Showing that the L1 norm of a given sequence of functions diverges For $n=1,2,3,\ldots$ define $f_n:\mathbb R\to\mathbb R$ by  $$f_n(x) = \frac{\sin(x)\sin(nx)}{x^2}.$$ Then certainly each $f_n$ is integrable on the real line.

However, I have to show that the $L^1$ norm of $f_n$ goes to infinity as $n\to\infty$. I have no idea how to deal with this function.

Could anyone help me?
Make the change of variables $x=y/n$ to see $$\int_0^1\frac{|\sin x\sin nx|}{x^2}\,dx = n\int_0^n\frac{|\sin (y/n)\sin y|}{y^2}\,dy.$$

Now $\sin u \ge (2/\pi)u$ for $0\le u\le \pi/2.$ So the above is at least

$$n\int_0^n\frac{(2/\pi)(y/n)|\sin y|}{y^2}\,dy = (2/\pi)\int_0^n\frac{|\sin y|}{y}\,dy \to \infty.$$
I=2J=twice the integral from 0 to infinity. Now 
(sin x.sin nx)/x^2= (1/2x^2)(cos(n-1)x-1-(cos(n+1)x-1)). Suffices to show the integral diverges over (0,1). We can be satisfied if we replace cox-1 by x^2 and integrate and get divergent value owing to alternation  of sign. This we do and change the variable (n-1)x=u,(n+1)x=v. I did all of this and got divergence.