Skip to main content
11 events
when toggle format what by license comment
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
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
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
Dec 22, 2017 at 16:59 history answered esg CC BY-SA 3.0