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# Betta

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### Betta

1. 1. FunciΓ³n Beta 1 π‘ π₯β1 (1 β π‘) π¦β1 ππ‘ ; π½ π₯, π¦ = π₯>0 π¦>0 0 Si hacemos π‘ = π ππ2 π ππ‘ = 2 π ππ π cos π ππ Si reemplazamos limites π‘ = 0 β π = 0 π‘=1 β π= π 2 Reemplazamos π 2 π½ π₯, π¦ = 2 (π ππ2 π) π₯β1 1 β π ππ2 π¦β1 π ππ π cos π ππ 0 π 2 π½ π₯, π¦ = 2 π ππ2π₯β1 π β cos2yβ1 π ππ 0 π 2 1 π½ π₯, π¦ = 2 π ππ2π₯β1 π β cos2yβ1 π ππ 0 Si hacemos π‘= 1 1+ π’ ππ’ 1+ π’ ππ‘ = 0 π½ π₯, π¦ = β β β π½ π₯, π¦ = 0 β π½ π₯, π¦ = 0 2 π π π‘ = 0 β π’ = β 1 1+ π’ π₯β1 π¦ π π 1 1β 1+ π’ π‘=1β0=0 π¦β1 1 1+ π’ π₯β1 1 1+ π’ π’ π¦ β1 ππ’ β β π₯β1 1 + π’ π¦β1 1 + π’ 1+ π’β1 1+ π’ π’ π¦ β1 ππ’ π½ π₯, π¦ = 1 + π’ π₯β1+π¦β1+2 π’ π¦β1 ππ’ π½ π₯, π¦ = 1 + π’ π₯+π¦ π¦β1 ππ’ 1+ π’ ππ’ 1+ π’ 2 2 2
2. 2. Teorema π½ π₯, π¦ = ΞxΞy ; Ξ x+y π₯>0 π¦>0 Ejemplo π 2 tan π ππ 0 π/2 0 π ππ π cos π 1/2 ππ π/2 π ππ 1/2 π πππ  β1/2 π ππ 0 Comparando 1 π½ π₯, π¦ = 2 2π₯ β 1 = 1 2 2π¦ β 1 = β π 2 π ππ2π₯β1 π β cos2yβ1 π ππ 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 βΞ 1 4 4 Ξ 3+1 4 4 Ξ 3 βΞ 1 4 4 4 ; ππππ Ξ =Ξ 1 =1 4 4 Ξ 4 π= π π
3. 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 β1 0 1+π₯ Resolver ππ₯ Por definiciΓ³n π½ π₯, π¦ = π’ π¦ β1 ππ’ 1+π’ π₯ +π¦ Comparando y-1=p-1 x+y=1 y=p x=1βp Reemplazamos β 0 π₯ πβ1 ππ₯ = π½ 1 β π, π 1+ π₯ = π½ π, 1 β π
4. 4. = Ξ p Ξ 1βp Ξ p+1βp = Ξ p Ξ 1βp Aplicamos teorema de gamma = π π ππ ππ Resolver β ββ π 2π₯ π 3π₯ + 1 2 ππ₯ π’ = π 3π₯ β ln π’ = ln π 3π₯ β ln π’ = 3π₯ π₯= 1 ln π’ β 3 1 ππ’ 3 ππ₯ ππ₯ = Evaluamos los lΓ­mites Cuando π₯ = β β β 0 1 = 3 π’=β 1 2β ln π’ π 3 π’+1 1 ππ’ 3 π’ 2 π 3 ln π’ β 0 β 2 π¦ π₯ = ββ β π’ π’+1 2 ππ’ Por propiedades de euler y logaritmos 1 = 3 1 = 3 β 0 β 0 2 π’3 β π’β1 ππ’ π’+1 2 β1 π’3 π’+1 2 ππ’ π’=0
5. 5. Si comparamos con π¦β1= β 1 3 π½ π₯, π¦ = β Reemplazamos 1 4 2 π½ , 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 βΞ Ξ 9 3 3 = 1 1 1 βΞ 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 ππ’ 1+π’ π₯ +π¦ π¦= β π₯+ π¦ =2 β π₯ =2β = π’ π¦ β1 1 π +1β π= 3 π β π= π π
