APPLICATIONS OF WALLIS THEOREM Mihály Bencze1, Florentin Smarandache2 1 Department of Mathematics, Áprily Lajos College, Braşov, Romania; Chair of Department of Math & Sciences, University of New Mexico, Gallup, NM 87301, USA 2 Abstract: In this paper we present theorems and applications of Wallis theorem related to trigonometric integrals. Let’s recall Wallis Theorem: Theorem 1. (Wallis, 1616-1703) π π 2 2 2n +1 2n+1 ∫ sin xdx = ∫ cos xdx = 2 ⋅ 4 ⋅ ...⋅ (2n) . 1⋅ 3⋅ ...⋅ (2n + 1) 0 0 Proof: Using the integration by parts, we obtain π I n = ∫ sin 2 0 π 2n +1 π xdx = ∫ sin x sin xdx = − cos x ⋅ sin 2nx 2 2n ( 0 2 0 + ) +2n ∫ sin 2n +1 x 1 − sin 2 x dx = 2nI n−1 − 2nI n 2 0 from where: π In = 2n I n −1 . 2n + 1 By multiplication, we obtain the statement. We prove in the same manner for cos x . Theorem 2. π π 2 2 0 0 2n 2n ∫ sin xdx = ∫ cos xdx = 1 ⋅ 3 ⋅ ... ⋅ (2n − 1) π ⋅ . 2 ⋅ 4 ⋅ ... ⋅ (2n) 2 Proof: Same as the first theorem. Theorem 3. If f (x) = ∑ a2 k x 2 k , then ∞ π ∫ 2 0 k=0 π f (sin x)dx = ∫ f (cos x)dx = 2 0 1 π 2 a0 + ∑a 2 π ∞ k =1 2k 1 ⋅ 3 ⋅ ... ⋅ (2k − 1) . 2 ⋅ 4 ⋅ ... ⋅ (2k) Proof: In the function f (x) = ∑ a2 k x 2 k we substitute x by sin x and then ∞ integrate from 0 to π k=0 , and we use the second theorem. Theorem 4. If g(x) = ∑ a2 k +1 x 2 k +1 , then 2 ∞ k=0 π π ∫ g(sin x)dx = ∫ g(cos x)dx =a1 + ∑ a2 k +1 2 ∞ 2 Theorem 5. If h(x) = ∑ ak x k , then 0 π π 2 2 k=0 ∫ h(sin x)dx = ∫ h(cos x)dx = 2 ⋅ 4 ⋅ ... ⋅ (2 k ) ⎞ . 1 ⋅ 3 ⋅ ... ⋅ (2 k + 1) ⎟⎠ Application 1. 0 0 + a2 k +1 π π 2 2 k =1 0 ∞ ∞ 1 ⋅ 3 ⋅ ... ⋅ (2k − 1) ⎛π a0 +a1 + ∑ ⎜ a2 k + ⎝ 2 2 ⋅ 4 ⋅ ... ⋅ (2k) k =1 2 π ∫ sin(sin x )dx = ∫ sin(cos x )dx =∑ (−1) ∞ k k=0 ∞ 2 k π π 2 2 0 0 ∫ cos(sin x )dx = ∫ cos(cos x )dx = Proof: We use that cos x = ∑ (−1)k ∞ k=0 Application 3. 1 1 ⋅ 3 ⋅ ... ⋅ (2 k + 1)2 2 x 2 k +1 Proof: We use that sin x = ∑ (−1) . (2k + 1)! k=0 Application 2. 0 0 (−1)k . ∑ 2 k = 0 4 k ( k !)2 π ∞ x2k . (2k)! π π 2 2 ∞ 0 k=0 ∫ sh(sin x )dx = ∫ sh(cos x )dx =∑ 1 0 2 ⋅ 4 ⋅ ... ⋅ (2k) . 1 ⋅ 3 ⋅ ... ⋅ (2k + 1) Proof: We use that shx = ∑ x 2 k +1 k = 0 (2k + 1)! ∞ Application 4. 2 2 1 . ⋅ 3 ⋅ ... ⋅ (2 k + 1)2 2 π π 2 2 ∫ ch(sin x )dx = ∫ ch(cos x )dx = Proof: We use that chx = ∑ 0 0 ∞ ∑4 2 π ∞ k=0 k 1 . ( k !)2 2k x . k = 0 (2k)! Application 5. 