Upcoming SlideShare
×

# Calculo superior para ingenieros Gamma Beta

1,344 views

Published on

ejercicios de gamma y beta para ingenieros, con calculo superior

0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Es un placer poderles servir..

Are you sure you want to  Yes  No
• muchas gracias me has salvado

Are you sure you want to  Yes  No
• Be the first to like this

Views
Total views
1,344
On SlideShare
0
From Embeds
0
Number of Embeds
43
Actions
Shares
0
32
2
Likes
0
Embeds 0
No embeds

No notes for slide

### Calculo superior para ingenieros Gamma Beta

1. 1. Calculo superior para ingenieros β π₯ πβ1 ln π₯ ππ₯ 0 1+ π₯π¦= π₯πln π¦ = πππ₯ πln π¦ = π πππ₯π πππ = π π ππππ¦= ππ πππππ¦ = ππ πππ β πππ₯ππππ₯ π = π₯ π πππ₯ππReemplazamos β π₯ π ln π₯ ππ₯ 0 π₯ 1+ π₯ β 1 π π₯π ππ₯ 0 π₯ 1+ π₯ ππ β π π₯π ππ₯ 0 ππ π₯ 1+ π₯ β π π₯π ππ₯ππ 0 π₯ 1+ π₯ β π π₯ πβ1 ππ₯ππ 0 (1 + π₯)Comparado con π’ π¦β1 ππ’π½ π₯, π¦ = 1 + π’ π₯+π¦π¦β1= πβ1 β π¦= π π₯+ π¦ =1 β π₯ =1β π¦
2. 2. Ξ 1βy Ξ pπ½ π₯, π¦ = Ξ 1βy+yComo y=p Ξ 1βp Ξ pπ½ π₯, π¦ = Ξ 1βp+p Ξ 1βp Ξ pπ½ π₯, π¦ = Ξ 1π½ π₯, π¦ = Ξ 1 β p Ξ pPor teorema de gamma tenemos que πΞ x Ξ 1βx = π ππππ₯Por lo tanto β π π₯ πβ1 ππ₯ππ 0 1+ π₯ π π= ππ π ππ ππ π= π csc ππ ππ= π β π (β csc ππ β cot ππ)= β π 2 csc ππ β cot ππ
3. 3. β ππ₯ ββ π₯ 2 + 2ππ₯ + π 2π₯ 2 + 2ππ₯ + π 2 + π2 β π2π₯ 2 + 2ππ₯ + π2 + π 2 β π2 2 π₯+ π + π 2 β π2 β ππ₯ ββ π₯ + π 2 + π 2 β π2Hacemos 1 π 2 β π2 2 π¦= π₯+ π π 2 β π2 1/2 ππ¦ = ππ₯Reemplazamos β π2 β π2 1/2 ππ¦ ββ π 2 β π2 π¦ 2 + π 2 β π2 1 β π2 β π2 2 ππ¦ ββ π 2 β π2 π¦ 2 + 1 β 1 ππ¦ π 2 β π2 1/2 ββ (π¦ 2 + 1)Hacemos un corrimiento hacia la derecha para hacer un traslado a lafunciΓ³n beta β 2 ππ¦ π 2 β π2 1/2 0 π¦2 + 1 1Sustituimos π€ = π¦ 2 β π€ 1/2 = π¦ π€ β1/2 ππ€ = ππ¦ 2 β 2 1 π€ β 1/2 1 β ππ€ 2 π€+1 π 2 β π2 2 0Por definiciΓ³n tenemos que
4. 4. π’ π¦β1 ππ’π½ π₯, π¦ = 1 + π’ π₯+π¦Hacemos la analogΓ­a y: 1π₯β1= β π₯+ π¦=1 2 1 1π₯= π¦= 2 2Luego entonces 1 1 1 1 π½ , 2 2 π 2 β π2 2 1 1 1 Ξ Ξ 2 2 1 1 1 π 2 β π2 2 Ξ + 2 2 1 1 1 1 Ξ Ξ 2 2 π 2 β π2 2 1 1 πβ π π2 β π2 2 π= 1 π2 β π2 2β¦Rta
