beta and gama
داﻟﺘﻲ ﺟﺎﻣﺎ وﺑﯿﺘﺎ
ﺗﻌﺮﯾﻒ داﻟﺘﻲ ﺟﺎﻣﺎ وﺑﯿﺘﺎ
ﺗﻌﺮف داﻟﺔ ﺟﺎﻣﺎ ﻛﺎﻵﺗﻲ:
∞
Γ(n) = ∫ e−t t n−1 dt
)(n > 0
)(1
0
وﻧﻼﺣﻆ أن اﻟﺪاﻟﺔ e − t t n−1ﻋﻨﺪ ∞ ﺗﺆول إﻟ ﻰ اﻟﺼ ﻔﺮ ﻷي ﻗﯿﻤ ﺔ ﻋﺪدﯾ ﺔ ﻟ ـ nوﻋﻠﯿ ﮫ ﻻﺗﻜ ﻮن ھﻨ ﺎك ﺻ ﻌﻮﺑﺎت
ﻋﻠﻤﯿ ﺔ ﻋﻨ ﺪ ∞ → tﺑﯿﻨﻤ ﺎ ﻓ ﻲ اﻟﺠ ﺰء اﻟﺴ ﻔﻠﻲ ﺗﻮﺟ ﺪ t → 0ﻧﺠ ﺪ أن e −t ≅ 1وﻋﻠﯿ ﮫ ﺗﺼ ﺒﺢ اﻟﻤﻌﺎدﻟ ﺔ ) (1ﻋﻠ ﻰ
اﻟﺼﻮرة
∞
∞
c
1
Γ(n) = ∫ t dt + ∫ e t dt = t n + ∫ e −t t n−1dt
n 0 c
0
c
−t n −1
c
n−1
وﻋﻠﯿﮫ إذا ﻛﺎن اﻟﺠﺰء اﻷول ﻣﻦ اﻟﻨﺎﺗﺞ ﻋﺪد ﻓﯿﺠﺐ ﻋﻠﯿﻨﺎ اﺧﺘﯿﺎر n > 0ﺣﺘﻰ ﻻ ﯾﻨﻌﺪم اﻟﻤﻘﺎم .
ﺑﻌﺾ اﻟﻨﻈﺮﯾﺎت واﻟﺨﻮاص اﻟﮭﺎﻣﺔ
ﻧﻈﺮﯾﺔ :١
Γ (1) = 1
اﻟﺒﺮھﺎن :
ﻣﻦ ﺗﻌﺮﯾﻒ ) Γ (nﻧﻀﻊ n = 1ﻓﻨﺤﺼﻞ ﻋﻠﻰ
]
∞
[
∞
0
Γ(1) = ∫ e −t dt = − e −t
0
Γ (1) = 1
)(3
ﻧﻈﺮﯾﺔ :٢
)Γ (n + 1) = nΓ(n
اﻟﺒﺮھﺎن :
ﻓﻲ اﻟﻤﻌﺎدﻟﺔ ) (1ﺑﺘﺤﺮﯾﻚ ﻛﻞ n → n + 1ﻧﺤﺼﻞ ﻋﻠﻰ
∞
Γ( n + 1) = ∫ e −t t n dt
0
ﺑﺎﻟﺘﻜﺎﻣﻞ ﺑﺎﻟﺘﺠﺰيء ﺑﺄﺧﺬ u = t n
−t
e dr = dv,ﻧﺤﺼﻞ ﻋﻠﻰ
) n t n −1dt
−t
∞
] − ∫ (−e
∞ −t n
0
[
Γ ( n + 1) = − e t
0
ﻛﻤﺎ ﻧﻼﺣﻆ اﻧﻌﺪام اﻟﺠﺰء اﻷول ﻣﻦ اﻟﻨﺘﯿﺠﺔ اﻟﺴﺎﺑﻘﺔ ﻓﻨﺤﺼﻞ ﻋﻠﻰ
∞
Γ( n + 1) = n ∫ e −t t n−1 dt
0
)(4
) Γ ( n + 1) = n Γ ( n
١
ﻣﻼﺣﻈﺎت :
(١اﻟﻌﻼﻗﺔ ) Γ (n + 1) = n Γ( nﺗﺴﻤﻰ ﺑﺎﻟﺼﯿﻐﺔ اﻟﺘﻜﺮارﯾﺔ ﻟﺪاﻟﺔ ﺟﺎﻣﺎ .
(٢ﯾﻤﻜﻦ ﺗﻌﻤﯿﻢ داﻟﺔ ﺟﺎﻣﺎ ﻟﻘﯿﻢ ) n < 0ﻟﻸﻋﺪاد اﻟﺤﻘﯿﻘﯿﺔ ﻏﯿﺮ اﻟﺼﺤﯿﺤﺔ ( ﺑﺎﺳﺘﺨﺪام اﻟﺼﯿﻐﺔ اﻟﺴﺎﺑﻘﺔ ﻋﻠﻰ
اﻟﺼﻮرة :
)Γ (n + 1
= )Γ (n
n
ﺗﻌﺮﯾﻒ داﻟﺔ ﺟﺎﻣﺎ ﻟﺴﻌﺔ اﻟﻌﺪد اﻟﺴﺎﻟﺐ
ﺛﺒﺖ ﻣﻦ اﻟﺴﺎﺑﻖ اﻟﻨﻈﺮﯾﺔ اﻵﺗﯿﺔ:
1
)Γ(n) = Γ(n + 1
n
ﻧﻼﺣﻆ ﻋﻨﺪﻣﺎ n = 0ﻓﺈن اﻟﻤﻘﺎم ﯾﻨﻌﺪم ﺑﯿﻨﻤﺎ ﺗﻜﻮن اﻟﺪاﻟﺔ ) Γ (n + 1ﻓﻲ ھﺬه اﻟﺤﺎﻟﺔ ﻣﻌﺮﻓ ﺔ ﻋﻨ ﺪ n = 0وﺗﻜ ﻮن ﺳ ﻌﺔ
اﻟﺪاﻟ ﺔ ﻓ ﻲ ھ ﺬه اﻟﺤﺎﻟ ﺔ ﻣﻮﺟﺒ ﺔ ,ﺣﯿ ﺚ ﻋﻨ ﺪﻣﺎ n → 0ﻓ ﺈن , Γ (n + 1) = Γ(1) → 1وﻋﻠﯿ ﮫ ﻧﺠ ﺪ أن
1
n
ﻹﯾﺠﺎد ﻗﯿﻢ x > −1وﻋﻠﯿ ﮫ ﯾﻤﻜ ﻦ اﻟﻘ ﻮل أن ) Γ(nﻣﻌﺮﻓ ﺔ ﻟﻘ ﯿﻢ n > −1وﺗﻜ ﻮن ﺳ ﻌﺔ اﻟﻄ ﺮف اﻷﯾﻤ ﻦ ﻣﻌ ﺮف ﻟﻘ ﯿﻢ
∞ → ) Γ (n) = Γ (n + 1وﻋﻠﯿﮫ ﯾﻜﻮن اﻟﻄﺮف اﻷﯾﺴ ﺮ ﻣﻌ ﺮف ﻟﻘ ﯿﻢ n > 0وﻟﻜ ﻦ ﯾﻤﻜ ﻦ اﺳ ﺘﺨﺪام اﻟﻄ ﺮف اﻷﯾﻤ ﻦ
x + 1 > −1أو . x > −2وﺗﻜ ﻮن ﻗﯿﻤ ﺔ ) Γ(nﻣﻌﺮﻓ ﺔ ﻋﻨ ﺪ n > −2وھﻜ ﺬا وﯾﻤﻜ ﻦ ﺑﺎﻻﺳ ﺘﻤﺮار اﻟﻮﺻ ﻮل إﻟ ﻰ ھ ﺬا
اﻟﻤﻔﮭﻮم ∞ = )Γ (m
ﻋﻨ ﺪﻣﺎ ﺗﻜ ﻮن mﻣﺴ ﺎوﯾﺔ ﺻ ﻔﺮاً أو ﻋ ﺪد ﺻ ﺤﯿﺢ ﺳ ﺎﻟﺐ ,وھ ﺬا واﺿ ﺢ ﻋﻨ ﺪﻣﺎ m = 0ﻧﺠ ﺪ أن ∞ = ) , Γ (0ﻋﻨ ﺪﻣﺎ
n = −1ﻧﺠﺪ أن
1
= )Γ(−1
∞ = ) Γ (0
−1
1
= )Γ ( − 2
∞ = )Γ (−2 + 1
ﻛﺬﻟﻚ
−2
وھﻜﺬا .
