Integral test of convergence for $X \in{\displaystyle L^{2}}$

Question

I'm reading a script where it says that for $X\in{\displaystyle L^{2}}$ we have $$\mathbb{E}[X^2]=\int_0^{\infty}\mathbb{P}(X^2\geq s)\text{d}s<\infty$$ and from this we can conclude with the integral test of convergence that $$\mathbb{P}(X^2\geq s)=\mathcal{o}\left(\frac{1}{sln(s)}\right)$$ The formula with the expectation is clear to me but I don't know how to arrive at the statement regarding $\mathcal{o}\left((sln(s))^{-1}\right)$. Or more generally I'm guessing it would then hold that if we have decreasing functions $f,g\geq 0$ with $\int_0^{\infty}f(s)\text{d}s<\infty$ and $\int_0^{\infty}g(s)\text{d}s =\infty$ then $\frac{f(x)}{g(x)}\to 0$ as $x\to\infty$. But I don't know for certain if this is true and how I would prove it.

Answer 1

I expect the statement regarding $\mathcal{o}\left((s\ln(s))^{-1}\right)$ can fail because its counterpart for functions is fail. For instance, for each $x\ge 0$ let $f(x)=(s(x) \ln(s(x)))^{-1}$, where $s(x)$ is the smallest number $s$ from the sequence $(\exp(s^2))_{s=0}^\infty$ such that $s(x)\ge x$. Then $f(x)\not\in \mathcal{o}\left((s ln(s))^{-1}\right),$ but $$\int_0^\infty f(x)dx=\int_0^1 f(x)dx+\sum_{s=0}^\infty \int_{\exp(s^2)}^{ \exp((s+1)^2)} f(x) dx\le$$ $$1+\sum_{s=0}^\infty \int_{\exp(s^2)}^{ \exp((s+1)^2)} (\exp((s+1)^2\ln(\exp((s+1)^2))^{-1} dx\le$$ $$1+\sum_{s=0}^\infty \exp((s+1)^2) \exp (s+1)^{-2} (s+1)^{-2}=$$ $$1+\sum_{s=0}^\infty (s+1)^{-2}=1+\frac {\pi^2}6.$$