This is a special case of the more general formula: integral_0^1 x^{-tx} dx = 1/t sum_{k=1}^{infinity} (t/n)^n I don't know that this really addresses your question, but it might be relevant. Franklin T. Adams-Watters -----Original Message----- From: dasimov@earthlink.net To: math-fun <math-fun@mailman.xmission.com> Sent: Thu, 13 Apr 2006 13:02:04 -0400 (GMT-04:00) Subject: [math-fun] Mysterious but well-known identity This following identity has been floating around for years: integral_0^1 1/x^x dx = sum{k=1,oo} 1/k^k It's reasonably straightforward to prove, but the usual proof (which I'll leave as an exercise for anyone who hasn't seen it), though short, is not especially enlightening. Is there some "explanation" for the form of this identity (*) integral_0^1 f(x) dx = sum{k=1,oo} f(k) (where f is analytic), or is this merely a mathematical "coincidence" (which if it had no proof, might not have a proof) ? --Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ___________________________________________________ Try the New Netscape Mail Today! Virtually Spam-Free | More Storage | Import Your Contact List http://mail.netscape.com