Im trying to evaluate the improper integral $$\int_{0}^{\infty}\left( \frac{e^{i \omega t}+e^{-i \omega t}}{2}\right) e^{-st} dt$$, where $\omega$ and $s$ are real positive constants and $i=\sqrt{-1}$.
Using Euler's formula $e^{ix}=\cos x+i\sin x$, i get $$\int_{0}^{\infty}\cos(\omega t) e^{-st} dt$$ But im not sure where to go from here, or if thats the right direction?
On
Suppose $\operatorname{re} s > 0$ (so the integral exists).
Let \begin{eqnarray} I_L &=& \int_0^L \left( \frac{e^{i \omega t}+e^{-i \omega t}}{2}\right) e^{-st} dt \\ &=& {1 \over 2} \int_0^L \left( e^{-(s-i \omega) t} + e^{-(s+i \omega) t} \right) dt \\ &=& {1 \over 2} \left( {1 \over -(s-i \omega) } e^{-(s-i \omega) t} \Big|_0^L + { 1 \over -(s+i \omega) } e^{-(s+i \omega) t}\Big|_0^L \right) \end{eqnarray}
Now let $L \to \infty$, this gives $I = {1 \over 2} \left( {1 \over s-i \omega } + { 1 \over s+i \omega } \right) = {s \over s^2+\omega^2}$.
Why not just integrate in the exponential form? If not, then you can use integration by parts (applied twice):
\begin{align} u = \cos(\omega t) && du = -\omega\sin(\omega t)dt \\ dv = e^{-st}dt && v = -\frac{1}{s}e^{-st} \end{align}
$$ \int\limits_0^\infty \cos(\omega t)e^{-st}dt = \left.-\frac{\cos(\omega t)}{s}e^{-st}\right|_0^\infty - \frac{\omega}{s}\int\limits_0^\infty\sin(\omega t)e^{-st}dt $$
Now do integration by parts again:
\begin{align} u = \sin(\omega t) && du = \omega\cos(\omega t)dt \\ dv = e^{-st}dt && v = -\frac{1}{s}e^{-st} \end{align}
$$ \int\limits_0^\infty \cos(\omega t)e^{-st}dt = \frac{1}{s} - \frac{\omega}{s}\left(\left.-\frac{\sin(\omega t)}{s}e^{-st}\right|_0^\infty + \frac{\omega}{s}\int\limits_0^\infty \cos(\omega t)e^{-st}dt\right) $$
Look, you have the same integral twice, so just collect it on one side:
$$ \left(1 + \frac{\omega^2}{s^2}\right)\int\limits_0^\infty \cos(\omega t)e^{-st}dt = \frac{1}{s} $$
Which now gives the integral:
$$ \int\limits_0^\infty \cos(\omega t)e^{-st}dt = \frac{s}{s^2 + \omega^2} $$