International J. of Math. Sci. & Engg. Appls. (IJMSEA) ISSN 0973-9424, Vol. 11 No. III (December, 2017), pp. 123-127 A NOTE ON BETA FUNCTION AND LAPLACE TRANSFORM PRADIP B. GIRASE Department of Mathematics, K. K. M. College, Manwath, Dist : Parbhani (M.S.)-431505, India Abstract In this article, we use the Laplace Transform and Inverse Laplace Transform to prove the identities of beta function. It is possible that this technique of proof may be applied to solve the other problems involving beta function. 1. Introduction The beta function [1], also called the Euler integral of the first kind, is a special function R 1 p−1 q−1 defined by B(p; q) = 0 x (1 − x) dx ,for Re(p) > 0, Re(q) > 0. The gamma function [2], also called the Euler integral of the second kind, is defined as 1 convergent improper integral Γ (n) = R e−xxn−1dx for Re(n) > 0. 0 Properties of beta function: 1. B(p; q) = B(q; p) R 1 xp−1 2. B(p; q) = 0 (1+x)p+q dx, Re(p) > 0, Re(q) > 0 π R 2 2p−1 2q−1 3. B(p; q) = 2 0 sinθ cosθ dx, Re(p) > 0, Re(q) > 0 Key Words : Beta function, Gamma function, Laplace Transform, Convolution Theorem. 2010 AMS Subject Classification : 33B15, 44A10. c http: //www.ascent-journals.com University approved journal (Sl No. 48305) 123 124 PRADIP B. GIRASE Γ(p)Γ(q) The key property of beta function is B(p; q) = Γ(p+q) , which is proved by using Laplace transform by Charng-Yih Yu in [3]. Identities of beta and gamma function: 1. B(p; q) = B(p + 1; q) + B(p; q + 1), p; q > 0 q q 2. B(p; q + 1) = p B(p + 1; q) = p+q B(p; q), p; q > 0 3. Γ (p + 1) = pΓ(p), p > 0 The proof of identities of beta function requires the following definitions and results. Definition 1.1 [4] : If f(t) is defined for all positive values of t, then the Laplace R 1 −st transform of f(t) is defined as the integral, L[f(t)] = F (s) = 0 e f(t)dt, provided that integral exists for complex parameter s, and f(t) = L−1[F (s)] is called as the inverse Laplace transform of F (s). Lemma 1.2 [4] : Convolution Theorem: L−1[F (s)] = f(t), and L−1[G(s)] = g(t), then −1 R t L [F (s)G(s)] = 0 f(x)g(t − x)dx = f(t) ∗ g(t), where ∗ denotes the convolution of f(t) and g(t). R 1 p−1 q−1 Γ(p)Γ(q) Lemma 1.3 [3] : If p > 0, q > 0, then 0 (1 − x) x dx = Γ(p+q) . 