Timeline for Integrate $1/(x_1x_2\cdots x_n)^k$ for $1\le x_i \le a$, where product of coordinates satisfies $ b\le x_1\cdots x_n\le c$
Current License: CC BY-SA 4.0
21 events
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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.
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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
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S May 22, 2019 at 11:43 | history | suggested | user64494 | CC BY-SA 4.0 |
The title and body and TeX are improved.
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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 |