Algebraically Decaying Function with a Convergent Integral I want to show that the function $w_s(z) = \left(\frac{1}{1+||z||^2}\right)^{s/2}$ where $z=(z_1,z_2,...,z_{2d})\in \mathbb{R}^{2d}$ and $||z||^2 = \sum_{i=1}^{2d}z_i^2$ has a convergent integral $\iint_{\mathbb{R}^{2d}}w_s$ whenever $s>2d.$ Anybody care to check if my sketch below is correct?
We can write $$\iint_{\mathbb{R}^{2d}}w_s =\\ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}...\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \left(\frac{1}{1+z_1^2+z_2^2+...+z_{2d-1}^2+z_{2d}^2} \right)^{s/2}dz_1dz_2...dz_{2d-1}dz_d.$$
So we can group any pair of $z_i$ and $z_j$ such that $j\neq i$ and do polar coordinate transform on it. That is, we can let $r^2=z_1^2+z_2^2$ and $dz_1dz_2 = rdrd\theta$ so we have:
\begin{align}
=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}...\left[\int_{0}^{\infty}\int_{0}^{2\pi}\left( \frac{1}{1+z_3^2+z_4^2...+z_{2d-1}+z_{2d}+r^2} \right)^{s/2}rdrd\theta\right]dz_3dz_4...dz_{2d-1}dz_{2d}.
\end{align}
Evaluating the partial $r$ and $\theta$ integral above, we find (we used the fact that $s>2d$ here, so that $\frac{s}{2}-1>0)$:
\begin{align}
=\frac{\pi}{(\frac{s}{2}-1)}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}...\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \left(\frac{1}{1+z_3^2+z_4^2+...+z_{2d-1}^2+z_{2d}^2} \right)^{s/2}dz_3dz_4...dz_{2d-1}dz_d.
\end{align}
We can repeat this process $d-1$ more times and end up with $$\frac{\pi^d}{(\frac{s}{2}-1)(\frac{s}{2}-2)...(\frac{s}{2}-d)}< \infty.$$
Am I correct? Is this 'rigorous' enough to pass as a proof?
 A: $$\Psi(s,d)=\iint_{\mathbb{R}^{2d}}w_s =\\ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}...\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \left(\frac{1}{1+z_1^2+z_2^2+...+z_{2d-1}^2+z_{2d}^2} \right)^{s/2}dz_1dz_2...dz_{2d-1}dz_d.$$
\begin{align}
=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}...\left[\int_{0}^{\infty}\int_{0}^{2\pi}\left( \frac{1}{1+z_3^2+z_4^2...+z_{2d-1}+z_{2d}+r^2} \right)^{s/2}r dr d\theta \right  ]dz_3dz_4...dz_{2d-1}dz_{2d}.
\end{align}
\begin{align}
=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}...\left[\int_{0}^{\infty} -\frac{\pi}{\frac{s}{2}-1}\frac{d}{dr}( \left ( \frac{1}{1+z_3^2+z_4^2...+z_{2d-1}+z_{2d}+r^2} \right)^{s/2-1} )dr \right  ]dz_3dz_4...dz_{2d-1}dz_{2d}.
\end{align}
\begin{align}
= \frac{\pi}{\frac{s}{2}-1}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}...\int_{0}^{\infty} \left ( \frac{1}{1+z_3^2+z_4^2...+z_{2d-1}+z_{2d}} \right)^{(s-2)/2} dz_3dz_4...dz_{2d-1}dz_{2d}.
\end{align}
$= \frac{\pi}{\frac{s}{2}-1} \Psi(s-2, d-1)=....=$
$=\frac{\pi^d}{ \prod_{k=0}^{d}\frac{(s-2k)}{2}-1}$
when $\frac{(s-2k)}{2}-1>0$ for any $k\in \{0,...d\}$ so for $s>2(d+1)$
