Proving that $x^{\alpha}(1+\Vert x\Vert^{2})^{-k}$ belongs to $L^{2}(\mathbb{R}^{n})$

Question

Let $\alpha\in\mathbb{N}^{n}$ be a multi-index, i.e. $\alpha=(\alpha_{1},\dots,\alpha_{n})$ such that $x^{\alpha}:=\prod_{i=1}^{n}x^{\alpha_{i}}_{i}$. The modulus of a multi-index is defined as the quantity $\vert\alpha\vert:=\sum_{i=1}^{n}\alpha_{i}$. Let $\Vert x\Vert$ denote the euclidean norm of $x$, i.e. $\Vert x\Vert:=\sqrt{\sum_{i=1}^{n}x^{2}_{i}}$. Define the following function, for a fixed $\alpha\in\mathbb{N}^{n}$ and a fixed $k\in\mathbb{N}_{0}$

$$f_{\alpha,k}:\mathbb{R}^{n}\to\mathbb{C}:x\mapsto\frac{x^{\alpha}}{(1+\Vert x\Vert^{2})^{k}}$$

I want to prove that $f_{\alpha,k}\in L^{2}(\mathbb{R}^{n})$ (the space of square integrable functions over $\mathbb{R}^{n}$) if and only if $2k>\vert\alpha\vert+\tfrac{n}{2}$. It is possible that this bound is false. In that case, do not hesitate to correct it. Anyway, here is my attempt:

\begin{align*} \int_{\mathbb{R}^{n}}\vert f_{\alpha,k}(x)\vert^{2}\text{d}x &= \int_{\mathbb{R}^{n}}\left\vert \frac{x^{\alpha}}{(1+\Vert x\Vert^{2})^{k}}\right\vert^{2}\text{d}x \\ &= \int_{\mathbb{R}^{n}}\left\vert \frac{x^{2\alpha}}{(1+\Vert x\Vert^{2})^{2k}}\right\vert\text{d}x\\ \end{align*}

Since $\vert x^{2\alpha}\vert\le C(1+\Vert x\Vert^{2})^{\vert\alpha\vert}$, for a constant $C>0$, we have

\begin{align*} \int_{\mathbb{R}^{n}}\vert f_{\alpha,k}(x)\vert^{2}\text{d}x &= \int_{\mathbb{R}^{n}}\left\vert \frac{x^{2\alpha}}{(1+\Vert x\Vert^{2})^{2k}}\right\vert\text{d}x\\ &\le C\int_{\mathbb{R}^{n}}\left\vert \frac{1}{(1+\Vert x\Vert^{2})^{2k-\vert\alpha\vert}}\right\vert\text{d}x\\ &=C\int_{\mathbb{R}^{n}}\frac{1}{(1+\Vert x\Vert^{2})^{2k-\vert\alpha\vert}}\text{d}x \end{align*}

But I'm stuck here. I don't know how to get rid of this integral. Any hint is appreciated. Thank you.

Answer 1

You can use spherical coordinates. In this way, the terms in $\cos \theta_i$ and $\sin\theta_i$ only appear in the numerator (the denominator depends only on $r$). Consequently, the considered integral is convergent if and only if $I:=\int_0^{+\infty}r^{n-1+2|\alpha|}/(1+r^2)^{2k}\mathrm dr$ converges. Since the only problem is at infinity, $I$ is convergent if and only if $\int_0^{+\infty} r^{-(4k-n-2|\alpha|+1)}\mathrm dr$ converges. This is equivalent to $$4k-n-2|\alpha| \gt 0.$$