Integral over the Japanese bracket I want to show that
$$\int \langle x\rangle^{-1-\epsilon} dx = \int (1+|x|^2)^{\frac{-1-\epsilon}{2}}dx$$
converges for $\epsilon>0$. Assume we're on $\mathbb{R}$. Because of symmetry we can integrate from $0$ to $\infty$ modulo a constant:
$$\int_0^{\infty} (1+r^2)^{\frac{-1-\epsilon}{2}}dr$$
Now I'm a bit helpless because I haven't found a way to calculate this integral just yet. Please provide me some hint how to proceed (subsitution, partial integration, etc.)

There are multiple ways to show that the integral converges, as suggested by the comments. We can substitute $t=r^2$ and obtain (replace $dr$ by $(4t)^{-1/2}dt$)
$$ \frac{1}{2}\int_0^{\infty} (1+t)^{-(1+\epsilon)/2}t^{-1/2}dt$$
Another substitution $(1+t)^{-1} =s $ yields
$$ I= \frac{1}{2} \int_0^1 \frac{s^{\epsilon/2 - 1}}{(1-s)^{1/2}}ds$$
(the minus sign from the derivative of the map $t\mapsto (1+t)^{-1}$ is reversed by flipping the transformed integral boundaries $1$ and $0$). This integral is known as the Beta function $\frac{1}{2} B(\epsilon/2,1/2)$ which is known to converge.
Another approach is to split the integral into to parts:
$$\int_0^1 (1+r^2)^{-(1+\epsilon)/2} dr + \int_1^{\infty} (1+r^2)^{-(1+\epsilon)/2} dr$$
Since $(1+r^2)$ is never zero, the integrand of the first integral is bounded and since the integral is over a compact set it is finite. For the second integral observe that $(1+r^2) > r^2$ for all $r$ and so
$$ (1+r^2)^{-(1+\epsilon)/2} \leq r^{-(1+\epsilon)}$$
Now
$$\int_1^{\infty} r^{-(1+\epsilon)} dr = r^{-\epsilon}|_1^{\infty} $$
which is finite (In hindsight the problem was really easy. Apparently I'm a really lazy person, apologies).