5. 5. β π ππ₯ π+1 π₯ ππ₯ ββ ππ + ππ’ = π (π+1)π₯ ln π’ = ln π (π +1)π₯ ln π’ 1 ππ’ln π’ = π+1 π₯ β = π₯ β ππ₯ = π+1 π+1 π’Reemplazamos los nuevos valores en la integral y los limitescorrespondientes ln π’ β π π π +1 1 ππ’ 0 ππ’ + π π+1 π’ π ln π’ 1 β π π +1 β π’ β1 0 ππ’π +1 ππ’ +πPor propiedades de los logaritmos y euler π 1 β1 = β π+1 π+1 π β 1 π’ π+1 β π’β1 ππ’π+1 0 ππ’ + π β1 β 1 π’ π+1 ππ’π+1 0 ππ’ + πRealizamos otra sustituciΓ³n de tal manera que ππ πππ = ππ’ β = π’ β ππ = ππ’ π π β1 ππ π+1 β 1 π π β πππ+1 0 ππ + π π β1 ππ π+1 β1 β 1 π ππ+1 β πππ+1 0 π π+1 π
6. 6. β1 β1 π π+1 β π π+1 π β1 β β π 1 π π+1 πππ+1 0 π π+1 β1 β1 +1 π π+1 β π π+1 β1 β +1 1 π π+1 ππ π+1 0 π π+1 β1 π π π+1 β π π+1 π β 1 π π+1 ππ π+1 0 π π+1Trasponemos tΓ©rminos β1 π β 1 π π+1 β π π+1 π πππ+1 0 π π+1 β π π+1 β1 π β β1 1 π π+1 β π π+1 π ππ π π+1 β π+1 0 π+1 β1 1 β β 1 π π+1 β π π +1 π ππ π π+1 β π+1 0 π+1 β1 β 1 π π+1 π 1 ππ 0 π+1 π π+1 β π π+1 β (π + 1)Si comparamos con π’ π¦β1 ππ’π½ π₯, π¦ = π’ + 1 π₯+π¦ 1 1 ππ¦β1=β β π¦=β +1 β π¦ = π+1 π+1 π+1
7. 7. π 1 π₯+ π¦=1 β π₯ =1β π¦ β π₯ =1β β π₯= π+1 π+1 1 1 π π 1 π½ , π+1 π+1 π π+1 β π π+1 β (π + 1) 1 π 1 Ξ Ξ π+1 π+1 π 1 1 π π π+1 β π π+1 β (π + 1) Ξ π + 1 + π + 1 1 π 1 Ξ Ξ π+1 π+1 π 1 π+1 π π+1 β π π+1 β π+1 Ξ π+1 1 1 π π 1 Ξ Ξ π+1 π+1 π π+1 β π π+1 β π+1 1 1 1 π 1 Ξ Ξ 1β π+1 π+1 π π+1 β π π+1 β (π + 1)Aplicamos ahora el teorema de gamma de tal manera que πΞ x Ξ 1βx = π ππππ₯ 1 π π 1 β π ππ‘π β¦ π π+1 β π π+1 β (π + 1) π ππ π+1
8. 8. 1 π π π₯ ln π₯ ππ₯0βπ’ = ln π₯π βπ’ = π₯βπ βπ’ ππ’ = ππ₯π βππ’ = π₯ ππ π π₯ = 0 β π’ = β π π π₯ = 1 β π’=0 0 π βππ’ β π’ π β π βπ’ ππ’β 0β π βπ’(π +1) β π’ π ππ’ β ππ‘π‘ = π’ π+1 β ππ‘ = π + 1 ππ’ β = ππ’ π+1 π‘ π π‘π = π’ β π’ =π+1 (π + 1) π β βπ‘ π‘π ππ‘β π β π β 0 π+1 π+1 β 1β π βπ‘ β π‘ π ππ‘ (π + 1)(π + 1) π 0 β 1β π βπ‘ β π‘ π ππ‘ (π + 1) π+1 0π₯β1= π β π₯ = π+1 1β π+1 Ξ π+1 π+1 β1 π π!= ππ‘π β¦ God bless (π +1) π +1