Estimate on an integral involving the Japanese bracket

Question

I'm currently reading the paper Well-posedness for the Zakharov system with the periodic boundary condition by Takaoka. In the proof of Lemma 2.3 about the integral $I_1$ one needs to establish the estimate

$$\int_{-\infty}^{\infty} \frac{d\tau'}{\langle n^2 + |\tau'+\alpha n^2|\rangle^{2a}} \leq C\frac{1}{\langle n\rangle^{4a-2}},$$

here $\langle \cdot \rangle := (1+|\cdot|^2)^{1/2}$ denotes the usual Japanese bracket, $a \in (1/2,3/4)$, $\alpha\neq 0$ is a real parameter and $n$ is an integer.

This estimate is not explicitly mentioned in the proof but an intermediate step in order to arrive at equation (2.1). It is clear to me that the order of the exponent of the RHS is correct, since the Japanese bracket has order one and one expects to gain one order by integration. If $a$ were an integer I'm sure that by substitution one would have to use some $\arctan$ properties.

How can I establish this estimate?

Answer 1

I assume that you mean $a>1/2$, for $a=1/2$ the integral diverges.

Denote $\tau=\tau'+\alpha n^2$ and use the relations $\langle x\rangle\asymp \max(1,|x|)$, $(x+y)^a\asymp \max(x^a,y^a)$ for $x,y>0$ (here $A\asymp B$ means $c_1 b\leqslant a\leqslant c_2 b$ for universal constants $c_1,c_2$). Your integral becomes $$ \asymp\int_{-\infty}^\infty\frac{d\tau}{\max(\langle n\rangle^{4a},|\tau|^{2a})}= \int_{-\langle n\rangle^2}^{\langle n\rangle^2}\frac{d\tau}{\langle n\rangle^{4a}}+2\int_{\langle n\rangle^2}^\infty\frac{d\tau}{\tau^{2a}}=\left(2+\frac2{2a-1}\right)\langle n\rangle^{2-4a}. $$