Funcion beta

contiene ejercicios de la función beta para ingenieros y carreras a fines

1. Función Beta
?????? �,� = �
�−1
1
0
(1−�)
�−1
�� ; �>0 �>0
Si hacemos �=���
2
?????? ��=2 ��� ??????cos?????? �??????
Si reemplazamos limites �=0 → ??????=0 �=1 → ??????=
??????
2

Reemplazamos
?????? �,� =2 (���
2
??????)
�−1
??????
2
0
1−���
2

�−1
��� ??????cos?????? �??????
?????? �,� =2 ���
2�−1
??????∗
??????
2
0
cos
2y−1
?????? �??????
1
2
?????? �,� = ���
2�−1
??????∗
??????
2
0
cos
2y−1
?????? �??????
Si hacemos
�=
1
1+�
��=
��
1+�
2
�?????? �=0→�=∞ � �?????? �=1→0=0
?????? �,� =−
1
1+�

�−1

0
∞
1−
1
1+�

�−1
��
1+�
2

?????? �,� =
1
1+�

�−1

∞
0

1+�−1
1+�

�−1
��
1+�
2

?????? �,� =
1
1+�
�−1
∗
∞
0
�
�−1
1+�
�−1
∗
��
1+�
2

?????? �,� =
�
�−1
��
1+�
�−1+�−1+2

?????? �,� =
�
�−1
��
1+�
�+�

2. Teorema
?????? �,� =
Γ x Γ y
Γ x+y
; �>0 �>0

Ejemplo
tan??????
??????
2
0
�??????

��� ??????
cos??????

1/2??????/2
0
�??????
���
1/2
?????? ���
−1/2
??????
??????/2
0
�??????
Comparando
1
2
?????? �,� = ���
2�−1
??????∗
??????
2
0
cos
2y−1
?????? �??????
2�−1=
1
2
→ 2�=
1
2
+1 → 2�=
3
2
→ �=
�
�

2�−1= −
1
2
→ 2�= −
1
2
+1 → 2�=
1
2
→ �=
�
�

Si aplicamos el teorema
=
1
2
∗
Γ
3
4
∗ Γ
1
4

Γ
3
4
+
1
4


=
1
2
∗
Γ
3
4
∗ Γ
1
4

Γ
4
4

;���� Γ
4
4
=Γ 1 =1

3. =
1
2
∗ Γ
3
4
∗ Γ
1
4

=
1
2
∗ Γ
1
4
∗ Γ 1−
1
4

Aplicamos teorema de gamma
=
1
2
∗
π
sen
π
4


=
1
2

??????
2
2

=
??????
2


Resolver
�
�−1
1+�
∞
0
��
Por definición
?????? �,� =
�
�−1
��
1+�
�+�

Comparando
y - 1 = p - 1
x + y = 1
y = p
x = 1 – p
Reemplazamos

�
�−1
1+�
∞
0
��= ?????? 1−�,�
= ?????? �,1−�

4. =
Γ p Γ 1−p
Γ p+1−p

=Γ p Γ 1−p
Aplicamos teorema de gamma
=
??????
��� �??????


Resolver

�
2�
�
3�
+1
2
��
∞
−∞

�=�
3�
→ln�=ln �
3�
→ ln� = 3�
�=
1
3
ln� → ��=
1
3
��
��

Evaluamos los límites
Cuando �=∞→ �=∞ � �=−∞ → �=0

�
2∗
1
3
ln�
�+1
2
∞
0
∗
1
3

��
�

=
1
3

�
2
3
ln�
� �+1
2
∞
0
��
Por propiedades de euler y logaritmos
=
1
3

�
2
3∗ �
−1
�+1
2
∞
0
��
=
1
3

�
−1
3
�+1
2
∞
0
��

5. Si comparamos con ?????? �,� =
�
�−1
��
1+�
�+�


�−1= −
1
3
→ �= −
1
3
+1→�=
�
�

�+�=2 → �=2−
2
3
→�=
�
�

Reemplazamos
1
3
??????
4
3
,
2
3

=
1
3
∗
Γ
4
3
Γ
2
3

Γ
4
3
+
2
3


=
1
3
∗

1
3
Γ
1
3
Γ
2
3

Γ
6
3


=
1
9
∗
Γ
1
3
Γ
2
3

Γ(2)

