Beta function and related applications, Lecture notes of Calculus

Definition of beta function and its usage in integral problems

Typology: Lecture notes

2023/2024

Uploaded on 04/09/2024

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Beta Fonksiyonu
Beta Fonksiyonu sa¼
daki ¸sekilde tan¬mlanabilir.
1) B(x; y) =
1
Z
0
tx1(1 t)y1dt
2) B(x; y) = 2
=2
Z
0
(sin )2x1(cos )2y1d
3) B(x; y) =
1
Z
0
ux1
(1 + u)x+ydu
4) B(x; y) = (x) (y)
(x+y)
5) B(x; 1x) = (x) (1 x) =
sin x
Örnek 1.
sa¼
daki integrallerin de¼
gerlerini hesaplay¬z.
a)
a
Z
0
x5paxdx b)
=6
Z
0q(sin 3)11 (cos 3)9d
Çözüm:
a) Bu integrali hesaplayabilmek için x=at dönü¸sümü yapal¬m. Buna göre,
a
Z
0
x5paxdx =a13=2
1
Z
0
t5(1 t)1
2dt
=a13=2B6;3
2
=a13=2 (6) 3
2
15
2
=a13=229
9009
bulunur.
1
pf3

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Beta Fonksiyonu

Beta Fonksiyonu a∏saºg¨daki ∏sekilde tan¨mlanabilir.

  1. B(x; y) =

Z^1

0

tx^1 (1 t)y^1 dt

  1. B(x; y) = 2

Z^ =^2

0

(sin )^2 x^1 (cos )^2 y^1 d

  1. B(x; y) =

Z^1

0

ux^1 (1 + u)x+y^

du

  1. B(x; y) = (x) (y) (x + y)

  2. B(x; 1 x) = (x) (1 x) =

sin x

÷rnek 1.

A∏saºg¨daki integrallerin deºgerlerini hesaplay¨n¨z.

a)

Z^ a

0

x^5

p a xdx b)

Z^ =^6

0

q (sin 3)^11 (cos 3)^9 d

«ˆz¸m:

a) Bu integrali hesaplayabilmek iÁin x = at dˆn¸∏s¸m¸ yapal¨m. Buna gˆre,

Z^ a

0

x^5

p a xdx = a^13 =^2

Z^1

0

t^5 (1 t)

(^12) dt

= a^13 =^2 B

= a^13 =^2

2

2

a^13 =^229 9009

bulunur.

b) 3  = t dˆn¸∏s¸m¸ yap¨l¨p, Beta fonksiyonunun 2) tan¨m¨kullan¨l¨rsa,

Z^ =^6

0

q (sin 3)^11 (cos 3)^9 d =

Z^ =^2

0

(sin t)

(^112) (cos t)

(^92) dt

B

elde edilir. 5) in kullan¨lmas¨yla

Z^ =^6

0

q (sin 3)^11 (cos 3)^9 d =

p 2 214

bulunur.

÷rnek 2.

n + (^12)

(2n)! p  22 n:n! (n = 0; 1 ; 2 :::) baºg¨nt¨s¨n¨n doºgruluºgunu gˆsterdikten sonra bu baºg¨nt¨-

dan yararlanarak a∏saºg¨daki e∏sitliklerin doºgruluºgunu gˆsteriniz.

a) B

n;

22 n: (n!)^2 n: (2n)! b) B

n +

(2n)! 22 n: (n!)^2

«ˆz¸m:

n +

n

n

n

n

n

n

n

n

(2n 1)(2n 3)::: 3 : 1 2 n

p 

= (2n)! p  22 n:n!

elde edilir.