Applications Wallis

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 ⋅ 4 ⋅ ...⋅ (2n) . 1⋅ 3⋅ ...⋅ (2n + 1) 0 0 Proof: Using the integration by parts, we obtain 2

2

2n +1 2n+1 ∫ sin xdx = ∫ cos xdx =

π

π

2

I n = ∫ sin

2n +1

π

2

xdx = ∫ sin x sin xdx = − cos x ⋅ sin 2nx 2n

0

2

0

+

0

π 2

(

)

+2n ∫ sin 2n +1 x 1 − sin 2 x dx = 2nI n−1 − 2nI n 0

from where:

2n I n −1 . 2n + 1 By multiplication, we obtain the statement. We prove in the same manner for cos x . In =

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 k=0

π 2

∫ 0

π 2

f (sin x)dx = ∫ f (cos x)dx = 0

1

π 2

a0 +

π

∞

∑a 2

2k

k =1

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 2

k=0

, and we use the second theorem. ∞

Theorem 4. If g(x) = ∑ a2 k +1 x 2 k +1 , then k=0

π

π

2

2

∞

∫ g(sin x)dx = ∫ g(cos x)dx =a + ∑ a

2 k +1

1

0

k =1

0

2 ⋅ 4 ⋅ ... ⋅ (2k) . 1 ⋅ 3 ⋅ ... ⋅ (2k + 1)

∞

Theorem 5. If h(x) = ∑ ak x k , then k=0

π

π

2

2

0

0

∫ h(sin x)dx = ∫ h(cos x)dx =

π

∞ 1 ⋅ 3 ⋅ ... ⋅ (2k − 1) ⎛π a0 +a1 + ∑ ⎜ a2 k + ⎝ 2 2 ⋅ 4 ⋅ ... ⋅ (2k) k =1 2

2 ⋅ 4 ⋅ ... ⋅ (2 k ) ⎞ . 1 ⋅ 3 ⋅ ... ⋅ (2 k + 1) ⎟⎠ Application 1. + a2 k +1 π

π

2

2

∞

∫ sin(sin x )dx = ∫ sin(cos x )dx =∑ (−1) 0

k

k=0

0

1 1 ⋅ 3 ⋅ ... ⋅ (2 k + 1)2 2

2

x 2 k +1 Proof: We use that sin x = ∑ (−1) . (2k + 1)! k=0 Application 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

π

(−1)k . ∑ 2 k = 0 4 k ( k !)2 ∞

x2k . (2k)!

Application 3. π

π

2

2

∞

∫ sh(sin x )dx = ∫ sh(cos x )dx =∑ 1 0

k=0

0

x 2 k +1 k = 0 (2k + 1)! ∞

Proof: We use that shx = ∑ Application 4.

2

2

1 . ⋅ 3 ⋅ ... ⋅ (2 k + 1)2 2

π

π

2

2

∫ ch(sin x )dx = ∫ ch(cos x )dx = 0

0 ∞

π

∞

∑4 2 k=0

k

1 . ( k !)2

2k

x . k = 0 (2k)!

Proof: We use that chx = ∑

Application 5. 1 π2 ∑ k2 6 k =1 ∞

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. ∞

Proof: In the expression of arcsin x = x + ∑

π

π

2

2

0

0

∫ sin x ctg( sin x)dx = ∫ cos x ctg( cos x)dx =

π 2

−

π

∞

Bk

∑ (k !) 2 k =1

2

where Bk is the k-th Bernoulli type number (see [1]). ∞

4 k Bk 2 k x . k =1 (2k)!

Proof: We use that xctgx = 1 − ∑ Application 7. π

π

2

2

∞

2 ⋅ 4 ⋅ ... ⋅ (2k)

∫ arctg(sin x)dx = ∫ arctg(cos x)dx = 1 +∑ (−1) 1 ⋅ 3 ⋅ ... ⋅ (2k − 1)(2k + 1) 0

k

k =1

0

x 2 k +1 Proof: We use that arctgx = ∑ (−1) . 2k + 1 k=0 Application 8. ∞

k

π

π

2

2

0

0

2 ⋅ 4 ⋅ ... ⋅ (2k ) . 2 k =1 1 ⋅ 3 ⋅ ... ⋅ (2k − 1)(2k + 1) ∞

∫ arg th(sin x)dx = ∫ arg th(cos x)dx = 1 +∑ x 2 k +1 . k = 0 2k + 1 ∞

Proof: We use that arg th x = ∑

3

2

.

Application 9. π

π

2

2

∞

∫ arg sh(sin x)dx = ∫ arg sh( cos x)dx = ∑ (−1) 0

k

k =1

0

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 ∞

k

Application 10. π

π

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 2

∞

2

2 2 k (4 k − 1)Bk 2 k −1 x . (2k)! k =1 ∞

Proof: We use that tg x = ∑ Application 11. π

π

(

)

2 ∞ 2 2 k −1 − 1 B sin x cos x π 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 = 1 + 2∑ x . Proof: We use that (2k)! sin x k =1

Application 12. π

π

(

)

2 ∞ 2 2 k −1 − 1 B sin x cos x π 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

∞

Ek , (k!)2 k =1 2

+ π∑

2 where Ek is the k-th Euler type number (see [1]). 0

0

2 k +1

∞

Ek 2 k x k =1 (2k)!

Proof: We use that sec x = 1 + ∑ Application 14. π

π

2

2

∫ sec h(sin x )dx = ∫ sec h(cos x )dx = 0

0 ∞

Proof: We use that sec h x = 1 + ∑ (−1)k 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.]

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