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.

Solutions Collecting From Web of "Improper integral $\int_0^\infty \frac{\sin(x)}{x}dx$ – Showing convergence."

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.

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.