How can I show that this sequence of integrals tends to the value $\infty$?

Question

For each $n \in \Bbb Z_+$, let $$I_n = \int_{-\infty}^\infty \frac{|\sin \left( \frac{x}{n} \right) \sin(x)|}{\frac{x}{n} x} \, dx.$$ I would like to show that $I_n \xrightarrow{n\to \infty} \infty.$

I have written the denominator in this odd form because I think it might be helpful.

I have tried using the inequality $|\sin(x)| \geq \sin^2(x)$, but the resulting integral converges independently of $n$. Any ideas?

Answer 1

For $0\lt x\le1$, $\frac{\sin(x)}x$ is decreasing,therefore, $\frac{\sin(x)}x\ge\frac{\sin(1)}1$. Thus, for $0\lt x\le1$, $$ \sin(x)\ge x\sin(1)\tag{1} $$ Since $\left|\,\sin\left(\frac xn\right)\sin(x)\,\right|\le1$, we have $$ \begin{align} n\int_n^\infty\left|\,\sin\left(\frac xn\right)\sin(x)\,\right|\,\frac{\mathrm{d}x}{x^2} &\le n\cdot\frac1n\\ &=1\\[6pt] &=O(1)\tag{2} \end{align} $$ Therefore, $$ \begin{align} n\int_{-\infty}^\infty\left|\,\sin\left(\frac xn\right)\sin(x)\,\right|\,\frac{\mathrm{d}x}{x^2} &=2n\int_0^\infty\left|\,\sin\left(\frac xn\right)\sin(x)\,\right|\,\frac{\mathrm{d}x}{x^2}\\ &=2n\int_0^n\left|\,\sin\left(\frac xn\right)\sin(x)\,\right|\,\frac{\mathrm{d}x}{x^2}+O(1)\\ &\ge2\sin(1)\int_0^n\left|\,\sin(x)\,\right|\,\frac{\mathrm{d}x}{x}+O(1)\tag{3} \end{align} $$ and by Monotone Convergence, $$ \begin{align} \lim_{n\to\infty}\int_0^n\left|\,\sin(x)\,\right|\,\frac{\mathrm{d}x}{x} &=\int_0^\infty\left|\,\sin(x)\,\right|\,\frac{\mathrm{d}x}{x}\\ &=\sum_{k=1}^\infty\int_{(k-1)\pi}^{k\pi}\left|\,\sin(x)\,\right|\,\frac{\mathrm{d}x}{x}\\ &\ge\sum_{k=1}^\infty\frac2{k\pi}\\[9pt] &=\infty\tag{4} \end{align} $$

Answer 2

Note that

$$\int_0^{n\pi/3} \frac{|\sin(x/n)|}{x/n} \frac{|\sin x|}{x} \, dx \geqslant \sum_{k=1}^{n-1}\int_{k\pi/3}^{(k+1)\pi/3}\frac{|\sin(x/n)|}{x/n} \frac{|\sin x|}{x} \, dx \\ \geqslant \frac{3\sin(\pi/3)}{\pi}\sum_{k=1}^{n-1}\frac{1/2}{(k+1)\pi/3} \to \infty$$

With $0 \leqslant x \leqslant n\pi/3$ we have

$$1 \geqslant \frac{\sin(x/n)}{x/n} \geqslant \frac{\sin(\pi/3)}{\pi/3}. $$

Also

$$\int_{k\pi/3}^{(k+1)\pi/3} \frac{|\sin x|}{x} \, dx \geqslant \frac1{(k+1)\pi/3}\int_{k\pi/3}^{(k+1)\pi/3} \frac{|\sin x|}{x} \, dx = \frac{|\cos(k+1)\pi/3 - \cos k\pi/3|}{(k+1)\pi/3} \geqslant \frac{1/2}{(k+1)\pi/3}$$