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


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 $\alpha \in (n, n+1)$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.