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Calculo superior para ingenieros Gamma Beta
Calculo superior para ingenieros Gamma Beta
Calculo superior para ingenieros Gamma Beta
Calculo superior para ingenieros Gamma Beta
Calculo superior para ingenieros Gamma Beta
Calculo superior para ingenieros Gamma Beta
Calculo superior para ingenieros Gamma Beta
Calculo superior para ingenieros Gamma Beta
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Calculo superior para ingenieros Gamma Beta

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ejercicios de gamma y beta para ingenieros, con calculo superior

ejercicios de gamma y beta para ingenieros, con calculo superior

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  • 1. Calculo superior para ingenieros ∞ 𝑥 𝑝−1 ln 𝑥 𝑑𝑥 0 1+ 𝑥𝑦= 𝑥𝑝ln 𝑦 = 𝑙𝑛𝑥 𝑝ln 𝑦 = 𝑝 𝑙𝑛𝑥𝑒 𝑙𝑛𝑝 = 𝑒 𝑝 𝑙𝑛𝑝𝑦= 𝑒𝑝 𝑙𝑛𝑝𝑑𝑦 = 𝑒𝑝 𝑙𝑛𝑝 ∗ 𝑙𝑛𝑥𝑑𝑝𝑑𝑥 𝑝 = 𝑥 𝑝 𝑙𝑛𝑥𝑑𝑝Reemplazamos ∞ 𝑥 𝑝 ln 𝑥 𝑑𝑥 0 𝑥 1+ 𝑥 ∞ 1 𝑑 𝑥𝑝 𝑑𝑥 0 𝑥 1+ 𝑥 𝑑𝑝 ∞ 𝑑 𝑥𝑝 𝑑𝑥 0 𝑑𝑝 𝑥 1+ 𝑥 ∞ 𝑑 𝑥𝑝 𝑑𝑥𝑑𝑝 0 𝑥 1+ 𝑥 ∞ 𝑑 𝑥 𝑝−1 𝑑𝑥𝑑𝑝 0 (1 + 𝑥)Comparado con 𝑢 𝑦−1 𝑑𝑢𝛽 𝑥, 𝑦 = 1 + 𝑢 𝑥+𝑦𝑦−1= 𝑝−1 → 𝑦= 𝑝 𝑥+ 𝑦 =1 → 𝑥 =1− 𝑦
  • 2. Γ 1−y Γ p𝛽 𝑥, 𝑦 = Γ 1−y+yComo y=p Γ 1−p Γ p𝛽 𝑥, 𝑦 = Γ 1−p+p Γ 1−p Γ p𝛽 𝑥, 𝑦 = Γ 1𝛽 𝑥, 𝑦 = Γ 1 − p Γ pPor teorema de gamma tenemos que 𝜋Γ x Γ 1−x = 𝑠𝑒𝑛𝜋𝑥Por lo tanto ∞ 𝑑 𝑥 𝑝−1 𝑑𝑥𝑑𝑝 0 1+ 𝑥 𝑑 𝜋= 𝑑𝑝 𝑠𝑒𝑛 𝑝𝜋 𝑑= 𝜋 csc 𝑝𝜋 𝑑𝑝= 𝜋 ∗ 𝜋 (− csc 𝑝𝜋 ∗ cot 𝑝𝜋)= − 𝜋 2 csc 𝑝𝜋 ∗ cot 𝑝𝜋
