Skip to main content
21 events
when toggle format what by license comment
Sep 16, 2019 at 17:42 answer added J Bausch timeline score: 1
Jun 8, 2019 at 23:52 vote accept J Bausch
May 24, 2019 at 15:58 comment added J Bausch Gerald Edgar: Yes, the d... I omitted for brevity, thanks for the proposed change. Johannes Trost: yes, indeed, $k'=1-k$. I can't edit the comment unfortunately. Matt F.: I plugged some cases into Mathematica as well and the case distinction exploded, agreed. Terry Tao: the asymptotic behaviour is also of interest, so thank you for this proposal; I'll have a closer look and get back to you whether that gives a useful bound--if you're interested, this integral emerges from an estimate on the (weighted) divisor summatory function, where the prime factors are upper-bounded by $a$.
May 22, 2019 at 21:52 comment added Yemon Choi @user64494 Why? What is so basic or standard about this question that you believe it should go on MSE?
May 22, 2019 at 21:25 comment added user44143 Terry Tao's suggestion of taking $n \rightarrow \infty$ is reasonable...but I have downvoted because the question asks for calculations from a lot of parameters ($a,b,c,k,n$), with no apparent organizational principle.
May 22, 2019 at 18:48 comment added Terry Tao Also, these integrals may obey delay-differential equations similar to that obeyed by the Dickman or Buchstab functions. See for instance Exercise 39 of my lecture notes terrytao.wordpress.com/2014/11/23/…
May 22, 2019 at 18:41 comment added Terry Tao If you perform a Fourier expansion of the indicator function $1_{[b',c']}$ and use Fubini's theorem (which requires some preliminary smoothing of the indicator function to justify properly, but never mind that) you can convert the $n$-dimensional $z$-integral in the previous comment to a one-dimensional integral over the Fourier variable, which should be a suitable form for instance for working out asymptotics in various limiting regimes such as $n \to \infty$, if that is your application of interest.
May 22, 2019 at 16:50 answer added Max Alekseyev timeline score: 6
May 22, 2019 at 15:41 comment added user44143 I don't see much hope for a clean answer. Consider the simplest case, $k=0, n=2$. Then the integral we want is $f(c)-f(b)$ where $$f(c)=\int_1^a \max(1,\min(a,\frac cx))dx = \begin{cases} -c + c \log c + a \ \ \ \ \ \text{ if } 1 \le c \le a \\ c + c \log(a^2/c) - a \text{ if } a \le c \le a^2 \\ a^2 - a \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ if } c \ge a^2\end{cases}$$
May 22, 2019 at 15:36 comment added user64494 A right forum for such type questions is math.stackexchange.com .
S May 22, 2019 at 15:09 history suggested user64494 CC BY-SA 4.0
The title and body are improved.
May 22, 2019 at 13:57 comment added Johannes Trost @J Bausch I guess, it should be $k'=1-k$, right ?
May 22, 2019 at 13:45 review Suggested edits
S May 22, 2019 at 15:09
May 22, 2019 at 12:51 comment added Gerald Edgar The problem statement now is missing the $dx_1 dx_2 \cdots dx_n$. And note, when you change variables, you need to switch to $dz_1 dz_2\cdots dz_n$.
May 22, 2019 at 12:19 comment added J Bausch Good suggestion! That leads to a simpler expression $$ \iiint_0^{a'} \exp(k'(z_1+\ldots+z_n))*\begin{cases} 1 & b' \le z_1+\ldots+z_n \le c' \\ 0 & \text{otherwise,} \end{cases} $$ where $a'=\ln a$, $b'=\ln b$, $c'=\ln c$, and $k'=k+1$. It should suffice to solve this for $b'=0$, of course, as then we can just subtract two integrals off each other, and we can rescale to set $a'=1$. For $k'=0$, we appear to get a list of polynomials for various ranges of $c'$, with coefficients from [oeis.org/A188668]. Not a proof, just an observation. Thanks, that was already helpful!
May 22, 2019 at 11:45 history rollback J Bausch
Rollback to Revision 1
S May 22, 2019 at 11:43 history suggested user64494 CC BY-SA 4.0
The title and body and TeX are improved.
May 22, 2019 at 11:39 comment added Johannes Trost How about changing variables to $z_{i}:=\ln x_{i}$ ? The restriction then converts to an integration over a simplex area ($\ln b \le \sum z_{i} \le \ln c$), which might be easier to handle.
May 22, 2019 at 11:35 review Close votes
May 23, 2019 at 8:55
May 22, 2019 at 11:02 review Suggested edits
S May 22, 2019 at 11:43
May 22, 2019 at 9:38 history asked J Bausch CC BY-SA 4.0