وﯾﻤﻜﻦ ﻟﻠﺪارس ﺗﻤﺜﯿﻞ داﻟﺔ ﺟﺎﻣﺎ ﺑﯿﺎﻧﯿﺎً ﻛﺎﻵﺗﻲ:
ﻧﻈﺮﯾﺔ :٣
ﻷي ﻋﺪد ﺻﺤﯿﺢ ﻣﻮﺟﺐ
! Γ ( n + 1) = n
٢
اﻟﺒﺮھﺎن:
ﻣﻦ ﺧﻼل اﻟﻨﻈﺮﯾﺔ ) (٢و ) (4ﻣﻊ ﺗﻜﺮار ﺗﺤﺮﯾﻚ nﺑﺎﻟﻤﻘﺎدﯾﺮ n − 3, n − 2 ,n − 1وھﻜﺬا ﻧﺠﺪ أن
)Γ( n + 1) = n(n − 1)Γ(n − 1
)= n(n − 1)(n − 2)Γ(n − 2
)= n(n − 1)(n − 2)(n − 3) L 2 ⋅1Γ(1
وﺑﺎﺳﺘﺨﺪام ) (3ﻧﺤﺼﻞ ﻋﻠﻰ اﻟﻌﻼﻗﺔ
)(5
ﻧﺘﯿﺠﺔ :١
إذا ﻛﺎن n = 0ﻓﺈن
!Γ ( n + 1) = n
0 ! = Γ ( 0 + 1) = Γ (1) = 1
ﻧﻈﺮﯾﺔ :٤
∞
Γ ( n) = 2 ∫ e − t t 2 n −1dt
2
0
اﻟﺒﺮھﺎن:
t =u
ﺑﺎﺳﺘﺨﺪام اﻟﺘﻌﻮﯾﺾ
dt = 2u du
و ﻣﻨﮫ ﻧﺠﺪ أن
و ﻋﻠﻰ ذﻟﻚ ﻓﺈن
2
∞
) Γ( n) = ∫ e −u (u 2 ) n −1 ( 2udu
2
0
وﺑﺎﻟﺘﺎﻟﻲ ﻧﺤﺼﻞ ﻋﻠﻰ
∞
Γ( n) = 2 ∫ e −u u 2 n−1du
2
)(6
0
وھﺬه اﻟﻨﻈﺮﯾﺔ ﻣﻔﯿﺪة ﻓﻲ اﻟﻌﻤﻠﯿﺎت اﻹﺣﺼﺎﺋﯿﺔ اﻟﻤﺴﺘﺨﺪم ﻓﯿﮭﺎ ﻣﺎ ﯾﺴﻤﻰ ﺑﺪاﻟﺔ اﻟﻜﺜﺎﻓﺔ وﻏﯿﺮھ ﺎ .ﻣﺜ ﺎل ﻋﻠ ﻰ ذﻟ ﻚ ﻋﻨ ﺪ
1
وﺿﻊ
2
= nﻧﺠﺪ أن
∞
)(7
2
1
Γ( ) = 2 ∫ e −u du
2
0
ﻧﻈﺮﯾﺔ :٥
ﯾﻤﻜﻦ اﻟﺮﺑﻂ ﺑﯿﻦ اﻟﺪوال اﻟﻤﺜﻠﺜﯿﺔ وداﻟﺔ ﺟﺎﻣﺎ ﻛﺎﻵﺗﻲ:
)Γ ( n ) Γ ( m
⋅
) 2Γ ( n + m
2 n −1
= θ⋅ sin 2 m −1θ dθ
π
2
∫ cos
0
اﻟﺒﺮھﺎن:
ﺑﺎﺳﺘﺨﺪام اﻟﻨﻈﺮﯾﺔ ) (4ﻧﻼﺣﻆ أن
٣
∞
Γ( n) = 2 ∫ e −u u 2 n−1du
2
0
∞
Γ( m) = 2 ∫ e −v v 2 n−1dv
2
0
: وﻋﻞ ذﻟﻚ ﻓﺈن
∞
Γ ( n )Γ ( m ) = 4 ∫ e u
−u 2
du ∫ e −v v 2 m−1dv
0
∞∞
∞
2 n −1
2
0
= 4 ∫ ∫ e −(u +v ) u 2n−1 v 2 m−1 du dv
2
2
0 0
v
r
θ
u
uv − plane
ﻧﺠﺪ أنdudv = rdrdθ ﻣﻊ ﺗﻄﺒﯿﻖ ﻧﻈﺮﯾﺔ اﻟﺠﺎﻛﻮﺑﯿﺎنu = r cos θ , v = r sin θ ﺑﺎﺳﺘﺨﺪام اﻹﺣﺪاﺛﯿﺎت اﻟﻘﻄﺒﯿﺔ
∞
π
2
Γ(n)Γ( m) = 4∫ ∫ e − r (r cos θ ) 2 n−1 ( r sin θ ) 2 m −1 r dr dθ
2
0 0
∞
π
2
= 4 ∫ ∫ e −r r 2 n+2 m −1 cos 2 n −1θ sin 2 m −1θ dr dθ
2
0 0
(٤) وﻟﻜﻦ ﻣﻦ ﻧﻈﺮﯾﺔ
∞
Γ( n + m) = 2 ∫ e −r r 2( n + m )−1dr
2
0
وﻋﻠﯿﮫ ﻧﺠﺪ أن
π
2
Γ(n)Γ(m) = 2Γ(n + m) ∫ cos 2 n −1θsin 2 m −1θ dθ
0
وﻋﻠﯿﮫ ﻧﺠﺪ أن
π
2
∫ cos
2 n −1
θsin 2 m −1θ dθ =
0
٤
Γ ( n) Γ ( m )
2Γ(n + m)
(8)
ﻣﻼﺣﻈﺔ :
1
