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 $T\to \infty$, we see $a\to 0$ and $b\to T$. 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 $0<a<\frac{T}{2}$ 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 $b\to T$, 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?
