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Damped Exponential Cosine Integral

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

Integrate by parts with
u\equiv e^{-\omega T}\qquad dv=\cos(\omega t)\,d\omega
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du\equiv -Te^{-\omega T}\,d\omega \qquad v={1\over t}\sin (\omega t),
\end{displaymath} (3)


\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
\end{displaymath} (4)

by parts. Let
u=e^{-\omega T}\qquad dv=\sin(\omega t)\,d\omega
\end{displaymath} (5)

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

\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
\end{displaymath} (7)


\int e^{\omega T} \cos(\omega t)\,d\omega = {1\over t}e^{-\o... t)-{T^2\over t^2} \int e^{-\omega T}\cos(\omega t)\,d\omega
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\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]
\end{displaymath} (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)]
\end{displaymath} (10)

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

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

© 1996-9 Eric W. Weisstein