Integrating $\int\frac{\cos(\omega t)\gamma e^{-\gamma t}}{\omega}dt$

Question

How to integrate $$\int\frac{\cos(\omega t)\gamma e^{-\gamma t}}{\omega}dt,$$ where $\gamma, \omega \neq 0$.

I tried using substitution $u=\omega t$, $du=\omega dt$ and got $\frac{1}{\omega^2} \int \cos(u) \gamma e^{-\frac{\gamma u}{\omega}} du$ and then integrating by parts, but so far it seems like an endless cycle of substitutions and ind integration by parts, and I can't get to something meaningful. Is there any easy way to compute this?

Edit: Using one of the hints, I arrived at $\frac\gamma\omega\mathrm{Re}\left[\frac{\mathrm{e}^{(i\omega-\gamma)t}}{i\omega-\gamma}+C\right]$, but not sure how do go from here. Have I made any errors here (I am not sure about the denominator)?

Answer 1

In general for $a,b$

$$ \begin{align} I=\int\cos(ax)e^{bx}&={\sin(ax)\over a}e^{bx}-{b\over a}\int\sin(ax)e^{bx}\\ &={\sin(ax)\over a}e^{bx}+{b\cos(ax)\over a^2}-{b^2\over a^2}\int \cos(ax)e^{bx}\\&={e^{bx}(a\sin(ax)+b\cos(ax))\over a^2}-{b^2\over a^2}I \end{align} $$

The on rearrangement gives

$$ \int\cos(ax)e^{bx}={e^{bx}\over a^2+b^2}(b\cos(ax)+a\sin(ax))+C $$

Edit: $$\int{\cos(\omega t)\gamma e^{-\gamma t}\over \omega}\\ ={\gamma\over\omega}\int\cos(\omega t)e^{-\gamma t}\\ ={\gamma\over\omega}{e^{-\gamma x}\over \omega^2+\gamma^2}(\omega\sin(\omega x)-\gamma\cos(\omega x))$$

Answer 2

We firstly take off the constant and get $I=\frac{\gamma}{\omega} \int \cos (\omega t) e^{-\gamma t} d t.$ $$ \begin{aligned}\int \cos (\omega t) e^{-\gamma t} d t = & \Re \int e^{\omega t i} e^{-\gamma t} d t \\ = & \Re \int e^{(\omega i-\gamma) t} d t \\ = & \Re\left(\frac{e^{(\omega i-\gamma)t}}{\omega i-\gamma}+ C\right) \end{aligned} $$

where $C $ is a complex constant.

By rationalisation, we get $$ \begin{aligned} \frac{e^{(\omega i-\gamma)t}}{\omega i-\gamma} & =-\frac{e^{(\omega i-\gamma)t}}{\gamma-\omega i} \cdot \frac{\gamma+\omega i}{\gamma+\omega i} \\ & =-\frac{e^{-\gamma t}}{\gamma^2+\omega} e^{\omega i}(\gamma+\omega i) \end{aligned} $$ Hence we get$$ \Re\left(\frac{e^{(\omega i-\gamma)t}}{\omega i-\gamma}\right)=-\frac{e^{-\gamma t}}{\gamma^2+\omega^2}(\gamma \cos \omega t-\omega \sin \omega t) $$ and arrive at $$ \boxed{I=\frac{\gamma e^{-\gamma t}}{\omega(\gamma^2+\omega^2)}(\omega \sin \omega t-\gamma \cos \omega t)+k} $$ where $k=\Re {(C)}$ is a real constant.