An upper bound for an integral

Question

I saw many references using the following estimate but I couldn't prove it.

Given $T>0$ and $0 < b \leq \frac{1}{2}$, exist $C(b)$ constant that depends only on $b$ such that \begin{equation} \int_{\sqrt{1+s^2}<\frac{1}{T}} \frac{1}{\sqrt{1+s^2}^{2b}}ds \leq C(b) T^{2b-1}. \end{equation}

Any help is appreciated.

Answer 1

Here is an elementary solution: When $T>1$ the domain of integration is empty, so the integral is zero i.e the inequality is trivial. Hence, suppose that $0<T\leqslant 1$. Since $\sqrt{1+s^2}\geqslant \vert s \vert$, we have that $\{\sqrt{1+s^2}<T^{-1}\} \subset (-T^{-1},T^{-1})$. Hence, \begin{align*} \int_{\sqrt{1+s^2}<T^{-1}} \frac 1{(1+s^2)^b} \, ds &\leqslant \int_{-T^{-1}}^{T^{-1}}\frac 1{(1+s^2)^b} \, ds \\ &= 2 \int_0^1\frac 1{(1+s^2)^b} \, ds + 2 \int_1^{T^{-1}}\frac 1{(1+s^2)^b} \, ds. \end{align*} Using that $(1+s^2)^{-b} \leqslant 1$ in the first integral and that $(1+s^2)^{-b}\leqslant s^{-2b}$ in the second, we obtain \begin{align*} \int_{\sqrt{1+s^2}<T^{-1}} \frac 1{(1+s^2)^b} \, ds &\leqslant 2 + 2 \int_1^{T^{-1}}\frac {ds}{s^{2b}} \\ &= 2 + \frac 2{1-2b} \big ( T^{2b-1}-1 \big ) \\ &\leqslant \bigg ( 2 + \frac 2{1-2b} \bigg ) T^{2b-1} \end{align*} provided that $0<b < \frac12$. For the last inequality, I used that $T^{2b-1}\geqslant 1$. When $b=\frac12$, you instead get that \begin{align*} \int_{\sqrt{1+s^2}<T^{-1}} \frac 1{(1+s^2)^b} \, ds &\leqslant 2 +2\log(T^{-1}) =2\big(1 +\vert \log(T)\vert\big). \end{align*} Though haven't proven it here, you can check that when $b=1/2$ the integral behaves like $\vert \log T \vert$ asymptotically as $T\to 0^+$ which means the your inequality is false in this case and the above inequality is the best you can do.

Answer 2

If $T \ge 1$, there is nothing to show, so suppose $T$ is a dyadic number $< 1$, and consider the level sets $E_\lambda = \{s:\sqrt{1+s^2}\sim \lambda\}$. Let $I$ be the value of the integral, so we have \begin{align*} I \sim \sum_{\substack{\lambda\ \text{dyadic}\\1< \lambda < T^{-1}}}\lambda^{-2b}|E_\lambda|. \end{align*} It's straightforward to check that for $\lambda > 1$, $E_\lambda$ contains and is contained in the union of two intervals of length $\sim\lambda$, hence $|E_\lambda|\sim \lambda$, so overall $I\sim \sum_{1<\lambda < T^{-1}\ \text{dyadic}}\lambda^{-2b+1}$. This sum is a geometric series, so as long as $b<\frac12$, it's majorized by the contribution from the largest value of $\lambda$, namely $\lambda\sim T^{-1}$, so we conclude $$ I \sim_b T^{2b-1}. $$ If $b = 1/2$, this approach shows $I\sim \log(1/T)$. The assumption that $T$ be dyadic is superficial since we are looking for an estimate within a constant factor of the true size.