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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.)

Cornman
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Jakob Elias
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    As you only want to show that it converges, it helps to split into $0\leq r \leq 1$ and $r>1$ and treat them separately. – Severin Schraven May 10 '21 at 07:27
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    @Cornman Explain to me why this is low quality. I provided a clear statement and my considerations how to solve it as well as the nudge I need to solve it on my own. There are more sophisticated questions for sure. – Jakob Elias May 10 '21 at 07:32
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    @JakobElias Clearly, the integral converges on $0\leq r\leq 1$. For $r>1$, use $$ \frac{1}{{(1 + r^2 )^{\frac{{1 + \varepsilon }}{2}} }} \le \frac{1}{{(r^2 )^{\frac{{1 + \varepsilon }}{2}} }} = \frac{1}{{r^{1 + \varepsilon } }}$$ and integrate the last expression explicitly. – Gary May 10 '21 at 07:38
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    I have retracted my vote to close, because this is not of low quality. – Cornman May 10 '21 at 07:38
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    @Gary that works, thanks. Should I post my solution as an answer to this question? – Jakob Elias May 10 '21 at 07:41
  • @JakobElias You are welcome to post your own answer. – YNK May 10 '21 at 07:44
  • @JakobElias Feel free to post your answer. – Gary May 10 '21 at 07:45
  • Make the change $t=r^2$ and then $s=\frac{1}{1+t}$. You get $I=\frac{1}{2}\int_0^1s^{\epsilon/2-1}(1-s)^{-1/2}ds=\frac{1}{2}B(\frac{\epsilon}{2};\frac{1}{2})=\frac{\sqrt\pi}{2}\frac{\Gamma(\epsilon/2)}{\Gamma(1/2+\epsilon/2)}$. Integral converges at $\epsilon>0$ and $I\to\frac{1}{2}\frac{2}{\epsilon}=\frac{1}{\epsilon}$ at $\epsilon\to0$ – Svyatoslav May 10 '21 at 09:09

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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).

Jakob Elias
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