Question

Bestimmtes Integral berechnen:

\( \int \limits_{0}^{\infty} \cos (a t) \cdot e^{-t} d t \quad a>0 \)

Comments

Versuch's mal mit partieller Integration.

Answer 1

Wir lassen erstmal die Integrationsgrenzen weg

u' = e^-x -> u = - e^-x

v = cos(a*x) -> v' = -  a*sin(a*x)

 ∫e^-x *cos(a*x) dx = u*v - ∫(u*v')dx

-> - e^-x *cos(a*x) - ∫((-e^-x)*(- a*sin(a*x)) dx = - e^-x *cos(a*x) - ∫e^-x *a*sin(a*x) dx = ∫e^-x *cos(a*x) dx

Nochmals partielle Integration für  ∫e^-x *a*sin(a*x) dx

u' = e^-x -> u = - e^-x

v = a*sin(a*x) -> v' =  a²*cos(a*x)

-> ∫e^-x *cos(a*x) dx = - e^-x *cos(a*x)  - (-e^-x *a*sin(a*x) -  ∫(-e^-x )*a²*cos(a*x) dx)

-> ∫e^-x *cos(a*x) dx = - e^-x *cos(a*x)  + e^-x *a*sin(a*x) -  ∫e^-x *a²*cos(a*x) dx)     | + ∫e^-x *a²*cos(a*x) dx

-> ∫e^-x *cos(a*x) dx + ∫e^-x *a²*cos(a*x) dx = - e^-x *cos(a*x)  + e^-x *a*sin(a*x)  

-> ∫e^-x *cos(a*x) dx + a² ∫e^-x *cos(a*x) dx = e^-x *(-cos(a*x) + a*sin(a*x))

-> ∫e^-x *cos(a*x) dx*(1 + a²) = e^-x *(-cos(a*x) + a*sin(a*x)) = (1/e^x)*(-cos(a*x) + a*sin(a*x))

=>  ∫e^-x *cos(a*x) dx = [(1/e^x)*(-cos(a*x) + a*sin(a*x))]/(1 + a²)

-> |[(1/e^x)*(-cos(a*x) + a*sin(a*x))]/(1 + a²)|₀^oo  = (1/e^oo)*(-cos(a*oo) + a*sin(a*88))/(1 + a²) - ((1/e⁰)*(-cos(a*0) + a*sin(a*0))/(1 + a²)) = 0 - 1*(-1 +0)/(1 + a²) = 1/(1 + a²)