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 :