ﻋﻨﺪﻣﺎ = n = mﻧﺤﺼﻞ ﻋﻠﻰ
2
) Γ( 12 )Γ( 12 ) Γ( 12 )Γ( 12
=
)2Γ(1
2
π
2
= ∫ dθ
0
ﺑﺈﺟﺮاء اﻟﺘﻜﺎﻣﻞ
}) (π {Γ
=
2
2
1
Γ( 2 ) = π
2
)(9
1
2
وأھﻤﻠ ﺖ اﻹﺷ ﺎرة اﻟﺴ ﺎﻟﺒﺔ − πﻣ ﻦ اﻟﺘﻌﺮﯾ ﻒ ﺑﺤﯿ ﺚ أن اﻟﻌ ﺪد nﻣﻮﺟ ﺐ ﻓﺘﻜ ﻮن ) Γ (nﻣﻮﺟﺒ ﺔ وﺑﮭ ﺬه اﻟﻘﺎﻋ ﺪة
اﻟﮭﺎﻣﺔ ﻟﻠﻌﻼﻗﺔ ) (9ﯾﺘﻢ إﺛﺒﺎت أن داﻟﺔ اﻟﻜﺜﺎﻓﺔ ﻣﻦ اﻟﻤﻌﺎدﻟﺔ ) (7ﻟﮭﺎ اﻟﻘﯿﻤﺔ
π
⋅
2
)(10
= du
−u 2
∞
∫e
0
ﻣﺜﺎل:١
اﺣﺴﺐ
) 6Γ( 83
) 5Γ( 23
)(iv
)Γ(3)Γ( 2.5
)Γ(5.5
) Γ( 52
) Γ( 12
)(iii
)(ii
)Γ ( 6
)2Γ(3
)(i
اﻟﺤﻞ:
!5
!5 ⋅ 4 ⋅ 3 ⋅ 2
)Γ(6
= 30
=
=
!2Γ(3) 2 ⋅ 2
!2 ⋅ 2
Γ( 52 ) Γ( 32 + 1) 32 Γ( 32 ) 32 Γ( 12 + 1) 32 ⋅ 12 Γ( 12 ) 3
=
=
)(ii
=
=
=
) Γ( 12
) Γ( 12
) Γ( 12
) Γ( 12
4
) Γ( 12
)(i
ﺑﺎﻟﻤﺜﻞ
ﻧﻈﺮﯾﺔ :٦
)2!(1.5)(0.5)Γ(0.5
!2
16
)Γ(3)Γ(2.5
= = 9 7 5
.
=
(4.5)(3.5)(2.5)(1.5)(0.5)Γ(0.5) ( 2 )( 2 )( 2 ) 315
)Γ(5.5
)(iii
6Γ( 83 ) 6 ⋅ ( 53 )( 23 )Γ( 23 ) 4
=
= .
) 5Γ( 23
) 5Γ( 23
3
)(iv
π
sin πn
= )Γ( n)Γ(1 − n
ﻧﺘﯿﺠﺔ :
1
ﻋﻨﺪﻣﺎ = nﻓﺈن :
2
π
1 1
=π
= Γ Γ
2 2 sin π
2
٥
و ﺑﺎﻟﺘ ﺎﻟﻲ ﻓﺈن
Γ( 12 ) = π
وھﻲ ﻧﻔﺲ اﻟﻨﺘﯿﺠﺔ اﻟﺘﻲ ﺣﺼﻠﻨﺎ ﻋﻠﯿﮭﺎ ﺳﺎﺑﻘﺎً
ﻣﺜﺎل: ٢
أوﺟﺪ ﻗﯿﻤﺔ
) (ii) Γ( −25
) (i) Γ( −21
اﻟﺤﻞ :
)Γ( n + 1
n
−1
ﺑﻮﺿﻊ
2
= )Q Γ( n
=n
= −2Γ( 12 ) = −2 π
−5
أﯾﻀﺎً ﺑﻮﺿﻊ
2
)Γ( −21 + 1
−1
2
= ) (i ) Γ( −21
=n
) Γ( −23
) ( −25
= ) (ii) Γ( −25
) Γ( −21
) ( −23
, Γ( −21 ) = −2 π
= ) ∴ Γ( −23
وﻋﻠﯿﮫ ﻧﺠﺪ أن
8
π
15
= ) Γ( −25 ) = ( −52 )( −32 )(−2 π
ﻣﺜﺎل:٣
اﺣﺴﺐ اﻟﺘﻜﺎﻣﻼت اﻵﺗﯿﺔ:
y e − y dy
3
∞
∫
dx
)(iii
∞
∫x e
6 −2 x
0
dx
)(ii
dx
− ln x
dx
0
)(i
0
1
∫
∫x e
3 −x
0
)( v
∞
−4 x2
∞
∫3
)(iv
0
اﻟﺤﻞ:
∞
!dx = ∫ x 4−1e − x dx = Γ ( 4) = 3
0
dx,
∞
∫x e
3 −x
0
∞
∫x e
6 −2 x
0
ﻧﺄﺧﺬ اﻟﻔﺮض
du = 2 dx , u = 2 xو ﻋﻠﻰ ﻓﺈن
٦
)(i
)(ii
∞
∞
∞
u 6 −u du
1
6! 45
6 −2 x
6 −u
∫0 x e dx = ∫0 ( 2 ) e 2 = (2) 7 ∫0 u e du = 27 = 8 .
