin x \ge 0$, $n \ge 1$ and $ \dfrac{1}{\pi\left(2n+1\right)}\le \dfrac{1}{x + 2n\pi} \le \dfrac{1}{2n\pi} \,\, \forall\,x\in [0,\pi]$ we have

$$ \underbrace{\dfrac{\sin x}{\pi\left(2n+1\right)}\le \dfrac{\sin x}{x + 2n\pi}}_{\text{(I)}} \le \dfrac{\sin x}{2n\pi} \,\, \forall\,x\in [0,\pi] $$

Integrating (I) from $0$ to $\pi$, we get $$ \int_{0}^{\pi} \frac{\sin x}{x + 2n\pi}\,\mathrm{d}x \ge \frac{2}{\pi} \frac{1}{2n+1} $$

Adding all the terms in $n$ and noting that the series $\displaystyle{\sum_{n=1}^{\infty} \frac{1}{2n+1}}$ diverges, we conclude that the term between the first brackets blows up as $n \to \infty$. A similar procedure may be used to see what happens to the term between the other brackets, which will diverge also. Then, I think we can conclude more convincingly that the integral diverges. Am I right?

  • 3
    $\begingroup$ You have interchanged order of summation and integration. This in general requires justification. $\endgroup$ Aug 12, 2014 at 2:21
  • $\begingroup$ Looks good, although it can be refined here and there. For examples, why the series diverge implies that the integrals diverge. $\endgroup$ Aug 12, 2014 at 2:21
  • $\begingroup$ You can have a slightly different proof using the mean value theorem for integrals $\endgroup$ Aug 12, 2014 at 6:52

2 Answers 2


Set up intervals $I_1 = [\pi, 2\pi]$, $I_2 = [2\pi, 3\pi]$, $\dotsc$, $I_n = [n\pi, (n+1)\pi]$, $\dotsc$.

Then an underestimate of $\int_\pi^\infty \left\lvert \frac{\sin(x)}{x} \right\rvert dx$ is obtained by drawing isosceles triangles with bases $I_n$ and heights $$\left\lvert \frac{\sin\left(n\pi + \pi/2\right)}{n\pi + \pi/2} \right\rvert = \frac1{n\pi + \pi/2} = \frac{2}{(2n+1)\pi}.$$


$$\int_\pi^\infty \left\lvert \frac{\sin(x)}{x} \right\rvert dx > \sum_{n=1}^\infty \frac12 \cdot \pi \cdot \frac{2}{(2n+1)\pi} = \sum_{n=1}^\infty \frac1{2n+1} = \frac13 + \frac15 + \frac17 + \dotsb,$$

which diverges.


Your proof is correct but you can shortened it as follow

$$\int_1^\infty \bigg|\frac{\sin x}{x}\ \bigg| dx\ge \int_π^\infty \Bigg|\frac{\sin x}{x}\ \bigg|\\= \lim_{n\to \infty}\int_π^{nπ} \bigg|\frac{\sin x}{x}\ \bigg|dx= \sum_{n=1}^{\infty}\int_{nπ}^{(n+1)π} \bigg|\frac{\sin x}{x}\ \bigg|dx\\ \ge \sum_{n=1}^{\infty}\frac{1}{ (n+1)π} \int_{nπ}^{(n+1)π} |\sin x|dx =\color{red}{\sum_{n=1}^{\infty}\frac{2}{ (n+1)π}=\infty} $$

indeed for all $n$ $$\color{blue}{\int_{nπ}^{(n+1)π} |\sin x|dx=2}$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .