For $t,T>0$, let $f(t):=-t^{frac{8}{3}}+t^{frac{2}{3}}T^2-1$. By analyzing the derivative $f'(t)=-frac{8}{3}t^{frac{5}{3}}+frac{2}{3}t^{-frac{1}{3}}T^2$, we see on $(0,infty)$, $f(t)$ reaches its maximum $cT^{frac{8}{3}}$ at $t=frac{T}{2}$. If $T$ is large enough, we can also see $f(t)$ has two positive roots, denoted $a<b$, both depending on $T$ and satisfying

$$0<a<frac{T}{2}<b<T.$$

It is easy to see that $f(t)$ is positive on $(a,b)$.

As $Tto infty$, we see $ato 0$ and $bto T$. Moreover, for $a$, since $-a^{frac{8}{3}}+a^{frac{2}{3}}T^2-1=0$ ($a$ is a root!),

$$a^{frac{2}{3}}=frac{1}{T^2-a^2}.$$

Since $0<a<T$ from above, we see from this identity that $a asymp frac{1}{T^3}$ as $T>>1$.

Here is my question (arise from research): how to estimate (no need to compute explicitly)

$$A(T):=int_a^b frac{[f(t)]^{frac{3}{2}}}{t^5}dt$$

in terms of $T$? This integral is hard since there is a "fractional polynomial" under the square root.

The only thing I could do is to perform a Cauchy-Schwarz inequality on the integral, which yields

$$sqrt{int_a^b frac{[f(t)]^{3}}{t^{10}}dt int_a^b 1 dt }.$$

This integrand has explicit anti-derivative and can be computed. By expanding $[f(t)]^{3}$, using the facts that $a asymp frac{1}{T^3}$ and $bto T$, we have

$$sqrt{int_a^b frac{[f(t)]^{3}}{t^{10}}dt int_a^b 1 dt}asymp T^{12}.$$

This gives an upper bound for $A(T)$. Now I want to show that actually

$$A(T)asymp T^{12}.$$

Does anyone have anything to suggest for the more precise estimate?