  • 3. ∞ 𝑑𝑥 −∞ 𝑥 2 + 2𝑎𝑥 + 𝑏 2𝑥 2 + 2𝑎𝑥 + 𝑏 2 + 𝑎2 − 𝑎2𝑥 2 + 2𝑎𝑥 + 𝑎2 + 𝑏 2 − 𝑎2 2 𝑥+ 𝑎 + 𝑏 2 − 𝑎2 ∞ 𝑑𝑥 −∞ 𝑥 + 𝑎 2 + 𝑏 2 − 𝑎2Hacemos 1 𝑏 2 − 𝑎2 2 𝑦= 𝑥+ 𝑎 𝑏 2 − 𝑎2 1/2 𝑑𝑦 = 𝑑𝑥Reemplazamos ∞ 𝑏2 − 𝑎2 1/2 𝑑𝑦 −∞ 𝑏 2 − 𝑎2 𝑦 2 + 𝑏 2 − 𝑎2 1 ∞ 𝑏2 − 𝑎2 2 𝑑𝑦 −∞ 𝑏 2 − 𝑎2 𝑦 2 + 1 ∞ 1 𝑑𝑦 𝑏 2 − 𝑎2 1/2 −∞ (𝑦 2 + 1)Hacemos un corrimiento hacia la derecha para hacer un traslado a lafunción beta ∞ 2 𝑑𝑦 𝑏 2 − 𝑎2 1/2 0 𝑦2 + 1 1Sustituimos 𝑤 = 𝑦 2 → 𝑤 1/2 = 𝑦 𝑤 −1/2 𝑑𝑤 = 𝑑𝑦 2 ∞ 2 1 𝑤 − 1/2 1 ∗ 𝑑𝑤 2 𝑤+1 𝑏 2 − 𝑎2 2 0Por definición tenemos que
  • 4. 𝑢 𝑦−1 𝑑𝑢𝛽 𝑥, 𝑦 = 1 + 𝑢 𝑥+𝑦Hacemos la analogía y: 1𝑥−1= − 𝑥+ 𝑦=1 2 1 1𝑥= 𝑦= 2 2Luego entonces 1 1 1 1 𝛽 , 2 2 𝑏 2 − 𝑎2 2 1 1 1 Γ Γ 2 2 1 1 1 𝑏 2 − 𝑎2 2 Γ + 2 2 1 1 1 1 Γ Γ 2 2 𝑏 2 − 𝑎2 2 1 1 𝜋∗ 𝜋 𝑏2 − 𝑎2 2 𝜋= 1 𝑏2 − 𝑎2 2…Rta
  • 5. ∞ 𝑒 𝑛𝑥 𝑛+1 𝑥 𝑑𝑥 −∞ 𝑎𝑒 + 𝑏𝑢 = 𝑒 (𝑛+1)𝑥 ln 𝑢 = ln 𝑒 (𝑛 +1)𝑥 ln 𝑢 1 𝑑𝑢ln 𝑢 = 𝑛+1 𝑥 → = 𝑥 → 𝑑𝑥 = 𝑛+1 𝑛+1 𝑢Reemplazamos los nuevos valores en la integral y los limitescorrespondientes ln 𝑢 ∞ 𝑛 𝑒 𝑛 +1 1 𝑑𝑢 0 𝑎𝑢 + 𝑏 𝑛+1 𝑢 𝑛 ln 𝑢 1 ∞ 𝑒 𝑛 +1 ∗ 𝑢 −1 0 𝑑𝑢𝑛 +1 𝑎𝑢 +𝑏Por propiedades de los logaritmos y euler 𝑛 1 −1 = − 𝑛+1 𝑛+1 𝑛 ∞ 1 𝑢 𝑛+1 ∗ 𝑢−1 𝑑𝑢𝑛+1 0 𝑎𝑢 + 𝑏 −1 ∞ 1 𝑢 𝑛+1 𝑑𝑢𝑛+1 0 𝑎𝑢 + 𝑏Realizamos otra sustitución de tal manera que 𝑗𝑏 𝑏𝑗𝑏 = 𝑎𝑢 → = 𝑢 → 𝑑𝑗 = 𝑑𝑢 𝑎 𝑎 −1 𝑗𝑏 𝑛+1 ∞ 1 𝑎 𝑏 ∗ 𝑑𝑗𝑛+1 0 𝑗𝑏 + 𝑏 𝑎 −1 𝑗𝑏 𝑛+1 −1 ∞ 1 𝑎 𝑏𝑛+1 ∗ 𝑑𝑗𝑛+1 0 𝑏 𝑗+1 𝑎
