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


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!),


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?



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