I understand how to show that $$\frac{\sin x}{x}$$ is improperly integrable, but I'm having troubles with showing that it is not absolutely integrable.

Here is my work:

$$\frac{|\sin x|}{x}\leq \frac{1}{x}$$

and I know $\int_{1}^{\infty}\frac{1}{x}dx=\lim_{d\to \infty} \log(d)\to \infty$

but that doesn't prove that $\frac{|\sin x|}{x} \to \infty$ since it is less than $\frac{1}{x}$.

I was thinking about doing squeeze theorem


Hint: $$\int_{(k-1)\pi}^{k\pi}\frac{|\sin x|}{x}\,dx \ge\int_{(k-1)\pi}^{k\pi}\frac{|\sin x|}{k\pi}\,dx =\frac1{k\pi}\left|\int_{(k-1)\pi}^{k\pi}\sin x\,dx\right| =\frac2{k\pi}\ .$$


If we define $$ F(t) = \int_{0}^{t}\left|\sin x\right|\,dx $$ it is not difficult to show that $F(t)=\frac{2t}{\pi}+O(1)$, for instance by exploiting the Fourier cosine series of $\left|\sin x\right|$. By integration by parts it follows that

$$ \int_{1}^{N}\frac{\left|\sin x\right|}{x}\,dx = \frac{F(N)}{N}-\frac{F(1)}{1}+\int_{1}^{N}\frac{F(t)}{t^2}\,dt = O(1)+\frac{2}{\pi}\int_{1}^{N}\frac{dt}{t}$$ hence the LHS diverges like $\frac{2}{\pi}\log N$.


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