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when toggle format what by license comment
May 24, 2020 at 3:05 review Close votes
May 24, 2020 at 16:08
May 19, 2020 at 17:12 history edited Martin Sleziak
added top-level tag; https://meta.mathoverflow.net/questions/1457/why-are-mo-tags-formatted-as-they-are
May 19, 2020 at 15:17 review Close votes
May 19, 2020 at 17:02
S May 19, 2020 at 15:04 history suggested jimjim CC BY-SA 4.0
Improved the title to be descriptive and not vague
May 19, 2020 at 13:28 review Suggested edits
S May 19, 2020 at 15:04
Feb 17, 2018 at 18:35 vote accept Jerry Leung
Feb 17, 2018 at 18:35 vote accept Jerry Leung
Feb 17, 2018 at 18:35
Dec 24, 2017 at 0:20 vote accept Jerry Leung
Feb 17, 2018 at 18:35
Dec 22, 2017 at 16:59 answer added esg timeline score: 14
Dec 10, 2017 at 18:47 answer added Fedor Petrov timeline score: 12
Dec 10, 2017 at 14:29 answer added Shahrooz timeline score: 7
Dec 10, 2017 at 12:49 history edited Joe Silverman CC BY-SA 3.0
Improved formatting
Dec 10, 2017 at 12:31 history reopened Fedor Petrov
Todd Trimble
Dec 10, 2017 at 9:46 review Reopen votes
Dec 10, 2017 at 12:33
Dec 9, 2017 at 14:07 history closed Gerald Edgar
Michael Renardy
coudy
Peter Humphries
David Handelman
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Dec 9, 2017 at 13:42 history edited Jerry Leung CC BY-SA 3.0
added 204 characters in body
Dec 9, 2017 at 13:38 comment added Jerry Leung $\int_0^1 \int_0^1 \cdots \int_0^1\frac{x_{1}^2+x_{2}^2+\cdots+x_{n}^2}{x_{1}+x_{2}+\cdots+x_{n}}dx_{1}\, dx_{2}\cdots \, dx_{n}=n\int_{0}^{+\infty }\frac{2 - e^{-t}\left ( 2 + 2t+t^2 \right )}{t^3}\left ( \frac{1-e^{-t}}{t} \right )^{n-1}dt$ ok,I make a mistake.
Dec 9, 2017 at 13:16 history edited Jerry Leung CC BY-SA 3.0
added 4 characters in body
Dec 9, 2017 at 12:13 review Close votes
Dec 9, 2017 at 14:11
Dec 9, 2017 at 12:01 comment added Jerry Leung I wanna evaluate this integral
Dec 9, 2017 at 11:58 comment added Martin Sleziak As you can see, this search engine found this question which is about this specific integral: Convergence in probability (limit of integrals): $\lim_{n \to \infty} \int_0^1 \int_0^1 \cdots \int_0^1 \frac{x_1^2+x_2^2+ \cdots +x_n^2}{x_1+x_2+ \cdots +x_n} dx_1 dx_2 \cdots dx_n = \frac23$
Dec 9, 2017 at 11:58 comment added Martin Sleziak I have tried to search in Approach0 both for the whole integral and the fraction in the integral.
Dec 9, 2017 at 11:42 history edited Jerry Leung
edited tags
Dec 9, 2017 at 11:42 review First posts
Dec 9, 2017 at 11:44
Dec 9, 2017 at 11:36 history asked Jerry Leung CC BY-SA 3.0