How can I show that this sequence of integrals tends to the value $\infty$? 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?
 A: 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
$$\int_{k\pi/3}^{(k+1)\pi/3} \frac{|\sin x|}{x} \, dx \geqslant \frac1{(k+1)\pi/3}\int_{k\pi/3}^{(k+1)\pi/3} \frac{|\sin x|}{x} \, dx =  \frac{|\cos(k+1)\pi/3 - \cos k\pi/3|}{(k+1)\pi/3} \geqslant \frac{1/2}{(k+1)\pi/3}$$
A: 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,
$$
\begin{align}
\lim_{n\to\infty}\int_0^n\left|\,\sin(x)\,\right|\,\frac{\mathrm{d}x}{x}
&=\int_0^\infty\left|\,\sin(x)\,\right|\,\frac{\mathrm{d}x}{x}\\
&=\sum_{k=1}^\infty\int_{(k-1)\pi}^{k\pi}\left|\,\sin(x)\,\right|\,\frac{\mathrm{d}x}{x}\\
&\ge\sum_{k=1}^\infty\frac2{k\pi}\\[9pt]
&=\infty\tag{4}
\end{align}
$$
