INTEGRAL EULERIANA SOLUCIONES 3
  INTEGRALES GAMMA
∫ ∞ (p-1=5/2)
1) e-x x5/2dx = Γ(7/2) = 5/2·3/2·1/2·Γ(1/2) = 15/8 √π
0
∫ ∞ -x3 Cambio Var. ∫ ∞ (p-1=0)
2) e x2 dx (x3 = t) 1/3 e-t dt = 1/3 · Γ(1) = 1/3
0 0
∫ ∞ -x2 Cambio Var. ∫ ∞ (p-1=1/2)
3) e x2 dx (x2 = t) 1/2 e-t t1/2 dt = 1/2 · Γ(3/2) = 1/2·1/2·Γ(1/2) = 1/4 √π
0 0
∫ ∞ Cambio Var. ∫ ∞ (p-1=8)
4) e-4x x8 dx (4x = t) 1/49 e-t t8 dt = 1/49·Γ(9) = 8! / 49
0 0
∫ ∞ Cambio Var. ∫ ∞ (p-1=7/2) 7·5·3·1 Γ(1/2) 35 · √π
5) e-3x x7/2 dx (3x = t) 1/39/2 e-t t7/2 dt = 1/39/2·Γ(9/2) = =
2·2·2·2 39/2 16 · 37/2
0 0
√x ∫ ∞ (p-1=1/3)
∫ ∞ e Cambio Var. 2 e-t t1/3 dt = 2 · Γ(4/3) = 2/3 · Γ(1/3)
6)   (√x = t)
3√x 0
0
  INTEGRALES BETA
∫ 1 (p-1=2 , q-1=3) Γ(3) · Γ(4) 2 · Γ(4) 2 1
7) x2 (1-x)3 dx = β(3,4) = = = =
Γ(7) 6·5·4·Γ(4) 6·5·4 60
0
∫ 1 (p-1=1/3 , q-1=2) Γ(4/3) · Γ(3) Γ(4/3) · 2 27
8) x1/3 (1-x)2 dx = β(4/3,3) = = =
Γ(13/3) 10/3·7/3·4/3·Γ(4/3) 140
0
∫ 1 (p-1=3 , q-1=1/2) Γ(4) · Γ(3/2) 3! · Γ(3/2) 32
9) x3 √(1-x) dx = β(4,3/2) = = =
Γ(11/2) 9/2·7/2·5/2·3/2·Γ(3/2) 315
0
∫ 1 1-x (p-1=-1/2 , q-1=1) Γ(1/2) · Γ(2) Γ(1/2) 4
10) dx = β(1/2,2) = = =
√x Γ(5/2) 3/2·1/2·Γ(1/2) 3
0
∫ π/2 (2p-1=5 , 2q-1=2) Γ(3) · Γ(3/2) 2 · Γ(3/2) 8
11) sen5x cos2x  dx = 1/2 · β(3,3/2) = = =
2 · Γ(9/2) 2 ·7/2·5/2·3/2·Γ(3/2) 105
0
∫ π/2 (2p-1=0 , 2q-1=3/2) Γ(1/2) · Γ(5/4) √π · 1/4 · Γ(1/4) √π · Γ(1/4)
12) 2 √ cos3x dx = β(1/2,5/4) = = =
Γ(7/4) 3/4 · Γ(3/4) 3 · Γ(3/4)
0
∫ 1 Cambio Var. ∫ 1 (p-1=1/2 , q-1=-1/2) Γ(3/2)·Γ(1/2) π
13) x5 (1-x4)-1/2 dx (x4 = t) 1/4 t1/2(1-t)-1/2dt = 1/4·β(3/2,1/2) = =
4·Γ(2) 8
0 0
∫ 8 Cambio Var. ∫ 1 (p-1=1/2 , q-1=1/4) 211/4 · √π · Γ(1/4)
14) x-1/2(2-x1/3)1/4dx (x1/3=2t) 6·23/4 t1/2(1-t)1/4dt = 6·23/4·β(3/2,5/4) =
7 · Γ(3/4)
0 0
∫ 0 Cambio Var. ∫ 1 (p-1=6 , q-1=2) Γ(7)·Γ(3) Γ(7) · 2 1
15) (x+1)6 x2 dx (x+1 = t) t6 (1-t)2 dt = β(7,3) = = =
Γ(10) 9·8·7·Γ(7) 252
-1 0
∫ 1 ∫ π/2 ∫ π/2
16) xp-1 (1-x)q-1 dx = sen2p-2t · cos2q-2t · 2 · sen t · cos t · dt = 2  sen2p-1t  cos2q-1t  dt
0 0 0
x = sen2t ;  1-x = 1-sen2t ;  1-x = cos2t ;  dx = 2 sen t cos t dt  ; Si x=1 => t=π/2  ; Si x=0 => t=0
∫ 1 ∫ ∞   t p-1  1    dt ∫ ∞ t p-1
17) xp-1 (1-x)q-1 dx =   = dt
(1+t)p-1 (1+t)q-1 (1+t)2 (1+t)p+q
0 0 0
x = t / (1+t)  ;  x = 1-(1+t)-1  ;  1-x = (1+t)-1  ;  dx = (t+1)-2dt  ; Si x=1 => t=∞  ; Si x=0 => t=0