Upload
kevin-picoaga
View
214
Download
0
Embed Size (px)
DESCRIPTION
Flow
Citation preview
5. Evalué las siguientes integrales.
a) ∫0
∞ cosh axcosh x
dx
Sugerencia f ( z )= eaz
cosh z,/∁ :−R≤ x≤ R ,0≤ y ≤π
SOLUCION.
Integro f (z) en la curva C.
∫−R
R eax
cosh xdx+∫
0
π ea(R+iy )
cosh (R+iy )dy+∫
R
−R ea(x+ iπ)
cosh (x+ iπ )dx+∫
π
0 ea (−R+iy)
cosh (−R+iy)dy=0
∫−R
R eax
cosh xdx+∫
0
π ea(R+iy )
cosh (R+iy)dy+∫
R
−R−eax
cosh (x+ iπ)dx+∫
π
0 ea (−R+iy)
cosh (−R+iy)dy=0
∫−R
R eax
cosh xdx+∫
0
π ea(R+iy )
cosh (R+iy)dy+∫
R
−R−eax
cosh x coshiπ+senh x senhiπdx+∫
π
0 ea(−R+iy )
cosh (−R+iy)dy=0
∫−R
R eax
cosh xdx+∫
0
π ea(R+iy )
cosh (R+iy)dy+∫
R
−R−eax
cosh xdx+∫
π
0 ea (−R+iy )
cosh (−R+iy)dy=0
∫−R
R eax
cosh xdx+∫
0
π ea(R+iy)
cosh (R+iy)dy+∫
−R
R−ea (−x)
cosh¿¿¿
∫−R
R eax
cosh x+−ea (−x)
cosh ¿¿¿
12∫−R
R coshaxcosh x
dx=−¿
12∫−R
R coshaxcosh x
dx=−¿
12∫−R
R coshaxcosh x
dx=−¿
12∫−R
R coshaxcosh x
dx=−[∫0
π ea (R+iy )+e−a (R+iy )
cosh (R+iy )dy ]
14∫0R coshaxcosh x
dx=−[ 12∫0π cosha (R+iy)cosh (R+iy )
dy ]
14∫0R coshaxcosh x
dx=−[ 12∫0π ea (R+iy ) +e−a (R+ iy )
e (R+iy ) +e−(R+ iy ) dy ]
Igualando parte realcon la parte real :