Timeline for Integral $\int_0^1 \int_0^1 \cdots \int_0^1\frac{x_{1}^2+x_{2}^2+\cdots+x_{n}^2}{x_{1}+x_{2}+\cdots+x_{n}}dx_{1}\, dx_{2}\cdots \, dx_{n}=?$
Current License: CC BY-SA 3.0
11 events
when toggle format | what | by | license | comment | |
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Feb 17, 2018 at 18:35 | vote | accept | Jerry Leung | ||
Jan 16, 2018 at 19:29 | history | edited | esg | CC BY-SA 3.0 |
more typos corrected
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Jan 16, 2018 at 7:04 | comment | added | Fedor Petrov | @esg my method actually also allows to get an answer, I simply missed something obvious (that there are no off-integral terms in integrating by parts here), see my updated answer | |
Dec 24, 2017 at 6:54 | comment | added | Shahrooz | nice solution @esg. | |
Dec 24, 2017 at 0:20 | vote | accept | Jerry Leung | ||
Feb 17, 2018 at 18:35 | |||||
Dec 23, 2017 at 18:37 | history | edited | esg | CC BY-SA 3.0 |
typos corrected
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Dec 23, 2017 at 18:33 | comment | added | esg | Thank you. (1) I also like your method (but have been unable to extend it to cover the rational term) (2) The rational integral: you're right, that's the most natural way to do it. In the present context I was probably led to explain it via divided differences because pdf/cdf of sums of iid uniform variates arise via a differencing process. | |
Dec 23, 2017 at 11:12 | comment | added | Fedor Petrov | For integrating a rational function like $f(z)=z^{n-2}/\prod_{k=1}^n(z+k)$ against $[0,+\infty)$ we may simply expand it via elementary fractions $f(z)=\sum c_k/(z+k)$, where $c_k={\rm Res}_{z=-k}f(z)=(f(z)(z+k))|_{z=-k}=k^{n-2}(-1)^{n-k}/k!(n-k)!$. Now $\sum c_k=0$ (it is clear a priori from the asymptotics of $f$ for large $z$: $\sum c_k=\lim_{z\to \infty} zf(z)=0$); and we get $\int_0^\infty f=\int_0^\infty \sum c_k (1/(z+k)-1/z)=-\sum c_k\log k$. | |
Dec 22, 2017 at 21:12 | comment | added | Fedor Petrov | Very nice method! | |
Dec 22, 2017 at 17:30 | history | edited | esg | CC BY-SA 3.0 |
deleted 1 character in body
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Dec 22, 2017 at 16:59 | history | answered | esg | CC BY-SA 3.0 |