Damped Exponential Cosine Integral

$$\int_0^\infty e^{-\omega T}\cos(\omega t)\,d\omega.$$ (1)

Integrate by parts with

$$u\equiv e^{-\omega T}\qquad dv=\cos(\omega t)\,d\omega$$ (2)

$$du\equiv -Te^{-\omega T}\,d\omega \qquad v={1\over t}\sin (\omega t),$$ (3)

so

$$\int e^{-\omega T}\cos(\omega t)\,d\omega = {1\over t}e^{-\o... ...\omega t)+{T\over t}\int e^{-\omega T}\sin(\omega t)\,d\omega.$$

Now integrate

$$\int e^{-\omega T}\sin(\omega t)\,d\omega$$ (4)

by parts. Let

$$u=e^{-\omega T}\qquad dv=\sin(\omega t)\,d\omega$$ (5)

$$du=-Te^{-\omega T}\,d\omega \qquad v=-{1\over t}\cos(\omega t),$$ (6)

so

$$\int e^{-\omega t}\sin(\omega t)\,d\omega = -{1\over t}\cos(\omega t)-{T\over t} \int e^{-\omega T}\cos(\omega t)\,d\omega$$ (7)

and

$$\int e^{\omega T} \cos(\omega t)\,d\omega = {1\over t}e^{-\o... ...ga t)-{T^2\over t^2} \int e^{-\omega T}\cos(\omega t)\,d\omega$$ (8)

$$\left({1+{T^2\over t^2}}\right)\int e^{-\omega T}\cos(\omega... ...{{1\over t} \sin(\omega t)-{T\over t^2} \cos(\omega t)}\right]$$ (9)

$${t^2+T^2\over t^2} \int e^{-\omega T}\cos(\omega t)\,d\omega = {e^{-\omega t}\over t^2} [t\sin(\omega T) - T\cos(\omega t)]$$ (10)

Therefore,

$$\int_0^\infty e^{-\omega T}\cos(\omega t)\,d\omega= 0+{T\over t^2+T^2}={T\over t^2+T^2}.$$ (11)

See also Cosine Integral, Fourier Transform--Lorentzian Function, Lorentzian Function