Revision as of 02:09, 6 December 2019 edit Contribute.Math (talk \| contribs) 53 edits I precised that Wallis Integrals are used in the proof ← Previous edit		Revision as of 02:10, 6 December 2019 edit undo Contribute.Math (talk \| contribs) 53 edits I precised that the proof is rigorous Next edit →
Line 300: and this completes the proof. ==~~Another~~A rigorous proof using Wallis' integrals== We know that <math> \forall \ t\ \in \left[ 0,\frac{\pi }{2}\right] \ \ sin( t) \ \in [ 0,1] \ \ </math>

Basel problem: Difference between revisions