Algorithmic Proofs of Two Curious Integral Identities of George Gasper and Michael Schlosser By Shalosh B. Ekhad Gasper and Schlosser discovered and proved (and another human proof was given by Mizan Rahman) See: "Some curious q-series expansions and beta integral evaluations", Ramanujan J. 13 (1-3) (2007), 229-242 https://www.mat.univie.ac.at/~schlosse/GScbeta.html the following two elegant identities 1 / | 2 b (b - 1) | 2 GAMMA(2 b) (c - (a + 1) ) (c - a (a + t)) (c - (a + 1) (a + t)) | / 0 b (b - 1) / 2 2 (2 b) t (1 - t) / (GAMMA(b) (c - (a + t) ) ) dt = 1 / 1 / | 2 (b - 1) | GAMMA(2 b) (c - (a + 1) ) (c - a (a + t)) | / 0 (b - 1) (b - 1) (b - 1) (c - (a + 1) (a + t)) t (1 - t) (c - (a - t) (a + t)) / 2 2 (2 b) / (GAMMA(b) (c - (a + t) ) ) dt = 1 / Let's call these two integrals G1(b) and G2(b) respectively. Using the amazing Almkvist-Zeilberger algorithm , implemented in procedure AZd, in Doron Zeilberger's Maple package EKHAD.txt available from http://sites.math.rutgers.edu/~zeilberg/tokhniot/EKHAD.txt it is proved rigorously that G1(b) and G2(b) satisfy the following THIRD-ORDER recurrences 2 2 2 (2 b + 1) (a + 1) (a + a - c) 3 3 2 2 (2 a b + 3 a + 3 a b + 2 a b c + 5 a + a b + 3 a c + b c + 2 a + c) a 9 2 9 8 2 7 2 9 8 G1(b) + (-8 a b - 16 a b - 36 a b + 8 a b c - 6 a - 74 a b 7 2 7 6 2 5 2 2 8 7 - 62 a b + 16 a b c + 28 a b c + 8 a b c - 28 a - 131 a b 7 6 2 6 5 2 5 2 4 2 2 + 6 a c - 49 a b + 62 a b c + 26 a b c + 16 a b c + 20 a b c 3 2 3 7 6 6 5 2 5 5 2 - 8 a b c - 50 a - 106 a b + 24 a c - 15 a b + 69 a b c + 6 a c 4 2 4 2 3 2 2 3 3 2 2 3 6 - 5 a b c + 34 a b c + 6 a b c - 16 a b c - 12 a b c - 40 a 5 5 4 2 4 4 2 3 2 3 2 - 32 a b + 28 a c + a b + 2 a b c + 12 a c - 16 a b c - a b c 3 3 2 2 2 2 3 2 3 5 4 3 2 - 6 a c - 11 a b c - 22 a b c - 2 a b c - 10 a + 4 a b + a b 3 3 2 2 2 2 2 2 3 2 2 - 35 a b c - 2 a c - 5 a b c - 26 a b c - 8 a c - 5 a b c 3 2 3 4 3 3 2 2 2 - a b c + b c + 4 a + 3 a b - 18 a c - 14 a b c - 8 a c 2 3 3 2 7 2 7 - 7 a b c + 2 b c + 2 a - 8 a c) G1(b + 1) + (-2 a b - 5 a b 6 2 5 2 7 6 5 2 5 4 2 - 7 a b + 10 a b c - 2 a - 18 a b - 9 a b + 21 a b c + 25 a b c 3 2 2 6 5 5 4 2 4 3 2 + 10 a b c - 8 a - 24 a b + 8 a c - 5 a b + 56 a b c + 18 a b c 3 2 2 2 2 2 3 5 4 4 3 2 + 21 a b c + 15 a b c - 2 a b c - 12 a - 14 a b + 24 a c - a b 3 3 2 2 2 2 2 2 2 3 + 44 a b c + 8 a c + 2 a b c + 28 a b c + 3 a b c - 5 a b c 2 3 4 3 3 2 2 2 2 2 - b c - 8 a - 3 a b + 22 a c + 6 a b c + 8 a c - a b c + 2 a b c 3 2 2 3 3 2 2 2 - 2 a c - b c - 2 b c - 2 a + 4 a c - 3 a b c - 2 a c - 2 b c - 2 a c) G1(b + 2) + (b + 2) 3 3 2 2 (2 a b + a + 3 a b + 2 a b c + 2 a + a b + a