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How can I find the values of $\alpha$ for which the following integrals (in $\mathbb{R}^n $ ) converge ?

  1. $\int_{|\vec{x}|\geq 1 } \frac{ln(|\vec{x}|^3 )}{|\vec{x}|^\alpha} d\vec{x} $

  2. $\int_{\mathbb{R}^n } \frac{sin(|\vec{x}|)}{|\vec{x}|^\alpha} d\vec{x} $

I guess the thing is we need to calculate the limit form of these integrals, in some other coordinate system, where $r= ||\vec{x}|| $ ... I can't figure out what will be that Jacobian in this case and can't understand how to solve these questions

Will you please help me?

Thanks in advance

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Recall that $\def\vol{{\rm vol}}\vol_{n-1}(rS^{n-1}) = 2\frac{\pi^{n/2}}{\Gamma(n/2)} r^{n-1} =: \beta_{n-1} r^{n-1}$, we have $\def\abs#1{\left|#1\right|}$ \begin{align*} \int_{\abs x \ge 1} \frac{\log \abs x^3}{\abs x^\alpha}\, dx &= \int_{1}^\infty \int_{rS^{n-1}} \frac{\log r^3}{r^\alpha}\, dS(x)\, dr\\ &= \beta_{n-1} \int_1^\infty \frac{\log r^3}{r^\alpha} r^{n-1}\, dr\\ &= 3\beta_{n-1}\int_1^\infty {r^{ n - 1-\alpha}\cdot \log r}\, dr \end{align*} This converges if $n-1-\alpha < -1$, that is $\alpha > n$.

For the second case, arguing along the same lines, we have \begin{align*} \int_{\mathbb R^n} \frac{\sin\abs x}{\abs x^\alpha}\, dx &= \int_0^\infty \frac{\sin r}{r^\alpha}\beta_{n-1} r^{n-1}\, dr\\ &= \int_0^\infty \sin r \cdot r^{n-1-\alpha}\, dr \end{align*} This converges "at $\infty$" if $n-1-\alpha < -1$, that is $\alpha > n$, and "at 0" if $n-\alpha> -1$ (note that $\sin r\cdot r^{n-1-\alpha} = \frac{\sin r}r \cdot r^{n-\alpha}$, that is if $\alpha < n+1$. So we must have $\alpha \in (n, n+1)$.

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  • $\begingroup$ Hi @martini, thanks for your answer. I can't understand the equality: $ \int_{\abs x \ge 1} \frac{\log \abs x^3}{\abs x^\alpha}\, dx = \int_{1}^\infty \int_{rS^{n-1}} \frac{\log r^3}{r^\alpha}\, dS(x)\, dr\\...$ (i.e. - I can't understand how you passed from an integral over some region to two integrals , one over $1 ,\infty$ and the other over $rS^n $ (why $rS^n$?) Thanks ! $\endgroup$ – homogenity Dec 10 '13 at 15:10
  • $\begingroup$ Note that $\{x \in \mathbb R^n \mid \left| x\right| \ge 1\} = \biguplus_{r \ge 1} rS^{n-1}$, where $S^{n-1}$ denotes the unit sphere, i. e. the set of vectors with unit length. $\endgroup$ – martini Dec 10 '13 at 15:15
  • $\begingroup$ Hmmm... when writing $r\cdot S^{n-1} $ do you mean pointwise multiplication ? i.e. - $(xr| x\in S^{n-1} ) $ ? $\endgroup$ – homogenity Dec 10 '13 at 17:44
  • $\begingroup$ and why did you put $dS(x)$ ? why does this volume element depend on $x$ ? $\endgroup$ – homogenity Dec 10 '13 at 17:48
  • $\begingroup$ (1) Yes, $rS^{n-1} = \{rx \mid x \in S^{n-1}$. And by writing $dS(x)$ I tried to make explicit that the variable with respect to which we integrate is $x$. $\endgroup$ – martini Dec 10 '13 at 18:18

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