Expression for the quantile function (the inverse complementary CDF) for the case $E(X) = 0$:
Let $\overline{F}(s) = 1 - F(s)$. Then, $t = \overline{F}(s)$ and
$$\frac{d}{ds} E(X^2|X \geq s) = \frac{d}{ds} \Big( \frac{\int_s^{+\infty}x^2f(x)dx}{\int_s^{+\infty}f(x)dx} \Big) =$$ $$ = \frac{-s^2 f(s) \int_s^{+\infty}f(x)dx+f(s) \int_s^{+\infty}x^2f(x)dx}{\big(\int_s^{+\infty}f(x)dx\big)^2} = $$ $$ = \frac{-s^2 f(s) + f(s) W(\overline{F}(s))}{\overline{F}(s)}$$
On the other hand, $$ \frac{d}{ds} E(X^2|X \geq s) = \frac{d}{ds} W(\overline{F}(s)) = -W'(\overline{F}(s)) f(s)$$ where $W'$ is the derivative of $W$. Therefore, $$ \frac{s^2 - W(\overline{F}(s))}{\overline{F}(s)} = W'(\overline{F}(s))$$ or, using $t$, $$ \frac{(\overline{F}^{-1}(t))^2 - W(t)}{t} = W'(t) $$ and $$ \overline{F}^{-1}(t) = \sqrt{tW'(t) + W(t)}$$ ($\overline{F}^{-1}(t)$ should be positive in some neighborhood of $t = 0$). Then, since $\overline{F}^{-1}(t) = F^{-1}(1-t)$and $(F^{-1})'(1-t) = 1/f(F^{-1}(1-t))$, $$ f(F^{-1}(1-t)) = \frac{2 \sqrt{tW'(t) + W(t)}}{2W'(t)+tW''(t)} $$ or $$ f(sF^{-1}(x) = \frac{2 \sqrt{tW'(t) sqrt{(1-x)W'(1-x) + W(t)}}{2W'(t)+tW''(t)W(1-x)} \Big|_{t = \overline{F}(s)} $$
I hope that it's correct and that it can tell something about the decay of the distribution when $t$ tends to $+0$. Maybe it can also be extended to the case $E(X) \neq 0$.
