Timeline for A relation between a binomial sum and a trigonometric integral

Current License: CC BY-SA 3.0

14 events

when toggle format	what		by	license	comment
Apr 13, 2017 at 12:58	history	edited	CommunityBot		replaced http://mathoverflow.net/ with https://mathoverflow.net/
Nov 26, 2016 at 4:00	history	edited	T. Amdeberhan	CC BY-SA 3.0	added 7 characters in body
Nov 26, 2016 at 3:52	history	edited	T. Amdeberhan	CC BY-SA 3.0	added 512 characters in body
Nov 25, 2016 at 20:14	comment	added	Ira Gessel		What I had in mind is $$ \begin{aligned} \sum_{k=0}^n(-1)^k \frac{a}{k+a} \binom nk &= \sum_{k=0}^n (-1)^k a\binom nk\int_0^1 t^{k+a-1}\,dt\\ &=\int_0^1 a t^{a-1} \sum_{k=0}^n (-1)^k \binom nk t^k\,dt\\ &=a\int_0^1 t^{a-1}(1-t)^n\, dt\\ &=aB(a,n+1). \end{aligned} $$
Nov 25, 2016 at 18:25	history	edited	T. Amdeberhan	CC BY-SA 3.0	added 134 characters in body
Nov 25, 2016 at 18:01	history	edited	T. Amdeberhan	CC BY-SA 3.0	added 2 characters in body
Nov 25, 2016 at 17:56	history	edited	T. Amdeberhan	CC BY-SA 3.0	added 2 characters in body
Nov 25, 2016 at 17:50	history	edited	T. Amdeberhan	CC BY-SA 3.0	added 198 characters in body
Nov 25, 2016 at 17:48	comment	added	T. Amdeberhan		Yes! That's perhaps what Gessel had in mind from the above comment.
Nov 25, 2016 at 17:46	comment	added	Pietro Majer		You can also derive it directly from the first line: changing variable linearly $\int_{-1}^{1}(1-u^2)^n du=2^{2n+1}\int_{0}^{1}t^n(1-t)^ndt= 2^{2n+1} B(n+1,n+1)$
Nov 25, 2016 at 17:39	history	edited	T. Amdeberhan	CC BY-SA 3.0	added 787 characters in body
Nov 25, 2016 at 17:15	comment	added	W-t-P		Great! The only question remaining, where does the closed form come from?
Nov 25, 2016 at 17:14	vote	accept	W-t-P
Nov 25, 2016 at 17:03	history	answered	T. Amdeberhan	CC BY-SA 3.0