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.)
 A: 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).
