# Does the limit of c exist or not

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

• What variable(s) are you changing in taking this limit?? And are you asking about $c$ or $1/c$ (which you seem to interchange)?
– whuber
Commented Sep 16, 2019 at 12:01
• @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/… Commented Sep 16, 2019 at 12:58
• @Martijn The connection with the previous question is clear, but the references to "limit" mean something important has not yet been explained.
– whuber
Commented Sep 16, 2019 at 13:02
• 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. Commented Sep 16, 2019 at 13:26
• @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
– mhmt
Commented Sep 16, 2019 at 14:14