$\lim_{n\to\infty}\int_0^\infty \frac{f(x)}{x^2}\sin nx dx=\frac{\pi}{2}f'(0)$ Let $f$ be twice continuously differentiable on $[0,\infty)$, and bounded, with $f(0)=0$. Show that
$\lim_{n\to\infty}\int_0^\infty \frac{f(x)}{x^2}\sin nxdx=\frac{\pi}{2}f'(0).$
My attempts:

*

*Using $\int_0^\infty \frac{\sin t}{t}dt=\frac{\pi}{2}$ to find
$$\left|\int_0^\infty \frac{f(x)}{x^2}\sin nxdx-\frac{\pi}{2}f'(0)\right|
=\left|\int_0^\infty \left[\frac{f(t/n)-f(0)}{t/n}-f'(0)\right]\frac{\sin t}{t}dt\right|$$
For the part $t\leq A$ for fixed $A>0$ it is OK, but for $t>A$, any ideas?


*Another try is
$$\left|\int_0^\infty \frac{f(x)}{x^2}\sin nxdx-\frac{\pi}{2}f'(0)\right|
=\left|\int_0^\infty \frac{f(x)-xf'(0)}{x^2}\sin nxdx\right|$$
The function $\frac{f(x)-xf'(0)}{x^2}$ has limit $f''(0)/2$ as $x\to0^+$.
Oh, it seems that the assumptions are not sufficient, any help? What condition should we add to prove this above limit? And the proof?
 A: We have
$$\int\limits_0^\infty {\sin (nx)\over x}\,dx={\pi\over 2}$$ Thus
$$\displaylines{\int\limits_0^\infty \frac{f(x)}{x^2}\sin (nx)\,dx-\frac{\pi}{2}f'(0) \\ =
\int\limits_0^\infty \left [{f(x)-xf'(0)\over x^2}\right]{\sin (nx)}\,dx\\ =\int\limits_0^1 \left [{f(x)-xf'(0)\over x^2}\right]{\sin (nx)}\,dx\\ +\int\limits_1^\infty{f(x)\over x^2}{\sin (nx)}\,dx+f'(0)\int\limits_1^\infty{\sin (nx)\over x}\,dx}
$$
The second integral tends to $0$
by the Riemann-Lebesgue lemma, as the function $f(x)/x^2$ is absolutely integrable. The third integral tends to $0$ as well since
$$\int\limits_1^\infty{\sin (nx)\over x}\,dx=
\int\limits_{n}^\infty{\sin x\over x}\,dx $$
Concerning the first integral the function
$$g(x):={f(x)-xf'(0)\over x^2}$$ is continuous on $(0,1]$ and
by applying  l'Hospital's rule we get $$\lim_{x\to 0^+}g(x)=\lim_{x\to 0^+}{f(x)-xf'(0)\over x^2}={1\over 2}f''(0)$$ Hence $g$ extends to a continuous function on $[0,1].$ Again by the Riemann-Lebesgue lemma we get
that the limit of the first integral is equal $0.$
