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!
 A: 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.