∞
∫
(iii)
y e − y dy ,
3
0
وﻧﻼﺣﻆ أن ﺣﺪود اﻟﺘﻜﺎﻣﻞ ﺗﺒﻘﻰ ﻛﻤﺎ ھﻲ,3 y 2 dy = du , y 3 = u
∞
∫
ye
− y3
0
∞
∞
∞
∞
1 −1
1 1 −1
1 −2
π
u e ⋅ u 3 du = ∫ u 2 e −u du = ∫ u 2 e −u du =
30
30
3
3
dy = ∫
1
3
0
(iv)
ﺑﻮﺿﻊ
−4 x
∫ 3 dx
2
−u
,3−4 x = (e ln 3 ) −4 x
2
Q 3 = e ln 3
2
3−4 x = e −4 x
2
2
ln 3
0
(4 ln 3) x 2 = u
1
u2
x=
2 ln 3
∞
∫3
− 4 x2
0
1
∫
( v)
0
u
dx =
ﺑﻮﺿﻊ
− 12
4 ln 3
du
∞
u−2
1
π
−u − 2
du = 1
dx = ∫ e ⋅
e
u
du
=
⋅
∫
4
ln
3
4
ln
3
4
ln
3
0
0
∞
1
−u
dx
,
− ln x
− ln x = u , x = e −u
, dx = −e −u du ﺑﻮﺿﻊ
u = 0, x = 1 ﻋﻨﺪ, u = ∞ وﻋﻠﯿﮫln 0 = −∞ ﻧﺠﺪ أنx = 0 ﻋﻨﺪﻣﺎ
∞
dx
e −u du
−1
= − ∫ 1 = ∫ u 2 e −u du = π ⋅
2
− ln x
∞ u
0
1
0
∫
0
:٤ﻣﺜﺎل
اﺣﺴﺐ
∞
m − ax
∫ x e dx
, m, n > 0
n
0
:اﻟﺤﻞ
dx =
∞
∞
m −ax
−u
∫ x e dx = ∫ e ⋅ (
0
n
0
u
1
n
1
n
a
1
n
m
⋅
1 )
an
u
u
1
n
1
−1
n
na
1
n
−1
x =
du
du =
∞
1
na
m
n
٧
+
1
n
u
a
1
n
وﻋﻠﯿﮫax n = u ﺑﻮﺿﻊ
1
n
−u
∫e un
0
m
+ 1n −1
du =
1
na
m +1
n
Γ(
m +1
)⋅
n
ﻣﺜﺎل:٥
اﺣﺴﺐ اﻟﺘﻜﺎﻣﻞ اﻵﺗﻲ:
1
(ln x) n dx
m
∫x
0
اﻟﺤﻞ:
ﺑﻮﺿﻊ ln x = −uھﺬا اﻻﺧﺘﯿﺎر ﯾﺠﻌﻞ ﺣﺪود اﻟﺘﻜﺎﻣﻞ ﻋﻨﺪ x = 0
u n du
−( m+1) u
∞
∫e
0
n
∞ = uوﻋﻨﺪ x = 1
0
)(ln x) dx = ∫ (e ) (−u ) (e du ) = ( −1
−u
−u m
n
u = 0وﻋﻠﯿﮫ
n
∞
−u
x=e
1
m
∫x
0
ﺑﻮﺿﻊ ( m + 1)u = v
∞
∞
( −1) n
(−1) n
v n dv
−v n
∫0 x (ln x) dx = (−1) ∫0 e (1 + m ) ⋅ 1 + m = (1 + m)n+1 ∫0 e v dv = (1 + m) n+1 Γ(n + 1).
−v
1
n
n
m
وﻋﻠﯿﮫ ﻧﺠﺪ أن
( − 1) n
!⋅ n
(1 + m ) n + 1
1
m
n
= ∫ x (ln x) dx
0
داﻟﺔ ﺑﯿﺘﺎ :
ﺗﻌﺮف داﻟﺔ ﺑﯿﺘﺎ ﻛﺎﻵﺗﻲ:
1
)(10
)( n > 0, m > 0
β ( n, m) = ∫ t n−1 (1 − t ) m−1 dt
0
وواﺿﺢ ﻋﻨﺪ t → 0ﻻﺑﺪ ﻣﻦ وﺟﻮد ﻗﯿﻮد ﻋﻠﻰ tﺑﯿﻨﻤﺎ ﻋﻨ ﺪ t = 1ﻻ ﺗﻮﺟ ﺪ أي ﻗﯿ ﻮد ﻋﻠﯿﮭ ﺎ وﻋﻠﯿ ﮫ ﯾﺘﻀ ﺢ ﻣ ﻦ
اﻟﻤﻌﺎدﻟﺘﯿﻦ ) (10), (1أن اﻟﻘﯿﻮد اﻟﻤﻔﺮوﺿﺔ ﻋﻨﺪ ﻧﻘﻄﺔ اﻷﺻﻞ ﻋﻠﻰ داﻟﺘﻲ ) β( n, m), Γ (nﻗﺪ ﺗﺤﻜﻤﺖ ﻓﻲ ﺗﺤﺪﯾﺪ
ﻣﺎھﯿﺔ . n, m
اﻟﻌﻼﻗﺔ ﺑﯿﻦ داﻟﺘﻲ ﺑﯿﺘﺎ وﺟﺎﻣﺎ :
ﻧﻈﺮﯾﺔ :٦
ھﺬه اﻟﻨﻈﺮﯾﺔ ﺗﺮﺑﻂ ﺑﯿﻦ داﻟﺘﻲ ﺑﯿﺘﺎ وﺟﺎﻣﺎ اﻋﺘﻤﺎداً ﻋﻠﻰ إﺛﺒﺎت ﻧﻈﺮﯾﺔ ) (٥ﻛﺎﻵﺗﻲ:
)Γ (n)Γ (m
= )β( n, m
)(11
)Γ (n + m
اﻟﺒﺮھﺎن :
ﻣﻦ اﻟﺘﻌﺮﯾﻒ
1
β( n, m) = ∫ t n −1 (1 − t ) m −1 dt
0
π
ﺑﻔﺮض أن t = cos 2 θوﻋﻠﯿﮫ ﻋﻨﺪ t = 0ﻧﺠﺪ أن
2
= θوﻋﻨﺪ t = 1ﻧﺠﺪ أن θ = 0أﯾﻀ ﺎً dt = −2 sin θ cos θdθ
وﻣﻦ ذﻟﻚ ﻧﺠﺪ أن
٨
)Γ ( n)Γ (m
)Γ ( n + m
)( n > 0, m > 0
= dθ
sin 2 m−1 θ
π
2
β ( n, m) = 2 ∫ cos 2 n−1 θ
0
)ﺑﺎﺳﺘﺨﺪام ﻧﻈﺮﯾﺔ)( (٥
1
t
= xﻓﻲ ) (10ﻓﺈن dt
ﺑﻮﺿﻊ
2
t +1