Γ 2 =1!
=
1
9
∗Γ
1
3
Γ
2
3

=
1
9
∗Γ
1
3
Γ 1−
1
3

Aplicamos teorema de gamma
=
1
9
∗
π
sen
π
3

=
1
9
∗
π
3
2

=
2
9
∗
π
3

6. Resolver

��
�−1 3−�
3
1

�−1
−
1
2 3−�
−
1
2
3
1
��
Sea x – 1 = 2y  x = 2y+1  dx = 2dy
Cuando x = 1 y = 0 cuando x=3 y=1
= 2�
−
1
2 3− 2�+1
−
1
2
1
0
2��
=2 2
−
1
2 (�)
−
1
2 3− 2�+1
−
1
2
1
0
��
=
2
2
�
−
1
2 3−2�−1
−
1
2
1
0
��
=
2
2
�
−
1
2 2−2�
−
1
2
1
0
��
=
2
2
�
−
1
2 2 1−�
−
1
2
1
0
��
=
2
2
�
−
1
2 2
−
1
2 1−�
−
1
2
1
0
��
=
2
2 2
�
− 1/2
(1−�)
− 1/2
1
0
��
Sea x - 1 = - ½  x = ½ y – 1 = - ½  y= ½
Luego
??????
1
2
,
1
2
=
Γ
1
2
Γ
1
2

Γ
1
2
+
1
2


Γ
1
2
Γ
1
2

Γ(1)
 Γ
1
2
Γ
1
2

= ??????∗ ??????
= ?????? Rta

7. Ejercicio especial
Resolver

�
�−1
1−�
�−1
�+�
�+�
1
0
��
Sugerencia �=
�+1 �
�+�

� �+� = �+1 �
��+��= �+1 �
��= �+1 �−��
��= �+1−� �
�=
��
�+1−�

��=
� �+1 ��
�+1−�
2


Reemplazamos



��
�+1−�

�−1
1−
��
�+1−�

�−1

��
�+1−�
+�
�+�
1
0

� �+1
�+1−�
2
��

��
�−1
�+1−�
�−1

�+1−�−��
�+1−�

�−1

��+� �+1−�
�+1−�

�+�
1
0

� �+1
�+1−�
2
��

��
�−1
�+1−�
�−1

�+1−�−��
�−1
�+1−�
�−1
��+�
2
+�−��
�+�
�+1−�
�+�
1
0

�(�+1)
(�+1−�)
2
��

��
�−1
�+1−�−��
�−1
�
2
+�
�+1−�
�−1+�−1+2
�
2
+�
�+�
�+1−�
�+�
1
0
��
�(�+1−�)−��(−1)
� �+1−� +��
�
2
+�−��+��
�
2
+�
�(�+1)

Derivada de un cociente






�?????? �=0 → �=0
�?????? �=1
1=
��
�+1−�

�+1−�=��
�+1=��+�
�+1=� �+1
�+1
�+1
=�
1=�

8. ��
�−1
�+1−�−��
�−1
�
2
+�
�+1−�
�+�
�
2
+�
�+�
�+1−�
�+�
1
0
��

��
�−1
�+1−�−��
�−1
�
2
+� �+1−�
�+�
�+1−�
�+�
�
2
+�
�+�
1
0
��

��
�−1
�+1−�−��
�−1
�
2
+�
�
2
+�
�+�
1
0
��

��
�−1
�+1−�−��
�−1

�
2
+�
�+�−1
1
0
��

�
�−1
�
�−1
�+1−�−��
�−1

� �+1
�+�−1
1
0
��

�
�−1
�
�−1
�+1−�−��
�−1

�
�+�−1
�+1
�+�−1
1
0
��

�
�+�−1
∗ �
−�+1
= �
�+�−1−�+1
= �
�


�
�−1
�+1−�−��
�−1

�
�
�+1
�+�−1
1
0
��

Como m, n, r son constantes son sacadas de la integral
1
�
�
�+1
�+�−1
�
�−1
�+1−�−��
�−1
1
0
��
�+1−�−�� = �+1 −�(1+�)
= �+1 (1−�)
Nos queda entonces
1
�
�
�+1
�+�−1
�
�−1
�+1
�−1
(1−�)
�−1
1
0
��
�+1
�−1
�
�
�+1
�+�−1
�
�−1
1−�
�−1
1
0
��
1
�
�
�+1
�
�
�−1
(1−�)
�−1
1
0
��

9. Si comparamos con
?????? �,� = �
�−1
1
0
1−�
�−1
�� ; �>0 �>0
�−1=�−1 → �=�
�−1=�−1 → �=�
Reemplazamos los nuevos valores

=
1
�
�
�+1
�
??????(�,�) … Rta