# Obtaining an upper bound for an integral

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!

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.
• Oh actually I did a small mistake, we cannot take $R=0$ ... I corrected it. Also, where is this integral coming from? Sep 15 at 16:28