I've been trying to answer this question and provide a closed form analytical solution.
The equation $$ f(t) = e{^{\sigma t}}cos(\omega t) $$
$$ F(t)= \int_{0}^{t} |f(\tau)|d\tau $$
find F(t) any value of t given that $$ \sigma <0 , \omega>0 $$
I've been trying a few ways to solve this, firstly by plotting the graph, which should look something similar to this. enter image description here
Based on this I broke down the integral into interval where $$ \ |f(\tau)| $$ is positive and negative $$ \int_{0}^{\pi/2} e{^{\sigma t}}cos(\omega t) - \int_{\pi/2}^{3\pi/2} e{^{\sigma t}}cos(\omega t) + \int_{3\pi/2}^{5\pi/2} e{^{\sigma t}}cos(\omega t) - \int_{5\pi/2}^{7\pi/2} e{^{\sigma t}}cos(\omega t)... $$
$$ \int e{^{\sigma t}}cos(\omega t) dt = $$ $$ \ e{^{\sigma t}}(\sigma cos(\omega t) +\omega sin(\omega t)/ (\sigma ^2 + \omega ^2) $$
I'm really lost I think my approach makes sense, but I don't know how to generalize for all values of t
Any help would be greatly appreciated.
$$f(\tau)=e^{\sigma\tau}\cos(\omega\tau)$$ lets focus on the periodicity of $\cos(\omega\tau)$ since $e^{\sigma\tau}>0\forall\sigma,\tau\in\mathbb{R}$. We know that: $$|\cos(\omega\tau)|=\begin{cases}\cos(\omega\tau)&0<\omega\tau<\pi/2&\Rightarrow0<\tau<\pi/2\omega\\-\cos(\omega\tau)&\pi/2<\omega\tau<3\pi/2&\Rightarrow\pi/2\omega<\tau<3\pi/2\omega\\\cos(\omega\tau)&3\pi/2<\omega\tau<2\pi&\Rightarrow3\pi/2\omega<\tau<2\pi/\omega\end{cases}$$ so if we look at the simplest case, $t=2\pi$ we would have: $$\int_0^{2\pi}|f(\tau)|d\tau=\int_0^{\pi/2\omega}\exp(\sigma\tau)\cos(\omega\tau)d\tau-\int_{\pi/2\omega}^{3\pi/2\omega}\exp(\sigma\tau)\cos(\omega\tau)d\tau+\int_{3\pi/2\omega}^{2\pi/\omega}\exp(\sigma\tau)\cos(\omega\tau)d\tau$$