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
20 events
| when toggle format | what | by | license | comment | |
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| Jun 8, 2019 at 23:52 | vote | accept | J Bausch | ||
| Jun 5, 2019 at 20:33 | comment | added | J Bausch | I'll check and accept asap. Thanks so much!! | |
| Jun 1, 2019 at 3:45 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
SageMath code added
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| May 31, 2019 at 18:28 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
edited body
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| May 31, 2019 at 18:15 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
Formulae are corrected.
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| May 31, 2019 at 18:12 | comment | added | Max Alekseyev | @JBausch: Good catch! There was an error in my formulae, now corrected. Essentially, we should use $\frac{b}{a}$ instead of $1$ in $I(1,\cdot)$ and $c$ instead of $\infty$ in $I(\cdot,\infty)$. This also addresses the issue with $\infty$ you pointed out earlier. | |
| May 31, 2019 at 18:09 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
Formulae are corrected.
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| May 30, 2019 at 9:21 | comment | added | J Bausch | @MaxAlekseyev did you have a chance to have a look? I've compared this to a numerical integration and yours matches up when $n\rightarrow\infty$, but e.g. in the case I wrote just above there's some discrepancy. No rush!! I really appreciate your help with this - I've worked through your argument but can't really pinpoint a mistake, which is why I ask. | |
| May 27, 2019 at 22:30 | comment | added | J Bausch | Another edge case that doesn't work: take $k=3/2, l=0, n=2, a=10, b=1, c=2$, then $I_2^{3/2,0}(1,2)=-2(0-0-I^{3/2,0}_1(1,\infty)+I^{1,0}_1(1/5,\infty)/\sqrt 2)$; both of the latter integrals can be calculated exactly; the total epression turns out to be $\approx-0.52$, which cannot be since the integrand is strictly positive. | |
| May 27, 2019 at 21:43 | comment | added | J Bausch | You're right; treating $\infty$ as a symbolic large constant and defining its product with zero to be zero does the trick. Will check the rest and report back whether it matches a numeric integration! | |
| May 27, 2019 at 19:51 | comment | added | J Bausch | @MaxAlekseyev: I've just tried implementing it in MM, but there seems to be something amiss: for e.g. $a=20, b=1, c=40, n=3, l=0$ we have: $I_3^{k,0}(1,40)$ expands the rightmost term $I_{2}^{1,0}(2,\infty)$; the latter then includes a term that has $\log(\infty)I^{1,0}_1(\infty,\infty)$ in it. Am I missing a break condition? | |
| May 24, 2019 at 16:01 | comment | added | J Bausch | Very useful answer, much appreciated. I'll try to implement this in MM and cross-check with the original expression. | |
| May 23, 2019 at 15:36 | comment | added | Max Alekseyev | @MattF.: My estimate is that $I^{1,l}_n(b,c)$ has $O(n)$ terms and $I^{k,0}_n(b,c)$ has $O(n^2)$ terms. | |
| May 22, 2019 at 21:15 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
added a simpler formula; deleted 6 characters in body
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| May 22, 2019 at 20:24 | comment | added | user44143 | Do you have an estimate for the number of terms in this calculation? If the calculation of $I_n$ uses at least $4$ terms of $I_{n-1}$, and so it uses at least $4^n$ basic terms, then it'd be nicer to get something with polynomial growth. | |
| May 22, 2019 at 18:46 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
added 43 characters in body
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| May 22, 2019 at 18:41 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
added 43 characters in body
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| May 22, 2019 at 18:28 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
deleted 6 characters in body
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| May 22, 2019 at 18:21 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
added 926 characters in body
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| May 22, 2019 at 16:50 | history | answered | Max Alekseyev | CC BY-SA 4.0 |