# Estimate $\int_a^b \frac{[-t^{\frac{8}{3}}+t^{\frac{2}{3}}T^2-1]^{\frac{3}{2}}}{t^5}dt$ as $T\to \infty$ where $a,b$ are two roots of the numerator?

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 $$a, both depending on $$T$$ and satisfying

$$0

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 $$0 from above, we see from this identity that $$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?