Betta
- 1. FunciΓ³n Beta
1
π‘ π₯β1 (1 β π‘) π¦β1 ππ‘ ;
π½ π₯, π¦ =
π₯>0
π¦>0
0
Si hacemos π‘ = π ππ2 π
ππ‘ = 2 π ππ π cos π ππ
Si reemplazamos limites π‘ = 0 β π = 0
π‘=1 β π=
π
2
Reemplazamos
π
2
π½ π₯, π¦ = 2
(π ππ2 π) π₯β1 1 β π ππ2
π¦β1
π ππ π cos π ππ
0
π
2
π½ π₯, π¦ = 2
π ππ2π₯β1 π β cos2yβ1 π ππ
0
π
2
1
π½ π₯, π¦ =
2
π ππ2π₯β1 π β cos2yβ1 π ππ
0
Si hacemos
π‘=
1
1+ π’
ππ’
1+ π’
ππ‘ =
0
π½ π₯, π¦ = β
β
β
π½ π₯, π¦ =
0
β
π½ π₯, π¦ =
0
2
π π π‘ = 0 β π’ = β
1
1+ π’
π₯β1
π¦ π π
1
1β
1+ π’
π‘=1β0=0
π¦β1
1
1+ π’
π₯β1
1
1+ π’
π’ π¦ β1
ππ’
β
β
π₯β1
1 + π’ π¦β1 1 + π’
1+ π’β1
1+ π’
π’ π¦ β1 ππ’
π½ π₯, π¦ =
1 + π’ π₯β1+π¦β1+2
π’ π¦β1 ππ’
π½ π₯, π¦ =
1 + π’ π₯+π¦
π¦β1
ππ’
1+ π’
ππ’
1+ π’
2
2
2
- 2. Teorema
π½ π₯, π¦ =
ΞxΞy
;
Ξ x+y
π₯>0
π¦>0
Ejemplo
π
2
tan π ππ
0
π/2
0
π ππ π
cos π
1/2
ππ
π/2
π ππ 1/2 π πππ β1/2 π ππ
0
Comparando
1
π½ π₯, π¦ =
2
2π₯ β 1 =
1
2
2π¦ β 1 = β
π
2
π ππ2π₯β1 π β cos2yβ1 π ππ
0
β 2π₯ =
1
2
1
3
+ 1 β 2π₯ =
2
2
β 2π¦ = β
β
1
1
+ 1 β 2π¦ =
2
2
π=
π
π
β
Si aplicamos el teorema
=
1
β
2
1
= β
2
Ξ 3 βΞ 1
4
4
Ξ 3+1
4 4
Ξ 3 βΞ 1
4
4
4
; ππππ Ξ
=Ξ 1 =1
4
4
Ξ
4
π=
π
π
- 3. 1
3
1
= β Ξ
β Ξ
2
4
4
=
1
1
1
β Ξ
β Ξ 1β
2
4
4
Aplicamos teorema de gamma
=
1
Ο
β
Ο
2
sen
4
=
1
2
=
π
2
2
π
2
β π₯ π β1
0 1+π₯
Resolver
ππ₯
Por definiciΓ³n
π½ π₯, π¦ =
π’ π¦ β1
ππ’
1+π’
π₯ +π¦
Comparando
y-1=p-1
x+y=1
y=p
x=1βp
Reemplazamos
β
0
π₯ πβ1
ππ₯ = π½ 1 β π, π
1+ π₯
= π½ π, 1 β π
- 4. =
Ξ p Ξ 1βp
Ξ p+1βp
= Ξ p Ξ 1βp
Aplicamos teorema de gamma
=
π
π ππ ππ
Resolver
β
ββ
π 2π₯
π 3π₯ + 1
2
ππ₯
π’ = π 3π₯ β ln π’ = ln π 3π₯ β ln π’ = 3π₯
π₯=
1
ln π’ β
3
1 ππ’
3 ππ₯
ππ₯ =
Evaluamos los lΓmites
Cuando π₯ = β β
β
0
1
=
3
π’=β
1
2β ln π’
π 3
π’+1
1 ππ’
3 π’
2
π 3 ln π’
β
0
β
2
π¦ π₯ = ββ β
π’ π’+1
2
ππ’
Por propiedades de euler y logaritmos
1
=
3
1
=
3
β
0
β
0
2
π’3
β π’β1
ππ’
π’+1 2
β1
π’3
π’+1
2
ππ’
π’=0
- 5. Si comparamos con
π¦β1= β
1
3
π½ π₯, π¦ =
β
Reemplazamos
1
4 2
π½
,
3
3 3
4
2
1 Ξ 3 Ξ 3
= β
4 2
3
Ξ 3+3
1
1
2
Ξ 3 Ξ 3
1
= β 3
6
3
Ξ 3
1
2
1 Ξ 3 Ξ 3
= β
9
Ξ(2)
Ξ 2 = 1!
