Show that $\int_{0}^{\infty}|\frac{\sin x}{x}|\,dx=\infty $

Question

Problem: Show that $\int_{0}^{\infty}|\frac{ \sin x}{x}|=\infty $ .

Here is what i have tried :

I found out that for all $x>0$ , $x-\frac{x^3}{6} < \sin x <x $ . So $1-\frac{x^2}{6}<\frac{ \sin x}{x}<1$ . So applying integration to both side from $0$ to $M$ we have $M-\frac{M^3}{18}<\int_{0}^{M}|\frac{ \sin x}{x}|< M$ . Now we send $M \rightarrow \infty$ and observe that left-hand side goes to infinity . Hence proved . Please point out if there is anything wrong with my approach . I will also be delighted if could provide alternative solutions . Thank you .

Answer 1

Your solution is not good, because the left side, $M-\frac{M^3}{18}$, does not go to $\infty$, but rather to $-\infty$.

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An alternative solution would include

• the fact that on any interval $[(k-1)\pi, k\pi]$, you have $$\frac{|\sin x|}{x} \geq \frac{|\sin x|}{k \pi}$$
• The knowledge that $$\int_{(k-1)\pi}^{k\pi} |\sin x| dx$$ is independent of $k$, i.e. that it is always equal to the same nonzero constant $C$
• The knowledge that the series $$\sum_{k=1}^{\infty}\frac1k$$ diverges (even if multiplied by a nonzero constant).

Answer 2

$\displaystyle\int_0^{+\infty}=\int_0^1+\int_1^{+\infty}$.

$|\sin(x)|\ge \sin^2x=\frac{1}{2}(1-\cos(2x))$ hence

$$\dfrac{|\sin x|}{x}\ge \frac{1}{2x}-\frac{1}{2}\cdot\dfrac{\cos 2x}{x}$$

As $\displaystyle\int_1^{+\infty}\dfrac{\cos 2x}{x}dx$ converge and $\displaystyle \int_1^{+\infty}\dfrac{1}{x}dx=+\infty$