) (1 + t
= , dxﻋﻨﺪﻣﺎ x = 0ﻓ ﺈن ، t = 0وﻋﻨ ﺪﻣﺎ x = 1ﻓ ﺈن
∞ = tوﻋﻠﻰ ذﻟﻚ ﻓﺈن :
∞
t m−1
t n−1 1
dt
( ∫ = )B( m, n
) (1 −
)
(1 + t ) 2
1+ t
1+ t
0
∞
t m−1
dt
∫ = )B ( m, n
(1 + t ) m+n
0
)(12
ﺗﻤﺮﯾﻦ :
; 0 < P
ﺗﻌﺮﯾﻒ داﻟﺘﻲ ﺟﺎﻣﺎ وﺑﯿﺘﺎ
ﺗﻌﺮف داﻟﺔ ﺟﺎﻣﺎ ﻛﺎﻵﺗﻲ:
∞
Γ(n) = ∫ e−t t n−1 dt
)(n > 0
)(1
0
وﻧﻼﺣﻆ أن اﻟﺪاﻟﺔ e − t t n−1ﻋﻨﺪ ∞ ﺗﺆول إﻟ ﻰ اﻟﺼ ﻔﺮ ﻷي ﻗﯿﻤ ﺔ ﻋﺪدﯾ ﺔ ﻟ ـ nوﻋﻠﯿ ﮫ ﻻﺗﻜ ﻮن ھﻨ ﺎك ﺻ ﻌﻮﺑﺎت
ﻋﻠﻤﯿ ﺔ ﻋﻨ ﺪ ∞ → tﺑﯿﻨﻤ ﺎ ﻓ ﻲ اﻟﺠ ﺰء اﻟﺴ ﻔﻠﻲ ﺗﻮﺟ ﺪ t → 0ﻧﺠ ﺪ أن e −t ≅ 1وﻋﻠﯿ ﮫ ﺗﺼ ﺒﺢ اﻟﻤﻌﺎدﻟ ﺔ ) (1ﻋﻠ ﻰ
اﻟﺼﻮرة
∞
∞
c
1
Γ(n) = ∫ t dt + ∫ e t dt = t n + ∫ e −t t n−1dt
n 0 c
0
c
−t n −1
c
n−1
وﻋﻠﯿﮫ إذا ﻛﺎن اﻟﺠﺰء اﻷول ﻣﻦ اﻟﻨﺎﺗﺞ ﻋﺪد ﻓﯿﺠﺐ ﻋﻠﯿﻨﺎ اﺧﺘﯿﺎر n > 0ﺣﺘﻰ ﻻ ﯾﻨﻌﺪم اﻟﻤﻘﺎم .
ﺑﻌﺾ اﻟﻨﻈﺮﯾﺎت واﻟﺨﻮاص اﻟﮭﺎﻣﺔ
ﻧﻈﺮﯾﺔ :١
Γ (1) = 1
اﻟﺒﺮھﺎن :
ﻣﻦ ﺗﻌﺮﯾﻒ ) Γ (nﻧﻀﻊ n = 1ﻓﻨﺤﺼﻞ ﻋﻠﻰ
]
∞
[
∞
0
Γ(1) = ∫ e −t dt = − e −t
0
Γ (1) = 1
)(3
ﻧﻈﺮﯾﺔ :٢
)Γ (n + 1) = nΓ(n
اﻟﺒﺮھﺎن :
ﻓﻲ اﻟﻤﻌﺎدﻟﺔ ) (1ﺑﺘﺤﺮﯾﻚ ﻛﻞ n → n + 1ﻧﺤﺼﻞ ﻋﻠﻰ
∞
Γ( n + 1) = ∫ e −t t n dt
0
ﺑﺎﻟﺘﻜﺎﻣﻞ ﺑﺎﻟﺘﺠﺰيء ﺑﺄﺧﺬ u = t n
−t
e dr = dv,ﻧﺤﺼﻞ ﻋﻠﻰ
) n t n −1dt
−t
∞
] − ∫ (−e
∞ −t n
0
[
Γ ( n + 1) = − e t
0
ﻛﻤﺎ ﻧﻼﺣﻆ اﻧﻌﺪام اﻟﺠﺰء اﻷول ﻣﻦ اﻟﻨﺘﯿﺠﺔ اﻟﺴﺎﺑﻘﺔ ﻓﻨﺤﺼﻞ ﻋﻠﻰ
∞
Γ( n + 1) = n ∫ e −t t n−1 dt
0
)(4
) Γ ( n + 1) = n Γ ( n
١
ﻣﻼﺣﻈﺎت :
(١اﻟﻌﻼﻗﺔ ) Γ (n + 1) = n Γ( nﺗﺴﻤﻰ ﺑﺎﻟﺼﯿﻐﺔ اﻟﺘﻜﺮارﯾﺔ ﻟﺪاﻟﺔ ﺟﺎﻣﺎ .
(٢ﯾﻤﻜﻦ ﺗﻌﻤﯿﻢ داﻟﺔ ﺟﺎﻣﺎ ﻟﻘﯿﻢ ) n < 0ﻟﻸﻋﺪاد اﻟﺤﻘﯿﻘﯿﺔ ﻏﯿﺮ اﻟﺼﺤﯿﺤﺔ ( ﺑﺎﺳﺘﺨﺪام اﻟﺼﯿﻐﺔ اﻟﺴﺎﺑﻘﺔ ﻋﻠﻰ
اﻟﺼﻮرة :
)Γ (n + 1
= )Γ (n
n
ﺗﻌﺮﯾﻒ داﻟﺔ ﺟﺎﻣﺎ ﻟﺴﻌﺔ اﻟﻌﺪد اﻟﺴﺎﻟﺐ
ﺛﺒﺖ ﻣﻦ اﻟﺴﺎﺑﻖ اﻟﻨﻈﺮﯾﺔ اﻵﺗﯿﺔ:
1
)Γ(n) = Γ(n + 1
n
ﻧﻼﺣﻆ ﻋﻨﺪﻣﺎ n = 0ﻓﺈن اﻟﻤﻘﺎم ﯾﻨﻌﺪم ﺑﯿﻨﻤﺎ ﺗﻜﻮن اﻟﺪاﻟﺔ ) Γ (n + 1ﻓﻲ ھﺬه اﻟﺤﺎﻟﺔ ﻣﻌﺮﻓ ﺔ ﻋﻨ ﺪ n = 0وﺗﻜ ﻮن ﺳ ﻌﺔ
اﻟﺪاﻟ ﺔ ﻓ ﻲ ھ ﺬه اﻟﺤﺎﻟ ﺔ ﻣﻮﺟﺒ ﺔ ,ﺣﯿ ﺚ ﻋﻨ ﺪﻣﺎ n → 0ﻓ ﺈن , Γ (n + 1) = Γ(1) → 1وﻋﻠﯿ ﮫ ﻧﺠ ﺪ أن
1
n
ﻹﯾﺠﺎد ﻗﯿﻢ x > −1وﻋﻠﯿ ﮫ ﯾﻤﻜ ﻦ اﻟﻘ ﻮل أن ) Γ(nﻣﻌﺮﻓ ﺔ ﻟﻘ ﯿﻢ n > −1وﺗﻜ ﻮن ﺳ ﻌﺔ اﻟﻄ ﺮف اﻷﯾﻤ ﻦ ﻣﻌ ﺮف ﻟﻘ ﯿﻢ
∞ → ) Γ (n) = Γ (n + 1وﻋﻠﯿﮫ ﯾﻜﻮن اﻟﻄﺮف اﻷﯾﺴ ﺮ ﻣﻌ ﺮف ﻟﻘ ﯿﻢ n > 0وﻟﻜ ﻦ ﯾﻤﻜ ﻦ اﺳ ﺘﺨﺪام اﻟﻄ ﺮف اﻷﯾﻤ ﻦ
x + 1 > −1أو . x > −2وﺗﻜ ﻮن ﻗﯿﻤ ﺔ ) Γ(nﻣﻌﺮﻓ ﺔ ﻋﻨ ﺪ n > −2وھﻜ ﺬا وﯾﻤﻜ ﻦ ﺑﺎﻻﺳ ﺘﻤﺮار اﻟﻮﺻ ﻮل إﻟ ﻰ ھ ﺬا
اﻟﻤﻔﮭﻮم ∞ = )Γ (m
ﻋﻨ ﺪﻣﺎ ﺗﻜ ﻮن mﻣﺴ ﺎوﯾﺔ ﺻ ﻔﺮاً أو ﻋ ﺪد ﺻ ﺤﯿﺢ ﺳ ﺎﻟﺐ ,وھ ﺬا واﺿ ﺢ ﻋﻨ ﺪﻣﺎ m = 0ﻧﺠ ﺪ أن ∞ = ) , Γ (0ﻋﻨ ﺪﻣﺎ
n = −1ﻧﺠﺪ أن
1
= )Γ(−1
∞ = ) Γ (0
−1
1
= )Γ ( − 2
∞ = )Γ (−2 + 1
ﻛﺬﻟﻚ
−2
وھﻜﺬا .
