Computing the limit of a sinc-like function Compute $\displaystyle{\lim_{k\to∞} \int_0^\infty \frac{k\sin(x/k)}{x^{3/2}} dx}$ .
Let $f_k = \frac{k\sin(x/k)}{x^{3/2}}$. by L'Hopital, $\displaystyle{\lim_{k\to \infty} f_k(x)} = \frac{1}{x^{3/2}}$, which is not Lebesgue integrable on $(0, 1]$ since the improper Riemann integral diverges to infinity. So if I can 'push the limit in the integral', I should be able to conclude that the limit diverges to infinity. I suspect that $f_k$ do indeed converge to $\frac{1}{x^{3/2}}$ uniformly on $(0, 1]$, since the but I am having trouble proving it. Am I on the right track to finding this limit?

Note that $0\le \sin (x)\le x$ for $x\ge 0$.  So, $\lim_{k\to \infty}\frac{k\sin(x/k)}{x^{3/2}}=\frac1{x^{1/2}}$.
Aside, enforcing the substitution $x/k\mapsto x$, we find that
$$\int_0^\infty \frac{k\sin(x/k)}{x^{3/2}}\,dx=\sqrt{k}\int_0^\infty \frac{\sin(x)}{x^{3/2}}\,dx=\sqrt{2\pi k}\to \infty$$