  • 6. −1 −1 𝑗 𝑛+1 ∗ 𝑏 𝑛+1 𝑏 −1 ∗ ∞ 𝑎 1 𝑎 𝑛+1 𝑑𝑗𝑛+1 0 𝑏 𝑗+1 −1 −1 +1 𝑗 𝑛+1 ∗ 𝑏 𝑛+1 −1 ∞ +1 1 𝑎 𝑛+1 𝑑𝑗 𝑛+1 0 𝑏 𝑗+1 −1 𝑛 𝑗 𝑛+1 ∗ 𝑏 𝑛+1 𝑛 ∞ 1 𝑎 𝑛+1 𝑑𝑗 𝑛+1 0 𝑏 𝑗+1Trasponemos términos −1 𝑛 ∞ 1 𝑗 𝑛+1 ∗ 𝑏 𝑛+1 𝑛 𝑑𝑗𝑛+1 0 𝑏 𝑗+1 ∗ 𝑎 𝑛+1 −1 𝑛 ∞ –1 1 𝑗 𝑛+1 ∗ 𝑏 𝑛+1 𝑛 𝑑𝑗 𝑎 𝑛+1 ∗ 𝑛+1 0 𝑗+1 −1 1 ∞ − 1 𝑗 𝑛+1 ∗ 𝑏 𝑛 +1 𝑛 𝑑𝑗 𝑎 𝑛+1 ∗ 𝑛+1 0 𝑗+1 −1 ∞ 1 𝑗 𝑛+1 𝑛 1 𝑑𝑗 0 𝑗+1 𝑎 𝑛+1 ∗ 𝑏 𝑛+1 ∗ (𝑛 + 1)Si comparamos con 𝑢 𝑦−1 𝑑𝑢𝛽 𝑥, 𝑦 = 𝑢 + 1 𝑥+𝑦 1 1 𝑛𝑦−1=− → 𝑦=− +1 → 𝑦 = 𝑛+1 𝑛+1 𝑛+1
  • 7. 𝑛 1 𝑥+ 𝑦=1 → 𝑥 =1− 𝑦 → 𝑥 =1− → 𝑥= 𝑛+1 𝑛+1 1 1 𝑛 𝑛 1 𝛽 , 𝑛+1 𝑛+1 𝑎 𝑛+1 ∗ 𝑏 𝑛+1 ∗ (𝑛 + 1) 1 𝑛 1 Γ Γ 𝑛+1 𝑛+1 𝑛 1 1 𝑛 𝑎 𝑛+1 ∗ 𝑏 𝑛+1 ∗ (𝑛 + 1) Γ 𝑛 + 1 + 𝑛 + 1 1 𝑛 1 Γ Γ 𝑛+1 𝑛+1 𝑛 1 𝑛+1 𝑎 𝑛+1 ∗ 𝑏 𝑛+1 ∗ 𝑛+1 Γ 𝑛+1 1 1 𝑛 𝑛 1 Γ Γ 𝑛+1 𝑛+1 𝑎 𝑛+1 ∗ 𝑏 𝑛+1 ∗ 𝑛+1 1 1 1 𝑛 1 Γ Γ 1− 𝑛+1 𝑛+1 𝑎 𝑛+1 ∗ 𝑏 𝑛+1 ∗ (𝑛 + 1)Aplicamos ahora el teorema de gamma de tal manera que 𝜋Γ x Γ 1−x = 𝑠𝑒𝑛𝜋𝑥 1 𝜋 𝑛 1 ∗ 𝜋 𝑅𝑡𝑎 … 𝑎 𝑛+1 ∗ 𝑏 𝑛+1 ∗ (𝑛 + 1) 𝑠𝑒𝑛 𝑛+1
  • 8. 1 𝑚 𝑛 𝑥 ln 𝑥 𝑑𝑥0−𝑢 = ln 𝑥𝑒 −𝑢 = 𝑥−𝑒 −𝑢 𝑑𝑢 = 𝑑𝑥𝑒 −𝑚𝑢 = 𝑥 𝑚𝑠𝑖 𝑥 = 0 → 𝑢 = ∞ 𝑠𝑖 𝑥 = 1 → 𝑢=0 0 𝑒 −𝑚𝑢 – 𝑢 𝑛 – 𝑒 −𝑢 𝑑𝑢∞ 0− 𝑒 −𝑢(𝑚 +1) ∗ 𝑢 𝑛 𝑑𝑢 ∞ 𝑑𝑡𝑡 = 𝑢 𝑚+1 → 𝑑𝑡 = 𝑚 + 1 𝑑𝑢 → = 𝑑𝑢 𝑚+1 𝑡 𝑛 𝑡𝑛 = 𝑢 → 𝑢 =𝑚+1 (𝑚 + 1) 𝑛 ∞ −𝑡 𝑡𝑛 𝑑𝑡− 𝑒 ∗ 𝑛 ∗ 0 𝑚+1 𝑚+1 ∞ 1− 𝑒 −𝑡 ∗ 𝑡 𝑛 𝑑𝑡 (𝑚 + 1)(𝑚 + 1) 𝑛 0 ∞ 1− 𝑒 −𝑡 ∗ 𝑡 𝑛 𝑑𝑡 (𝑚 + 1) 𝑛+1 0𝑥−1= 𝑛 → 𝑥 = 𝑛+1 1− 𝑛+1 Γ 𝑛+1 𝑚+1 −1 𝑛 𝑛!= 𝑅𝑡𝑎 … God bless (𝑚 +1) 𝑛 +1

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