$\lim_{x\to0^{+}}xf(x)=0$ if f(x) is integrable and monotonic decreasing.

Question

let $f:(0,1]\to[0,\infty)$ decreasing monotonic function, and the improper integral $\int_{0}^{1}f(x)dx$ converges, how can I show that the $\lim_{x\to0^{+}}xf(x)=0$ I tried using the limit convergence test of improper integrals in order to prove that the limit must be 0, otherwise, we can infer from it that the improper integral $\int_{0}^{1}\frac{1}{x}dx$ does converge, which is untrue. But I can't figure out how to prove that the limit does exist (even in the broad sense), so I can't infer $\lim_{x\to0^{+}}\frac{f(x)}{\frac{1}{x}}$ does exist.