Let say m is dimension $\exists$ $f(x)$ $f$ is density function and

\begin{equation*} f(x) = \frac{c(m,a,b)}{\|x\|^a\left(\log\frac{e}{\|x\|}\right)^{b}}\mathbf{1}_{\|x\|\leq 1} \geq 0 \end{equation*} where $a$ and $b$ depend on $m$ dimension and also we know the density function is \begin{equation*} \int_{\mathbb{R}^m}f(x)dx = 1. \end{equation*} Let \begin{equation*} c_0 = \frac{1}{\int_{\|x\|\leq 1}\|x\|^a\left(\log\frac{e}{\|x\|}\right)^{b}dx} \end{equation*} Suppose that $\rho = \|x\| $, then we have \begin{equation*} c= \frac{2\pi^{m/2}}{\Gamma(m/2)}\int_0^1\rho^{m-1}\rho^{-a}\left(\log\frac{e}{\rho}\right)^{-b}\,d\rho. \end{equation*} So, my question is how do we know limit of $c$ exist or not. Thanks a lot in advance

Note: $\mathbf{1}$ is indicator

  • 2
    $\begingroup$ What variable(s) are you changing in taking this limit?? And are you asking about $c$ or $1/c$ (which you seem to interchange)? $\endgroup$
    – whuber
    Sep 16, 2019 at 12:01
  • $\begingroup$ @whuber this looks like the OP is expressing a density function of the form $$ f(x) \propto \frac{1}{\|x\|^a\left(\log\frac{e}{\|x\|}\right)^{b}}\mathbf{1}_{\|x\|\leq 1}$$ and wishes to compute some normalization coefficient but does not know whether the integral is finite. It is similar as this question: stats.stackexchange.com/questions/427352/… $\endgroup$ Sep 16, 2019 at 12:58
  • $\begingroup$ @Martijn The connection with the previous question is clear, but the references to "limit" mean something important has not yet been explained. $\endgroup$
    – whuber
    Sep 16, 2019 at 13:02
  • $\begingroup$ With the answers from that related question (substituting $u = 1 - log(x)$) the integral expression becomes: $$ \int_0^1 \frac{1}{x^a\left(\log\frac{e}{x}\right)^{b}} dx = e^{1-a} \int_1^\infty e^{(a-1)u} u^{-b} du$$ and it is a matter of where, for which $a$ and $b$, the integral converges. $\endgroup$ Sep 16, 2019 at 13:26
  • $\begingroup$ @whuber I edited the draft, i want to see the integral of c exists or not exists. As to limit. maybe we can consider for which $a$ and $b$, the integral converges where $a$ and $b$ depend on $m$ dimension $\endgroup$
    – mhmt
    Sep 16, 2019 at 14:14


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