Could you please give me some hint how to prove convergence of $\int_0^1f\left(\sqrt x \right)dx$
when f(x) is continuous for $0<x\le1$ and $\lim_{x\to0^+}x^3f^2(x)=1$ ?
I tried the usual way: $\int_0^1f\left(\sqrt x \right)dx=\left[y=\sqrt x \right]=2\int_0^1\frac{f(y)}y dy$.
From $\lim_{x\to0^+}x^3f^2(x)=1$ we can conclude that for all $0<x<\lambda<1$ $\frac 1{2x^3}\le f^2(x)\le \frac 3{2x^3}$, therefore $\int_0^1 f^2(x)dx$ diverges.But how divergence of $\int_0^1 f^2(x)dx$ is connected to convergence of $\int_0^1f\left(\sqrt x \right)dx$ ?
Thanks.