c + b c + a) c G1(b + 3) = 0 3 3 2 2 2 (2 a b + 2 a + 3 a b + 2 a b c + 4 a + a b + 4 a c + b c + 2 a + 2 c) a 2 2 9 2 9 8 2 (a + a - c) (2 b + 1) (a + 1) G2(b) + (-8 a b - 12 a b - 36 a b 7 2 9 8 7 2 7 6 2 + 8 a b c - 4 a - 58 a b - 62 a b + 4 a b c + 28 a b c 5 2 2 8 7 6 2 6 5 2 + 8 a b c - 20 a - 108 a b - 49 a b + 22 a b c + 26 a b c 5 2 4 2 2 3 2 3 7 6 6 + 28 a b c + 20 a b c - 8 a b c - 40 a - 93 a b + 4 a c 5 2 5 5 2 4 2 4 2 3 2 2 - 15 a b + 28 a b c + 12 a c - 5 a b c + 66 a b c + 6 a b c 3 3 2 2 3 6 5 5 4 2 4 - 20 a b c - 12 a b c - 38 a - 31 a b + 12 a c + a b - 7 a b c 4 2 3 2 3 2 3 3 2 2 2 2 3 + 28 a c - 16 a b c + 20 a b c - 8 a c - 11 a b c - 30 a b c 2 3 5 4 4 3 2 3 3 2 - 2 a b c - 14 a + 3 a b + 6 a c + a b - 30 a b c + 4 a c 2 2 2 2 2 3 2 2 3 2 3 4 - 5 a b c - 31 a b c - 12 a c - 5 a b c - 4 a b c + b c + 2 a 3 3 2 2 2 2 3 3 + 3 a b - 8 a c - 13 a b c - 18 a c - 13 a b c + 3 b c + 2 a 2 2 3 7 2 7 6 2 - 6 a c - 6 a c + 2 c ) G2(b + 1) + (-2 a b - 4 a b - 7 a b 5 2 6 5 2 5 4 2 3 2 2 + 10 a b c - 15 a b - 9 a b + 14 a b c + 25 a b c + 10 a b c 6 5 4 2 4 3 2 3 2 - 2 a - 21 a b - 5 a b + 41 a b c + 18 a b c + 28 a b c 2 2 2 2 3 5 4 4 3 2 3 + 15 a b c - 2 a b c - 6 a - 13 a b + 6 a c - a b + 36 a b c 3 2 2 2 2 2 2 2 3 2 3 4 + 16 a c + 2 a b c + 41 a b c + 3 a b c - 6 a b c - b c - 6 a 3 3 2 2 2 2 2 3 - 3 a b + 12 a c + 6 a b c + 22 a c - a b c + 7 a b c - 4 a c 2 2 3 3 2 2 2 3 - b c - 3 b c - 2 a + 4 a c - 3 a b c + 2 a c - 3 b c - 2 c 2 - 2 a c - 2 c ) G2(b + 2) + 3 2 2 (2 a b + 3 a b + 2 a b c + a + a b + 2 a c + b c + a + c) (b + 2) c G2(b + 3) = 0 The proof certificates are omitted here, but can be gotten by the reader by typing AZd(2*GAMMA(2*b)/GAMMA(b)^2*(c-a(a+1)^2)*(c-a*(a+t))^b*(c-(a+1)*(a+t))^(b-1)/(c-(a+t)^2)^(2*b)*t^b*(1-t)^(b-1),t,b,B); AZd(GAMMA(2*b)/GAMMA(b)^2*(c-(a+1)^2)*(c-a*(a+t))^(b-1)*(c-(a+1)*(a+t))^(b-1)/(c-(a+t)^2)^(2*b)*t^(b-1)*(1-t)^(b-1)*(c-(a-t)*(a+t)),t,b,B); in a Maple session where EKHAD.txt has been read to. On the other hand, the discrete function that is IDENTICALLY 1 also satisfies these recurrences. Indeed, replacing G1(b) and G2(b) with 1 and simplifying the algebra, we get that the left sides equal, respectively 0 0 It remains to prove the identities for b=0,b=1,b=2, but these are routine Calculus 1 integrals that are left to the readers. This proves the Gasper-Schlosser identities for all NON-NEGATIVE integers, b. By the usual "analytic continuation" and/or Carlson's theorem, this is true for all b where the integrals makes sense. ---------------------------------- This ends this article that took, 6.267, seconds to generate