6. 6. Resolver 3 ππ₯ π₯β1 1 3 3β π₯ 1 β 2 π₯β1 3β π₯ 1 β 2 ππ₯ 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 2ππ¦ 0 1 =2 1 β 2 2 1 (π¦)β2 3 β 2π¦ + 1 1 β 2 ππ¦ 0 = = = = 1 2 2 1 1 1 π¦ β2 2 β 2π¦ 1 β π¦ 2 1 1 β 2 2 1β π¦ 1 1 π¦ β2 2β2 ππ¦ 1 β 2 ππ¦ 1β π¦ 1 β 2 ππ¦ 0 1 2 = ππ¦ 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 ππ¦ 2 2 0 ο x=Β½ Sea x - 1 = - Β½ y β 1 = - Β½ ο  y= Β½ Luego π½ = = π 1 1 , 2 2 πβ 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. 7. Ejercicio especial Resolver 1 π₯ π β1 1 β π₯ πβ1 ππ₯ 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(β1) 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¦ π2 + π π₯= ππ₯ = π(π + 1) π π + 1 ππ¦ 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 β1 π π π₯ = 1 πβ1 π π+1 π+1β π¦ π +π π¦π π β1 π + 1 β π¦ β π¦π π+1β π¦ π + 1 β π¦ π β1 π +π π¦π + π π + 1 β π¦ π+1β π¦ 1= ππ¦ 2 π + 1 β π¦ = π¦π π + 1 = π¦π + π¦ πβ1 π π+1 π+1β π¦ 2 ππ¦ π¦π π β1 π + 1 β π¦ β π¦π πβ1 π β1 π(π + 1) π+1β π¦ π + 1 β π¦ πβ1 ππ¦ 2 + π β π¦π π +π π¦π + π (π + 1 β π¦)2 π + 1 β π¦ π +π π¦π π β1 π + 1 β π¦ β π¦π πβ1 π 2 + π π + 1 β π¦ π β1+πβ1+2 π 2 + π π +π π + 1 β π¦ π +π π¦π π+1β π¦ ππ¦ π+1= π¦ π+1 π+1 = π¦ π+1 1= π¦
8. 8. π¦π 1 π β1 π + 1 β π¦ β π¦π πβ1 π + 1 β π¦ π +π π 2 + π π +π π + 1 β π¦ π +π 0 1 π¦π π β1 π¦π π β1 π¦π π β1 π + 1 β π¦ β π¦π π 2 + π π +π 0 1 π + 1 β π¦ β π¦π 2 + π π +πβ1 π 0 1 π¦ π β1 π¦ π π β1 π πβ1 π π +πβ1 1 π¦ β π π β1 π 0 π = π π +π ππ¦ π +π ππ¦ πβ1 ππ¦ π + 1 β π¦ β π¦π π + 1 π +πβ1 βπ +1 π+1β π¦ π2 + π π + 1 β π¦ β π¦π π π + 1 π +πβ1 π +πβ1 π 0 πβ1 πβ1 0 1 ππ¦ π + 1 β π¦ β π¦π πβ1 π 2 + π π + 1 β π¦ π +π π 2 + π 0 1 π2 + π πβ1 ππ¦ πβ1 ππ¦ π +πβ1βπ +1 π + 1 β π¦ β π¦π π + 1 π +πβ1 = π π πβ1 ππ¦ Como m, n, r son constantes son sacadas de la integral π π 1 π+1 1 π¦ π +πβ1 π β1 π + 1 β π¦ β π¦π πβ1 ππ¦ 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 β1 π+1 πβ1 1 1β π¦ πβ1 0 1 π π¦ 0 (1 β π¦) πβ1 ππ¦ 0 π + 1 πβ1 π + 1 π +πβ1 1 π+1 π β1 π¦ π +πβ1 π β1 (1 β π¦) πβ1 ππ¦ ππ¦
9. 9. Si comparamos con 1 π‘ π₯β1 1 β π‘ π½ π₯, π¦ = 0 π₯β1= πβ1 β π₯ = π π¦β1= πβ1 β π¦ = π Reemplazamos los nuevos valores = 1 π π π+1 π π½(π, π) β¦ Rta π¦ β1 ππ‘ ; π₯>0 π¦>0