وﯾﻤﻜﻦ ﻟﻠﺪارس ﺗﻤﺜﯿﻞ داﻟﺔ ﺟﺎﻣﺎ ﺑﯿﺎﻧﯿﺎً ﻛﺎﻵﺗﻲ:
ﻧﻈﺮﯾﺔ :٣
ﻷي ﻋﺪد ﺻﺤﯿﺢ ﻣﻮﺟﺐ
! Γ ( n + 1) = n
٢
اﻟﺒﺮھﺎن:
ﻣﻦ ﺧﻼل اﻟﻨﻈﺮﯾﺔ ) (٢و ) (4ﻣﻊ ﺗﻜﺮار ﺗﺤﺮﯾﻚ nﺑﺎﻟﻤﻘﺎدﯾﺮ n − 3, n − 2 ,n − 1وھﻜﺬا ﻧﺠﺪ أن
)Γ( n + 1) = n(n − 1)Γ(n − 1
)= n(n − 1)(n − 2)Γ(n − 2
)= n(n − 1)(n − 2)(n − 3) L 2 ⋅1Γ(1
وﺑﺎﺳﺘﺨﺪام ) (3ﻧﺤﺼﻞ ﻋﻠﻰ اﻟﻌﻼﻗﺔ
)(5
ﻧﺘﯿﺠﺔ :١
إذا ﻛﺎن n = 0ﻓﺈن
!Γ ( n + 1) = n
0 ! = Γ ( 0 + 1) = Γ (1) = 1
ﻧﻈﺮﯾﺔ :٤
∞
Γ ( n) = 2 ∫ e − t t 2 n −1dt
2
0
اﻟﺒﺮھﺎن:
t =u
ﺑﺎﺳﺘﺨﺪام اﻟﺘﻌﻮﯾﺾ
dt = 2u du
و ﻣﻨﮫ ﻧﺠﺪ أن
و ﻋﻠﻰ ذﻟﻚ ﻓﺈن
2
∞
) Γ( n) = ∫ e −u (u 2 ) n −1 ( 2udu
2
0
وﺑﺎﻟﺘﺎﻟﻲ ﻧﺤﺼﻞ ﻋﻠﻰ
∞
Γ( n) = 2 ∫ e −u u 2 n−1du
2
)(6
0
وھﺬه اﻟﻨﻈﺮﯾﺔ ﻣﻔﯿﺪة ﻓﻲ اﻟﻌﻤﻠﯿﺎت اﻹﺣﺼﺎﺋﯿﺔ اﻟﻤﺴﺘﺨﺪم ﻓﯿﮭﺎ ﻣﺎ ﯾﺴﻤﻰ ﺑﺪاﻟﺔ اﻟﻜﺜﺎﻓﺔ وﻏﯿﺮھ ﺎ .ﻣﺜ ﺎل ﻋﻠ ﻰ ذﻟ ﻚ ﻋﻨ ﺪ
1
وﺿﻊ
2
= nﻧﺠﺪ أن
∞
)(7
2
1
Γ( ) = 2 ∫ e −u du
2
0
ﻧﻈﺮﯾﺔ :٥
ﯾﻤﻜﻦ اﻟﺮﺑﻂ ﺑﯿﻦ اﻟﺪوال اﻟﻤﺜﻠﺜﯿﺔ وداﻟﺔ ﺟﺎﻣﺎ ﻛﺎﻵﺗﻲ:
)Γ ( n ) Γ ( m
⋅
) 2Γ ( n + m
2 n −1
= θ⋅ sin 2 m −1θ dθ
π
2
∫ cos
0
اﻟﺒﺮھﺎن:
ﺑﺎﺳﺘﺨﺪام اﻟﻨﻈﺮﯾﺔ ) (4ﻧﻼﺣﻆ أن
٣
∞
Γ( n) = 2 ∫ e −u u 2 n−1du
2
0
∞
Γ( m) = 2 ∫ e −v v 2 n−1dv
2
0
: وﻋﻞ ذﻟﻚ ﻓﺈن
∞
Γ ( n )Γ ( m ) = 4 ∫ e u
−u 2
du ∫ e −v v 2 m−1dv
0
∞∞
∞
2 n −1
2
0
= 4 ∫ ∫ e −(u +v ) u 2n−1 v 2 m−1 du dv
2
2
0 0
v
r
θ
u
uv − plane
ﻧﺠﺪ أنdudv = rdrdθ ﻣﻊ ﺗﻄﺒﯿﻖ ﻧﻈﺮﯾﺔ اﻟﺠﺎﻛﻮﺑﯿﺎنu = r cos θ , v = r sin θ ﺑﺎﺳﺘﺨﺪام اﻹﺣﺪاﺛﯿﺎت اﻟﻘﻄﺒﯿﺔ
∞
π
2
Γ(n)Γ( m) = 4∫ ∫ e − r (r cos θ ) 2 n−1 ( r sin θ ) 2 m −1 r dr dθ
2
0 0
∞
π
2
= 4 ∫ ∫ e −r r 2 n+2 m −1 cos 2 n −1θ sin 2 m −1θ dr dθ
2
0 0
(٤) وﻟﻜﻦ ﻣﻦ ﻧﻈﺮﯾﺔ
∞
Γ( n + m) = 2 ∫ e −r r 2( n + m )−1dr
2
0
وﻋﻠﯿﮫ ﻧﺠﺪ أن
π
2
Γ(n)Γ(m) = 2Γ(n + m) ∫ cos 2 n −1θsin 2 m −1θ dθ
0
وﻋﻠﯿﮫ ﻧﺠﺪ أن
π
2
∫ cos
2 n −1
θsin 2 m −1θ dθ =
0
٤
Γ ( n) Γ ( m )
2Γ(n + m)
(8)
ﻣﻼﺣﻈﺔ :
1
