3
$\begingroup$

I'm trying to get an upper bound in the following way: $$ I_{\alpha} = \int_{\mathbb{R}} \frac{dx}{(|x^2 - \alpha| + 1)^{1+ \epsilon}} \leq C \frac{1}{\alpha^{\gamma}}, $$ where $\alpha >0$ and $0 < \varepsilon <<1$, for some $\gamma >0$. My attempt was to write

$$ I_{\alpha} = \int_{|x| > \sqrt{\alpha}} \frac{dx}{(|x^2 - \alpha| + 1)^{1+ \epsilon}} + \int_{|x| \leq \sqrt{\alpha}} \frac{dx}{(|x^2 - \alpha| + 1)^{1+ \epsilon}}. $$ However, it is not clear how to obtain an upper bound of form $C \frac{1}{\alpha^{\gamma}}$ for the second integral above. Any suggestions or ideas are welcome. Thanks!

$\endgroup$
2
$\begingroup$

So you can get $$ I_α < 6 \,\min(1,\alpha^{-\gamma}) \leqslant \frac{6}{\alpha^\gamma} $$ with $γ = \frac{\varepsilon}{2(2+\varepsilon)}$.


  • First notice that if $α$ is small, say $α \leqslant 1$, then $$ \int_{|x|>1} \frac{\mathrm d x}{(\,|x^2-\alpha|+1)^{1+\varepsilon}} \leqslant \int_{|x|>1} \frac{\mathrm d x}{|x|^{2(1+\varepsilon)}} = \frac{2}{1+2\,\varepsilon} < 2 \\ \int_{|x|<1} \frac{\mathrm d x}{(\,|x^2-\alpha|+1)^{1+\varepsilon}} \leqslant \int_{|x|<1} \mathrm d x = 2 $$ so in this case $$ I_α \leqslant 4. $$

  • Hence we can focus on the case $\alpha>1$. Let me start by doing the change of variable $x = \sqrt \alpha\, y$ to get $$ I_α = \int_{\mathbb R} \frac{\sqrt α\,\mathrm d y}{(\alpha\,|y^2-1|+1)^{1+\varepsilon}} = \int_0^\infty \frac{2\,\sqrt α\,\mathrm d y}{(\alpha\,|y^2-1|+1)^{1+\varepsilon}}. $$ Here we see clearly that the difficulty lies in the region where $y^2 = 1$. Let put this zone around $0$ by taking $z = y^2-1$. Then $$ I_α = \int_{-1}^\infty \frac{\sqrt α\,\mathrm d z}{\sqrt{1+z}\,(\alpha\,|z|+1)^{1+\varepsilon}}. $$ As said before, we need to isolate the region near $z=0$ and say that it is small. Hence we cut the integral in three parts: for $r,R\geqslant 0$ we have $$ \begin{align*} I_{α,1}=\int_{-1}^{-r} \frac{\sqrt α\,\mathrm d z}{\sqrt{1+z}\,(\alpha\,|z|+1)^{1+\varepsilon}} &\leqslant \frac{\sqrt α}{(\alpha\,r+1)^{1+\varepsilon}} \int_{0}^{1-r} \frac{\mathrm d w}{\sqrt{w}} = \frac{2\,\sqrt{(1-r)\,α}}{(\alpha\,r+1)^{1+\varepsilon}} \\ I_{α,2}=\int_{-r}^{R} \frac{\sqrt α\,\mathrm d z}{\sqrt{1+z}\,(\alpha\,|z|+1)^{1+\varepsilon}} &\leqslant \sqrt α \int_{1-r}^{1+R} \frac{\mathrm d w}{\sqrt{w}} = 2\,\sqrt{\alpha}\,(\sqrt{1+R}-\sqrt{1-r}) \\ I_{α,3} = \int_{R}^{\infty} \frac{\sqrt α\,\mathrm d z}{\sqrt{1+z}\,(\alpha\,|z|+1)^{1+\varepsilon}} &<\alpha^{-1/2-\varepsilon} \int_{R}^{\infty} \frac{\mathrm d z}{z^{3/2+\varepsilon}} < \frac{2}{(\alpha\,R)^{1/2+\varepsilon}} \end{align*} $$ Now we just have to optimize on $r$ and $R$ to get $I_{\alpha,2}$ small without making the other integrals too large. To get the best rate of decay for each part, we can take $R=\alpha^{-\frac{3}{3+2\varepsilon}}$ and $r=\alpha^{-\frac{1+\varepsilon}{2+\varepsilon}}$. This gives $$ \begin{align*} I_{α,1} &\leqslant \frac{\sqrt{2\alpha}}{(\alpha r)^{1+\varepsilon}} = \frac{\sqrt 2}{\alpha^\gamma} \\ I_{α,2} &\leqslant 2\,\sqrt{\alpha}\,(\sqrt{1-R}-\sqrt{1-r}) \leqslant \sqrt{\alpha}\,(r+R) = \alpha^{-\gamma} + \alpha^{-\beta} \\ I_{α,3} &< \frac{2}{\alpha^{\beta}}, \end{align*} $$ where $β = \frac{1}{2\,(3+2\varepsilon)} > \gamma$ if $\varepsilon$ is small.

$\endgroup$
5
  • $\begingroup$ LL3.14, thank you very much! Your solution was very enlightening to me. $\endgroup$
    – Marcelo Ng
    Sep 15 at 12:13
  • $\begingroup$ Oh actually I did a small mistake, we cannot take $R=0$ ... I corrected it. Also, where is this integral coming from? $\endgroup$
    – LL 3.14
    Sep 15 at 16:28
  • $\begingroup$ Dear @LL 3.14, thanks for the correction. This integral comes from a theory involving a perturbation of the nonlinear Schrödinger equation. $\endgroup$
    – Marcelo Ng
    Sep 17 at 18:28
  • 1
    $\begingroup$ Oh, interesting, thank you :) If your question is solved, you should indicate it as so. $\endgroup$
    – LL 3.14
    Sep 18 at 7:15
  • $\begingroup$ Dear @LL3.14. The perturbation of the Schrodinger equation mentioned above is known as the Schrödinger-Debye system :). $\endgroup$
    – Marcelo Ng
    2 days ago

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.