2. Main Results Lemma 2.1 [4] : Relation between Gamma function and Laplace transform: If s > 0, n Γ(n+1) then L[t ] = sn+1 . n R 1 −st n Proof : Consider L[t ] = 0 e t dt x dx n put st = x, t = s and dt = s (new limits are x = 0 to x = 1 ), therefore L[t ] = R 1 −x x n dx 1 R 1 −x (n+1)−1 Γ(n+1) 0 e ( s ) s = sn+1 0 e x dx = sn+1 2 Theorem 2.2 : If p > 0, q > 0, then 1. B(p; q) = B(p + 1; q) + B(p; q + 1) q q 2. B(p; q + 1) = p B(p + 1; q) = p+q B(p; q) Proof : 1. ConsiderR:H:S: = B(p + 1; q) + B(p; q + 1) (1) Z 1 Z 1 = x(p+1)−1(1 − x)q−1dx + xp−1(1 − x)(q+1)−1dx (2) 0 0 A NOTE ON BETA FUNCTION AND LAPLACE TRANSFORM 125 y dy put x = t ,dx = t , in above integrals. (new limits are y=0, y=t) Z t y y dy Z t y y dy R:H:S: = ( )(p+1)−1(1 − )q−1 + ( )p−1(1 − )(q+1)−1 (3) 0 t t t 0 t t t Z t y(p+1)−1(t − y)q−1 Z t yp−1(t − y)(q+1)−1 = p+q dy + p+q dy (4) 0 t 0 t Z t Z t 1 (p+1)−1 q−1 1 p−1 (q+1)−1 = p+q y (t − y) dy + p+q y (t − y) dy (5) t 0 t 0 By Lemma 1.2 1 1 R:H:S: = L−1[L[t(p+1)−1]L[tq−1]] + L−1[L[tp−1]L[t(q+1)−1]] (6) tp+q tp+q By Lemma 2.1 1 Γ(p + 1) Γ(q) 1 Γ(p) Γ(q + 1) R:H:S: = L−1[ ] + L−1[ ] (7) tp+q sp+1 sq tp+q sp sq+1 1 1 1 1 = Γ(p + 1) Γ (q) L−1[ ] + Γ(p)Γ(q + 1) L−1[ ] (8) tp+q sp+q+1 tp+q sp+q+1 −1 1 tn−1 Since L [ sn ] = Γ(n) , 1 tp+q 1 tp+q R:H:S: = Γ(p + 1) Γ (q) + Γ(p)Γ(q + 1) (9) tp+q Γ(p + q + 1) tp+q Γ(p + q + 1) Γ(p + 1) Γ (q) + Γ (p)Γ(q + 1) = (10) Γ(p + q + 1) By identity Γ (p + 1) = pΓ(p), (p + q)Γ (p)Γ(q) R:H:S: = (11) (p + q)Γ (p + q) Γ(p)Γ(q) = (12) Γ(p + q) = B(p; q) (13) 2. Z 1 ConsiderB(p; q + 1) = xp−1(1 − x)(q+1)−1dx (14) 0 126 PRADIP B. GIRASE y dy put x = t ,dx = t , in above integral. (new limits are y=0, y=t) Z t y y dy B(p; q + 1) = ( )p−1(1 − )(q+1)−1 (15) 0 t t t Z t yp−1(t − y)(q+1)−1 = p+q dy (16) 0 t Z t 1 p−1 (q+1)−1 = p+q y (t − y) dy (17) t 0 By Lemma 1.2 1 B(p; q + 1) = L−1[L[tp−1]L[t(q+1)−1]] (18) tp+q By Lemma 2.1 1 Γ(p) Γ(q + 1) B(p; q + 1) = L−1[ ] (19) tp+q sp sq+1 1 1 = Γ(p)Γ(q + 1) L−1[ ] (20) tp+q sp+q+1 −1 1 tn−1 Since L [ sn ] = Γ(n) , 1 tp+q B(p; q + 1) = Γ(p)Γ(q + 1) (21) tp+q Γ(p + q + 1) qΓ(p)Γ(q) = (22) (p + q)Γ (p + q) q = B(p; q) (23) (p + q) Γ(p)Γ(q + 1) Also; B(p; q + 1) = (24) Γ(p + q + 1) qpΓ(p)Γ(q) = (25) pΓ(p + q + 1) qΓ(p + 1) Γ (q) = (26) pΓ(p + 1 + q) q = B(p + 1; q) (27) p 2 A NOTE ON BETA FUNCTION AND LAPLACE TRANSFORM 127 3. Conclusion The Laplace transform technique can be used to solve problems involving Beta function instead of probability theory. References [1] http://en.wikipedia.org/wiki/Beta function. [2] http://en.wikipedia.org/wiki/Gamma function. [3] Charng-Yih Yu, A different proof of the beta function, Mathematica Aeterna, 4(7) (2014),737-740. [4] Grewal B. S. and Grewal J. S., Higher Engineering Mathematics, 40th Edition, Khanna Publishers, New Delhi (2007)..