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Fix $q>1$. Define the function

$$ f_q(c):=\int_e^\infty \frac{e^{-c r^2}r}{\log(r)^q}d r. $$

The problem is whether the following is true,

$$ \lim_{c\rightarrow 0} c \log(1/c)^q f_q(c) = C \in (0,\infty)? $$

If not, what is the right rate that $f_q(c)$ blows up at $c=0$?

Note that if $q=0$, then $f_q(c)=\frac{e^{-c e^2}}{2c}$ and the above limit is true with $C=1/2$.

It is hard to choose right tags for this question, I am sorry for anything improper.

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It converges for all $q$. You can separate the interval into the ranges $[e,1/\sqrt c]$ and $[1/\sqrt c,\infty)$. On the first range, the exponential term is essentially 1. On the second range, the logarithm is essentially constant.

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  • $\begingroup$ Thanks Anthony Quas. I don't see why it diverges for $q>1$. For me, it is rather clear that the term $e^{-c r^2}$ term makes the whole integrand converge. $\endgroup$
    – Anand
    Commented Oct 18, 2014 at 15:21
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    $\begingroup$ @Anand: I agree. I think I was integrating from 1 to $\infty$. I corrected the answer. $\endgroup$
    – Anthony Quas
    Commented Oct 18, 2014 at 16:24
  • $\begingroup$ Thanks a lot! I agree that it converges. The question is how fast it blows up as $c$ goes to zero. :-) $\endgroup$
    – Anand
    Commented Oct 18, 2014 at 16:49
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    $\begingroup$ @Anand: No I mean your expression $c\log(1/c)^qf_q(c)$ converges to something positive as $c\to 0$. $\endgroup$
    – Anthony Quas
    Commented Oct 18, 2014 at 17:01
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    $\begingroup$ @Anand: I don't think this is right. There are quite easy upper bounds of $1/(c(\log(1/c))^q)$. On $[e,c^{-1/4}]$, you can bound the integrand above by (a multiple of) $r$; on $[c^{-1/4},c^{-1/2}]$ you can bound above by a const. multiple of $r/(\log(1/c))^q$; on $[c^{-1/2},\infty)$, you can bound above by a const multiple of $re^{-cr^2}/(\log(1/c))^q$. Then you can integrate all of these. $\endgroup$
    – Anthony Quas
    Commented Oct 21, 2014 at 12:17

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