Calculate $\lim_{k\to\infty}\int_{(1,\infty)}\frac{k\sin(x/k)}{x^3}d\mathcal{L}^1(x)$ I'm trying to solve the following exercise: Show that $\lim_{k\to\infty}\int_{(1,\infty)}\frac{k\sin(x/k)}{x^3}d\mathcal{L}^1(x)$ exists and compute its value.

I have switched to the improper Riemann integral for $k$ fixed using dominated convergence since $|\frac{k\sin(x/k)}{x^3}| \le k\cdot 1/x^3\in\mathcal{L}^1((1,\infty])$ but now I don't know how to continue since (again for $k$ fixed) I have been unable to compute the value of the improper integral above.
Is there a different way to approaching this problem?
 A: Note that the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(t)= \frac{\sin(t)}{t}$ is a uniformly bounded (by $1$), thus
$$
 \left| \frac{k \sin(x/k)}{x^3} \right| = \left| \frac{\sin(x/k)}{x/k} \right| \frac{1}{x^2} \leq \frac{1}{x^2}, \qquad  k\geq 1. 
$$
Moreover, the standard limit $\lim_{t\to 0} \frac{\sin(t)}{t}=1$ implies that
$$ \lim_{k \to \infty} \frac{k \sin(x/k)}{x^3} = \frac{1}{x^2}. 
$$
Since $x \mapsto 1/x^2$ is integrable on $[1,\infty)$, we may invoke the dominated convergence Theorem, which gives
$$
\lim_{k \to \infty} \int_{1}^{\infty} \frac{k \sin(x/k)}{x^3} dx = \int_{1}^{\infty} \frac{1}{x^2} dx = 1.
$$
A: Dominated Convergence works, but here is an alternate approach.
$$
\begin{align}
\int_1^\infty\frac{k\sin(x/k)}{x^3}\,\mathrm{d}x
&=\frac1k\int_{1/k}^\infty\frac{\sin(x)}{x^3}\,\mathrm{d}x\tag{1a}\\
&=\frac1k\int_{1/k}^\infty\frac{x}{x^3}\,\mathrm{d}x\\
&+\color{#C00}{\frac1k\int_{1/k}^1\frac{\sin(x)-x}{x^3}\,\mathrm{d}x+\frac1k\int_1^\infty\frac{\sin(x)-x}{x^3}\,\mathrm{d}x}\tag{1b}\\
&=\frac1k\cdot k+\color{#C00}{O\!\left(\frac1k\right)}\tag{1c}
\end{align}
$$
Explanation:
$\text{(1a)}$: substitute $x\mapsto kx$
$\text{(1b)}$: break integrand into pieces
$\text{(1c)}$: $\left|\frac{\sin(x)-x}{x^3}\right|\le\min\left(\frac16,\frac1{x^2}+\frac1{x^3}\right)$
Thus,
$$
\lim_{k\to\infty}\int_1^\infty\frac{k\sin(x/k)}{x^3}\,\mathrm{d}x=1\tag2
$$
