1)Show that for all $n\in\mathbb{N}$ the following is true:

$$\int_{\pi}^{n\pi}|\frac{\sin(x)}{x}|dx\geq C\cdot \sum_{k=1}^{n-1}\frac{1}{k+1}$$

for a constant $C>0$ and conclude that the improper integral $\int_0^\infty \frac{\sin(x)}{x}dx$ isn't absolutely convergent.

2)Show that the improper integral $\int_0^\infty \frac{1-\cos(x)}{x^2}dx$ is absolutely convergent. (The integrand is to be expanded continuous at $x=0$.).

3)Using 2), show that the improper integral $\int_0^\infty \frac{\sin(x)}{x}dx$ is convergent.

We started discussing improper integrals in class and our prof showed us how some can be solved and some can't.


Here were my ideas so far:

1) I thought about to do the integral and seeing if what I get out of it gives me any idea to show the inequality. But I couldn't even solve the integral (not by hand nor with the help of an integral calculator). So I don't know what to do next.

2)To be hoenst I'm totally lost here. No idea how to approach it.

3)Well, since I didn't solve 2).

Sorry for my lack of work here, but this topic just doesn't want to stick with me.

  • $\begingroup$ There's a typo in (2)... $\endgroup$ – David C. Ullrich Jul 4 '15 at 21:43

1). Since $$ \int_{n\pi}^{(n+1)\pi}\left|\sin{x}\right|dx=2 $$ there is \begin{align} \int_{\pi}^{(n+1)\pi}\left|\dfrac{\sin{x}}{x}\right|dx&=\sum_{k=1}^n\int_{k\pi}^{(k+1)\pi}\left|\dfrac{\sin{x}}{x}\right|dx \\ &\geqslant\sum_{k=1}^n\dfrac1{(k+1)\pi}\int_{k\pi}^{(k+1)\pi}\left|\sin{x}\right|dx \\ &=\dfrac{2}{\pi}\sum_{k=1}^n\dfrac1{k+1} \end{align} So $\int_{0}^{\infty}\dfrac{\sin{x}}{x}dx$ diverges absolutely.

2). Since \begin{align} \int_{\pi}^{(n+1)\pi}\dfrac{1-\cos{x}}{x^2}dx&=\sum_{k=1}^n\int_{k\pi}^{(k+1)\pi}\dfrac{1-\cos{x}}{x^2}dx \\ &\leqslant\sum_{k=1}^n\dfrac1{k^2\pi^2}\int_{k\pi}^{(k+1)\pi}(1-\cos{x})dx \\ &=\dfrac1{\pi}\sum_{k=1}^n\dfrac1{k^2} \end{align} So $\int_{0}^{\infty}\dfrac{1-\cos{x}}{x^2}dx$ is absolutely convergent.

3).By partial integration, there is \begin{align} \int_{0}^{\infty}\dfrac{1-\cos{x}}{x^2}dx&=-\dfrac{1-\cos{x}}{x}\Bigg|_0^{\infty}+\int_{0}^{\infty}\dfrac{\sin{x}}{x}dx \\ &=-\dfrac{2\sin^2{\dfrac{x}{2}}}{x}\Bigg|_0^{\infty}+\int_{0}^{\infty}\dfrac{\sin{x}}{x}dx \\ &=\int_{0}^{\infty}\dfrac{\sin{x}}{x}dx \end{align} So $\int_{0}^{\infty}\dfrac{\sin{x}}{x}dx$ is convergent.

  • $\begingroup$ The first part looks like mine. $\endgroup$ – Mark Viola Jul 4 '15 at 23:04
  • $\begingroup$ It is pure coincidence. I did not know your proof before I submit my answer. $\endgroup$ – Math Wizard Jul 4 '15 at 23:31
  • $\begingroup$ No worry. Good solution! $\endgroup$ – Mark Viola Jul 4 '15 at 23:42
  • $\begingroup$ I took the liberty of correcting the integration by parts. I hope that you don't mind. $\endgroup$ – Mark Viola Jul 5 '15 at 0:46
  • $\begingroup$ @mathcraze Thanks, I understood it thanks to you and Dr. MV. And yes, you are right, I misspelled it in b). :D. $\endgroup$ – MigJakker Jul 5 '15 at 12:55


Fot the first part, write

$$\begin{align} \int_{\pi}^{n\pi}\left|\frac{\sin x}{x}\right|dx&=\sum_{k=1}^{n-1}\int_{k\pi}^{(k+1)\pi}\left|\frac{\sin x}{x}\right|dx\\\\ &\ge \sum_{k=1}^{n-1}\frac{1}{(k+1)\pi}\int_{k\pi}^{(k+1)\pi}|\sin x|\,dx\\\\ &=\frac{2}{ \pi}\sum_{k=1}^{n-1}\frac{1}{k+1} \end{align}$$

For the second part, note that $|1-\cos x|=|2\sin^2(x/2)|\le 2$. And

$$\left|\int_0^{\infty}\frac{1-\cos x}{x^2}dx\right|=2\int_0^{\infty}\frac{\sin^2(x/2)}{x^2}dx$$

There is a removable discontinuity at $x=0$, so the we need to analyze the convergence at the upper limit. Since


and the integrand is non-negative, the convergence is absolute.

For the third part, use integration by parts.


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