1
1
2
βΞ
Ξ
9
3
3
=
1
1
1
βΞ
Ξ 1β
9
3
3
Aplicamos teorema de gamma
=
=
=
1
Ο
β
9 sen Ο
3
1 Ο
β
9
3
2
2 Ο
β
9
3
2
3
ππ’
1+π’ π₯ +π¦
π¦= β
π₯+ π¦ =2 β π₯ =2β
=
π’ π¦ β1
1
π
+1β π=
3
π
β π=
π
π
- 6. Resolver
3
ππ₯
π₯β1
1
3
3β π₯
1
β
2
π₯β1
3β π₯
1
β
2
ππ₯
1
Sea x β 1 = 2y ο x = 2y+1 ο dx = 2dy
Cuando x = 1 y = 0
1
=
1
β
2
2π¦
cuando x=3 y=1
1
β
2
3 β 2π¦ + 1
2ππ¦
0
1
=2
1
β
2
2
1
(π¦)β2
3 β 2π¦ + 1
1
β
2
ππ¦
0
=
=
=
=
1
2
2
1
1
1
π¦ β2 2 β 2π¦
1
β
π¦ 2
1
1
β
2
2 1β π¦
1
1
π¦ β2 2β2
ππ¦
1
β
2
ππ¦
1β π¦
1
β
2
ππ¦
0
1
2
=
ππ¦
0
2
2
3 β 2π¦ β 1
0
2
2
1
β
2
0
2
2
1
β
π¦ 2
π¦ β 1/2 (1 β π¦)β 1/2 ππ¦
2 2
0
ο x=Β½
Sea x - 1 = - Β½
y β 1 = - Β½ ο y= Β½
Luego
π½
=
= π
1 1
,
2 2
πβ
π
Rta
1
=
1
Ξ 2 Ξ 2
1 1
Ξ 2 +2
1
1
Ξ 2 Ξ 2
ο
Ξ(1)
ο
Ξ
1
2
Ξ
1
2
- 7. Ejercicio especial
Resolver
1
π₯
π β1
1 β π₯ πβ1
ππ₯
π₯ + π π +π
0
π+1 π₯
π¦=
Sugerencia
π+π₯
π¦ π+ π₯ = π+1 π₯
Derivada de un cociente
π¦π + π¦π₯ = π + 1 π₯
π(π + 1 β π¦) β π¦π(β1)
π¦π = π + 1 π₯ β π¦π₯
π π + 1 β π¦ + π¦π
π¦π = π + 1 β π¦ π₯
π 2 + π β π¦π + π¦π
π¦π
π+1β π¦
π2 + π
π₯=
ππ₯ =
π(π + 1)
π π + 1 ππ¦
π+1β π¦ 2
π π π₯ = 0 β π¦ = 0
Reemplazamos
1
π¦π
π+1β π¦
0
1
0
1
0
1β
π¦π
π+1β π¦
π¦π
π+1β π¦+ π
0
1
π β1
π π π₯ = 1
πβ1
π π+1
π+1β π¦
π +π
π¦π π β1
π + 1 β π¦ β π¦π
π+1β π¦
π + 1 β π¦ π β1
π +π
π¦π + π π + 1 β π¦
π+1β π¦
1=
ππ¦
2
π + 1 β π¦ = π¦π
π + 1 = π¦π + π¦
πβ1
π π+1
π+1β π¦
2
ππ¦
π¦π π β1
π + 1 β π¦ β π¦π πβ1
π β1
π(π + 1)
π+1β π¦
π + 1 β π¦ πβ1
ππ¦
2 + π β π¦π π +π
π¦π + π
(π + 1 β π¦)2
π + 1 β π¦ π +π
π¦π
π β1
π + 1 β π¦ β π¦π πβ1 π 2 + π
π + 1 β π¦ π β1+πβ1+2
π 2 + π π +π
π + 1 β π¦ π +π
π¦π
π+1β π¦
ππ¦
π+1= π¦ π+1
π+1
= π¦
π+1
1= π¦
- 8. π¦π
1
π β1
π + 1 β π¦ β π¦π πβ1
π + 1 β π¦ π +π
π 2 + π π +π
π + 1 β π¦ π +π
0
1
π¦π
π β1
π¦π
π β1
π¦π
π β1
π + 1 β π¦ β π¦π
π 2 + π π +π
0
1
π + 1 β π¦ β π¦π
2 + π π +πβ1
π
0
1
π¦
π β1
π¦
π
π β1
π
πβ1
π
π +πβ1
1
π¦
β
π
π β1
π
0
π
=
π
π +π
ππ¦
π +π
ππ¦
πβ1
ππ¦
π + 1 β π¦ β π¦π
π + 1 π +πβ1
βπ +1
π+1β π¦
π2 + π
π + 1 β π¦ β π¦π
π π + 1 π +πβ1
π +πβ1
π
0
πβ1
πβ1
0
1
ππ¦
π + 1 β π¦ β π¦π πβ1 π 2 + π
π + 1 β π¦ π +π π 2 + π
0
1
π2 + π
πβ1
ππ¦
πβ1
ππ¦
π +πβ1βπ +1
π + 1 β π¦ β π¦π
π + 1 π +πβ1
=
π
π
πβ1
ππ¦
Como m, n, r son constantes son sacadas de la integral
π
π
1
π+1
1
π¦
π +πβ1
π β1
π + 1 β π¦ β π¦π
πβ1
ππ¦
0
π + 1 β π¦ β π¦π = π + 1 β π¦(1 + π)
= π + 1 (1 β π¦)
Nos queda entonces
π
π
π
π
π
π
1
π+1
1
π¦
π β1
π+1
πβ1
1
1β π¦
πβ1
0
1
π
π¦
0
(1 β π¦) πβ1 ππ¦
0
π + 1 πβ1
π + 1 π +πβ1
1
π+1
π β1
π¦
π +πβ1
π β1
(1 β π¦) πβ1 ππ¦
ππ¦
- 9. Si comparamos con
1
π‘ π₯β1 1 β π‘
π½ π₯, π¦ =
0
π₯β1= πβ1 β π₯ = π
π¦β1= πβ1 β π¦ = π
Reemplazamos los nuevos valores
=
1
π
π
π+1 π
π½(π, π)
β¦ Rta
π¦ β1
ππ‘ ;
π₯>0
π¦>0