1 π2 ∑ k2 6 k =1 ∞ Proof: In the expression of arcsin x = x + ∑ 1 ⋅ 3 ⋅ ... ⋅ (2k − 1) x 2 k +1 we substitute x k =1 2 ⋅ 4 ⋅ ... ⋅ (2 k )(2k + 1) π2 ∞ 1 =∑ . by sin x , and use theorem 4. It results that 8 k = 0 (2k + 1)2 Because: ∞ 1 1 ∞ 1 1 ∞ = + ∑ k 2 ∑ (2k + 1)2 4 ∑ k 2 k =1 k=0 k =1 we obtain: ∞ 1 π ∑ k2 = 6 . k =1 Application 6. π π 2 2 0 0 ∞ ∫ sin x ctg( sin x)dx = ∫ cos x ctg( cos x)dx = π 2 − ∑ (k !) 2 π ∞ k =1 Bk 2 Proof: We use that xctgx = 1 − ∑ where Bk is the k-th Bernoulli type number (see [1]). ∞ 4 k Bk 2 k x . k =1 (2k)! Application 7. π π 2 2 ∫ arctg(sin x)dx = ∫ arctg(cos x)dx = 1 +∑ (−1) 1 ⋅ 3 ⋅ ... ⋅ (2k − 1)(2k + 1) ∞ x 2 k +1 Proof: We use that arctgx = ∑ (−1) . 2k + 1 k=0 Application 8. 0 0 π π 2 2 ∞ k =1 2 ⋅ 4 ⋅ ... ⋅ (2k ) . 2 k =1 1 ⋅ 3 ⋅ ... ⋅ (2k − 1)(2k + 1) ∞ Proof: We use that arg th x = ∑ 0 2 ⋅ 4 ⋅ ... ⋅ (2k) k ∫ arg th(sin x)dx = ∫ arg th(cos x)dx = 1 +∑ 0 k x 2 k +1 . k = 0 2k + 1 ∞ 3 2 . Application 9. π π 2 2 ∫ arg sh(sin x)dx = ∫ arg sh( cos x)dx = ∑ (−1) ∞ k k =1 1 . (2k + 1)2 1⋅ 3 ⋅ ...⋅ (2k − 1)x 2 k +1 Proof: We use that arg shx = ∑ (−1) . 2 ⋅ 4 ⋅ ...⋅ (2k)(2k + 1) k=0 0 0 Application 10. π π 2 2 ∞ k 22 k −1 (4k − 1) Bk . 2 2 2 ∫0 tg( sin x)dx = ∫0 tg( cos x)dx = ∑ k =1 1 ⋅ 3 ⋅ ... ⋅ (2k − 1) k ∞ Proof: We use that tg x = ∑ 2 2 k (4 k − 1)Bk 2 k −1 x . (2k)! k =1 ∞ Application 11. π ( π ) 2 ∞ 2 2 k −1 − 1 B cos x π sin x k 2k 2 ∫0 sin(sin x) dx = ∫0 sin(cos x) dx = 2 + π ∑ 2 (k!) k =1 2 ( ) ∞ 2 2 k −1 − 1 B x k 2k Proof: We use that = 1 + 2∑ x . (2k)! sin x k =1 Application 12. π ( π ) 2 ∞ 2 2 k −1 − 1 B cos x π sin x k . 2k 2 ∫0 sh(sin x) dx = ∫0 sh(cos x) dx = 2 + π ∑ 2 (k!) k =1 2 ( ) ∞ 2 2 k −1 − 1 Bk 2 k x k Proof: We use that = 1 + 2∑ (−1) x . shx (2k)! k =1 Application 13. π π 2 2 ∫ sec(sin x)dx = ∫ sec(cos x)dx = + π∑ π ∞ Ek , (k!)2 k =1 2 2 where Ek is the k-th Euler type number (see [1]). 0 0 2 k +1 Proof: We use that sec x = 1 + ∑ ∞ Ek 2 k x k =1 (2k)! Application 14. π π 2 2 ∫ sec h(sin x )dx = ∫ sec h(cos x )dx = Proof: We use that sec h x = 1 + ∑ (−1)k 0 0 ∞ k =1 4 π 2 + π ∑ (−1)k ∞ k =1 Ek 2 k x . (2k)! Ek . 2 ( k !)2 2 k +1 REFERENCES [1] [2] Octav Mayer – Theoria funcţiilor de o variabilă complexă – Ed. Academiei, Bucharest, 1981. Mihály Bencze – About Taylor formula – (manuscript). [Published in “Octogon”, Vol. 6, No. 2, 117-120, 1998.] 5