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Definition of beta function and its usage in integral problems
Typology: Lecture notes
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Beta Fonksiyonu
Beta Fonksiyonu a∏saºg¨daki ∏sekilde tan¨mlanabilir.
0
tx ^1 (1 t)y ^1 dt
0
(sin )^2 x ^1 (cos )^2 y ^1 d
0
ux ^1 (1 + u)x+y^
du
B(x; y) = (x) (y) (x + y)
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)
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
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 =
0
(sin t)
(^112) (cos t)
(^92) dt
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.