# Improper integral $\int_0^\infty \frac{\sin(x)}{x}dx$ - Showing convergence.

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.

Anyways,

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.

• There's a typo in (2)... Commented Jul 4, 2015 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.

• The first part looks like mine. Commented Jul 4, 2015 at 23:04
• It is pure coincidence. I did not know your proof before I submit my answer. Commented Jul 4, 2015 at 23:31
• No worry. Good solution! Commented Jul 4, 2015 at 23:42
• I took the liberty of correcting the integration by parts. I hope that you don't mind. Commented Jul 5, 2015 at 0:46
• @mathcraze Thanks, I understood it thanks to you and Dr. MV. And yes, you are right, I misspelled it in b). :D. Commented Jul 5, 2015 at 12:55

HINTS:

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

$$\int_1^{\infty}\frac{1}{x^2}dx=1$$

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

For the third part, use integration by parts.