ﻋﻨﺪﻣﺎ = n = mﻧﺤﺼﻞ ﻋﻠﻰ
2
) Γ( 12 )Γ( 12 ) Γ( 12 )Γ( 12
=
)2Γ(1
2
π
2
= ∫ dθ
0
ﺑﺈﺟﺮاء اﻟﺘﻜﺎﻣﻞ
}) (π {Γ
=
2
2
1
Γ( 2 ) = π
2
)(9
1
2
وأھﻤﻠ ﺖ اﻹﺷ ﺎرة اﻟﺴ ﺎﻟﺒﺔ − πﻣ ﻦ اﻟﺘﻌﺮﯾ ﻒ ﺑﺤﯿ ﺚ أن اﻟﻌ ﺪد nﻣﻮﺟ ﺐ ﻓﺘﻜ ﻮن ) Γ (nﻣﻮﺟﺒ ﺔ وﺑﮭ ﺬه اﻟﻘﺎﻋ ﺪة
اﻟﮭﺎﻣﺔ ﻟﻠﻌﻼﻗﺔ ) (9ﯾﺘﻢ إﺛﺒﺎت أن داﻟﺔ اﻟﻜﺜﺎﻓﺔ ﻣﻦ اﻟﻤﻌﺎدﻟﺔ ) (7ﻟﮭﺎ اﻟﻘﯿﻤﺔ
π
⋅
2
)(10
= du
−u 2
∞
∫e
0
ﻣﺜﺎل:١
اﺣﺴﺐ
) 6Γ( 83
) 5Γ( 23
)(iv
)Γ(3)Γ( 2.5
)Γ(5.5
) Γ( 52
) Γ( 12
)(iii
)(ii
)Γ ( 6
)2Γ(3
)(i
اﻟﺤﻞ:
!5
!5 ⋅ 4 ⋅ 3 ⋅ 2
)Γ(6
= 30
=
=
!2Γ(3) 2 ⋅ 2
!2 ⋅ 2
Γ( 52 ) Γ( 32 + 1) 32 Γ( 32 ) 32 Γ( 12 + 1) 32 ⋅ 12 Γ( 12 ) 3
=
=
)(ii
=
=
=
) Γ( 12
) Γ( 12
) Γ( 12
) Γ( 12
4
) Γ( 12
)(i
ﺑﺎﻟﻤﺜﻞ
ﻧﻈﺮﯾﺔ :٦
)2!(1.5)(0.5)Γ(0.5
!2
16
)Γ(3)Γ(2.5
= = 9 7 5
.
=
(4.5)(3.5)(2.5)(1.5)(0.5)Γ(0.5) ( 2 )( 2 )( 2 ) 315
)Γ(5.5
)(iii
6Γ( 83 ) 6 ⋅ ( 53 )( 23 )Γ( 23 ) 4
=
= .
) 5Γ( 23
) 5Γ( 23
3
)(iv
π
sin πn
= )Γ( n)Γ(1 − n
ﻧﺘﯿﺠﺔ :
1
ﻋﻨﺪﻣﺎ = nﻓﺈن :
2
π
1 1
=π
= Γ Γ
2 2 sin π
2
٥
و ﺑﺎﻟﺘ ﺎﻟﻲ ﻓﺈن
Γ( 12 ) = π
وھﻲ ﻧﻔﺲ اﻟﻨﺘﯿﺠﺔ اﻟﺘﻲ ﺣﺼﻠﻨﺎ ﻋﻠﯿﮭﺎ ﺳﺎﺑﻘﺎً
ﻣﺜﺎل: ٢
أوﺟﺪ ﻗﯿﻤﺔ
) (ii) Γ( −25
) (i) Γ( −21
اﻟﺤﻞ :
)Γ( n + 1
n
−1
ﺑﻮﺿﻊ
2
= )Q Γ( n
=n
= −2Γ( 12 ) = −2 π
−5
أﯾﻀﺎً ﺑﻮﺿﻊ
2
)Γ( −21 + 1
−1
2
= ) (i ) Γ( −21
=n
) Γ( −23
) ( −25
= ) (ii) Γ( −25
) Γ( −21
) ( −23
, Γ( −21 ) = −2 π
= ) ∴ Γ( −23
وﻋﻠﯿﮫ ﻧﺠﺪ أن
8
π
15
= ) Γ( −25 ) = ( −52 )( −32 )(−2 π
ﻣﺜﺎل:٣
اﺣﺴﺐ اﻟﺘﻜﺎﻣﻼت اﻵﺗﯿﺔ:
y e − y dy
3
∞
∫
dx
)(iii
∞
∫x e
6 −2 x
0
dx
)(ii
dx
− ln x
dx
0
)(i
0
1
∫
∫x e
3 −x
0
)( v
∞
−4 x2
∞
∫3
)(iv
0
اﻟﺤﻞ:
∞
!dx = ∫ x 4−1e − x dx = Γ ( 4) = 3
0
dx,
∞
∫x e
3 −x
0
∞
∫x e
6 −2 x
0
ﻧﺄﺧﺬ اﻟﻔﺮض
du = 2 dx , u = 2 xو ﻋﻠﻰ ﻓﺈن
٦
)(i
)(ii
∞
∞
∞
u 6 −u du
1
6! 45
6 −2 x
6 −u
∫0 x e dx = ∫0 ( 2 ) e 2 = (2) 7 ∫0 u e du = 27 = 8 .
