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INTEGRAL
EULERIANA |
BETA |
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(Integral impropia de
2ª especie) |
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Dependen de dos parámetros p y q > 0 y no de la variable de integración, con la expresión: |
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∫ |
1 |
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Impropia: p y q < 1 (Racional discontinua con asintotas en 0 y
1) |
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| 1) |
β(p,q) |
= |
xp-1 (1-x)q-1 dx |
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| Propia…: p y q ≥ 1 (Polinomio
continuo en un intervalo finito) |
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0 |
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OTRAS EXPRESIONES DE LA
BETA |
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∫ |
π/2 |
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| 2) |
β(p,q) |
= |
2
sen2p-1t cos2q-1t dt |
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(Realizar el cambio x = sen2t en la función Beta) |
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0 |
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∫ |
∞ |
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| 3) |
β(p,q) |
= |
tp-1 / (1+t)p+q dt |
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(Realizar el cambio x = t / (1+t) en la
función Beta) |
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0 |
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PROPIEDADES |
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∫ |
1 |
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| 1) |
β(p,1) |
= |
1/p |
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xp-1 dx = |
xp/p |
1 |
= |
1/p |
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0 |
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0 |
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∫ |
1 |
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∫ |
1 |
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| 2) |
β(p,q) |
= |
β(q,p) |
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xp-1 (1-x)q-1 dx |
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(1-t)p-1 tq-1 dt |
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β(q,p) |
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dx= -dt |
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Simetria de la Beta |
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0 |
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0 |
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La integral definida representa el área en valor
absoluto. |
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∫ |
1 |
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∫ |
1 |
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| 3) |
β(p,q) |
= |
(q-1)/p · β(p+1,q-1) |
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xp-1 (1-x)q-1dx = |
xp (1-x)q-1 / p |
1 |
+ (p-1)/p |
xp(1-x)q-2dx |
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0 |
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si p,q>1 , p,q € R+ |
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0 |
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Ley de recurrencia |
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(1-x)q-1 |
= |
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(-1)(q-1)(1-x)q-2dx |
= |
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Por partes |
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de la Beta |
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xp-1 dx |
= |
dv |
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xp/p |
= |
v |
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∫ |
π/2 |
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| 4) |
β(1/2,1/2) |
= |
π |
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2 dt = |
2t |
π/2 |
= |
π |
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(Usar la segunda expresión de la Beta) |
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0 |
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0 |
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| 5) |
β(p,q) |
= |
Γ(p)·Γ(q)
/ Γ(p+q) |
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Si: p,q>0 |
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(Relación entre la
Gamma y la Beta) |
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| 6) |
Γ(1/2) |
= |
√π |
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β(1/2,1/2) |
= |
Γ(1/2)·Γ(1/2)
/ Γ(1/2+1/2) |
= |
Γ2(1/2) |
= |
π |
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EJERCICIOS |
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∫ |
1 |
x |
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∫ |
1 |
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(p-1=1/2 , q-1=-1/2) |
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(prop:5) |
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| 1) |
√( |
)dx |
= |
x1/2(1-x)-1/2 |
dx |
= β(3/2,1/2) = Γ(3/2) Γ(1/2) / Γ(2) = 1/2 Γ(1/2)
Γ(1/2) = π/2 |
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| 1-x |
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0 |
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0 |
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∫ |
3 |
3-x |
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∫ |
3 |
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∫ |
1 |
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∫ |
1 |
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(p-1=4 , q-1=5) |
(prop:5) |
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| 2) |
x4( |
)5dx = |
x4(1-x/3)5dx |
= |
(3t)4(1-t)53dt |
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35 |
t4(1-t)5dt = 35 β(5,6) = 27/140 |
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| 3 |
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0 |
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cambio de variables |
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limites |
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x/3 = t |
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3=3t |
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1/3 dx = dt |
, dx = 3dt |
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0=3t |
, t=0 |
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∫ |
1 |
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∫ |
1 |
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∫ |
1 |
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(p-1=-1/2 , q-1=-1/2) |
(prop:4) |
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| 3) |
x (1-x4)-1/2dx |
= |
t1/4(1-t)-1/2 1/4 t-3/4 dt = |
1/4 |
t-1/2(1-t)-1/2dt = 1/4
β(1/2,1/2) = π/4 |
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0 |
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0 |
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cambio de variables |
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limites |
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x4 = t |
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x =
t1/4 |
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1=t1/4 |
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4x3 dx = dt |
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dx = 1/4 t -3/4dt |
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0=t1/4 |
, t=0 |
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