∞
∫
(iii)
y e − y dy ,
3
0
وﻧﻼﺣﻆ أن ﺣﺪود اﻟﺘﻜﺎﻣﻞ ﺗﺒﻘﻰ ﻛﻤﺎ ھﻲ,3 y 2 dy = du , y 3 = u
∞
∫
ye
− y3
0
∞
∞
∞
∞
1 −1
1 1 −1
1 −2
π
u e ⋅ u 3 du = ∫ u 2 e −u du = ∫ u 2 e −u du =
30
30
3
3
dy = ∫
1
3
0
(iv)
ﺑﻮﺿﻊ
−4 x
∫ 3 dx
2
−u
,3−4 x = (e ln 3 ) −4 x
2
Q 3 = e ln 3
2
3−4 x = e −4 x
2
2
ln 3
0
(4 ln 3) x 2 = u
1
u2
x=
2 ln 3
∞
∫3
− 4 x2
0
1
∫
( v)
0
u
dx =
ﺑﻮﺿﻊ
− 12
4 ln 3
du
∞
u−2
1
π
−u − 2
du = 1
dx = ∫ e ⋅
e
u
du
=
⋅
∫
4
ln
3
4
ln
3
4
ln
3
0
0
∞
1
−u
dx
,
− ln x
− ln x = u , x = e −u
, dx = −e −u du ﺑﻮﺿﻊ
u = 0, x = 1 ﻋﻨﺪ, u = ∞ وﻋﻠﯿﮫln 0 = −∞ ﻧﺠﺪ أنx = 0 ﻋﻨﺪﻣﺎ
∞
dx
e −u du
−1
= − ∫ 1 = ∫ u 2 e −u du = π ⋅
2
− ln x
∞ u
0
1
0
∫
0
:٤ﻣﺜﺎل
اﺣﺴﺐ
∞
m − ax
∫ x e dx
, m, n > 0
n
0
:اﻟﺤﻞ
dx =
∞
∞
m −ax
−u
∫ x e dx = ∫ e ⋅ (
0
n
0
u
1
n
1
n
a
1
n
m
⋅
1 )
an
u
u
1
n
1
−1
n
na
1
n
−1
x =
du
du =
∞
1
na
m
n
٧
+
1
n
u
a
1
n
وﻋﻠﯿﮫax n = u ﺑﻮﺿﻊ
1
n
−u
∫e un
0
m
+ 1n −1
du =
1
na
m +1
n
Γ(
m +1
)⋅
n
ﻣﺜﺎل:٥
اﺣﺴﺐ اﻟﺘﻜﺎﻣﻞ اﻵﺗﻲ:
1
(ln x) n dx
m
∫x
0
اﻟﺤﻞ:
ﺑﻮﺿﻊ ln x = −uھﺬا اﻻﺧﺘﯿﺎر ﯾﺠﻌﻞ ﺣﺪود اﻟﺘﻜﺎﻣﻞ ﻋﻨﺪ x = 0
u n du
−( m+1) u
∞
∫e
0
n
∞ = uوﻋﻨﺪ x = 1
0
)(ln x) dx = ∫ (e ) (−u ) (e du ) = ( −1
−u
−u m
n
u = 0وﻋﻠﯿﮫ
n
∞
−u
x=e
1
m
∫x
0
ﺑﻮﺿﻊ ( m + 1)u = v
∞
∞
( −1) n
(−1) n
v n dv
−v n
∫0 x (ln x) dx = (−1) ∫0 e (1 + m ) ⋅ 1 + m = (1 + m)n+1 ∫0 e v dv = (1 + m) n+1 Γ(n + 1).
−v
1
n
n
m
وﻋﻠﯿﮫ ﻧﺠﺪ أن
( − 1) n
!⋅ n
(1 + m ) n + 1
1
m
n
= ∫ x (ln x) dx
0
داﻟﺔ ﺑﯿﺘﺎ :
ﺗﻌﺮف داﻟﺔ ﺑﯿﺘﺎ ﻛﺎﻵﺗﻲ:
1
)(10
)( n > 0, m > 0
β ( n, m) = ∫ t n−1 (1 − t ) m−1 dt
0
وواﺿﺢ ﻋﻨﺪ t → 0ﻻﺑﺪ ﻣﻦ وﺟﻮد ﻗﯿﻮد ﻋﻠﻰ tﺑﯿﻨﻤﺎ ﻋﻨ ﺪ t = 1ﻻ ﺗﻮﺟ ﺪ أي ﻗﯿ ﻮد ﻋﻠﯿﮭ ﺎ وﻋﻠﯿ ﮫ ﯾﺘﻀ ﺢ ﻣ ﻦ
اﻟﻤﻌﺎدﻟﺘﯿﻦ ) (10), (1أن اﻟﻘﯿﻮد اﻟﻤﻔﺮوﺿﺔ ﻋﻨﺪ ﻧﻘﻄﺔ اﻷﺻﻞ ﻋﻠﻰ داﻟﺘﻲ ) β( n, m), Γ (nﻗﺪ ﺗﺤﻜﻤﺖ ﻓﻲ ﺗﺤﺪﯾﺪ
ﻣﺎھﯿﺔ . n, m
اﻟﻌﻼﻗﺔ ﺑﯿﻦ داﻟﺘﻲ ﺑﯿﺘﺎ وﺟﺎﻣﺎ :
ﻧﻈﺮﯾﺔ :٦
ھﺬه اﻟﻨﻈﺮﯾﺔ ﺗﺮﺑﻂ ﺑﯿﻦ داﻟﺘﻲ ﺑﯿﺘﺎ وﺟﺎﻣﺎ اﻋﺘﻤﺎداً ﻋﻠﻰ إﺛﺒﺎت ﻧﻈﺮﯾﺔ ) (٥ﻛﺎﻵﺗﻲ:
)Γ (n)Γ (m
= )β( n, m
)(11
)Γ (n + m
اﻟﺒﺮھﺎن :
ﻣﻦ اﻟﺘﻌﺮﯾﻒ
1
β( n, m) = ∫ t n −1 (1 − t ) m −1 dt
0
π
ﺑﻔﺮض أن t = cos 2 θوﻋﻠﯿﮫ ﻋﻨﺪ t = 0ﻧﺠﺪ أن
2
= θوﻋﻨﺪ t = 1ﻧﺠﺪ أن θ = 0أﯾﻀ ﺎً dt = −2 sin θ cos θdθ
وﻣﻦ ذﻟﻚ ﻧﺠﺪ أن
٨
)Γ ( n)Γ (m
)Γ ( n + m
)( n > 0, m > 0
= dθ
sin 2 m−1 θ
π
2
β ( n, m) = 2 ∫ cos 2 n−1 θ
0
)ﺑﺎﺳﺘﺨﺪام ﻧﻈﺮﯾﺔ)( (٥
1
t
= xﻓﻲ ) (10ﻓﺈن dt
ﺑﻮﺿﻊ
2
t +1
) (1 + t
= , dxﻋﻨﺪﻣﺎ x = 0ﻓ ﺈن ، t = 0وﻋﻨ ﺪﻣﺎ x = 1ﻓ ﺈن
∞ = tوﻋﻠﻰ ذﻟﻚ ﻓﺈن :
∞
t m−1
t n−1 1
dt
( ∫ = )B( m, n
) (1 −
)
(1 + t ) 2
1+ t
1+ t
0
∞
t m−1
dt
∫ = )B ( m, n
(1 + t ) m+n
0
)(12
ﺗﻤﺮﯾﻦ :
; 0 < P