When deriving the line-shape for natural lifetime broadening this integral has to be solved.
Some sources give the answer as follows:
$$ I = -\left(\dfrac{1}{i\,(\omega_0-\omega)-\gamma}-\dfrac{1}{i\,(\omega_0+\omega)+\gamma}\right) $$
How do they get there? When I evaluate the Integral I only get:
$$ \dfrac{1}{i\,(\omega_0-\omega)-\gamma}\,\left[\exp\left\{-\gamma\, t + i\,( \omega_0-\omega)\,t\right\}\right]_{-\infty}^{+\infty} $$
Which somehow diverges ($\exp\{t\to\infty\}$). So where is my reasoning wrong?
You overlooked two things:
Following your source, we have
$$E(t) = E_0 e^{-\gamma t} \left( e^{i \omega_0 t} + e^{-i\omega_0t}\right)\tag{1}$$
Note that I added the complex conjugated.
Then:
Meaning there's an implicit heaviside function $\theta(t)$ in $E(t$), so that $E(t)$ is $0$ for $t < 0$.
The general formula is given by
$$a(\omega) = \int \limits_{-\infty}^{\infty}E(t) e^{-iwt}\mathrm{d}t$$
Plugging in $E(t)$
$$a(\omega) = \int \limits_{-\infty}^{\infty} \theta(t) E_0 e^{-\gamma t} \left( e^{i (\omega_0-\omega) t} + e^{-i(\omega_0+\omega)t}\right) \mathrm{d}t$$
Exploiting the heaviside function, the integral boundaries become
$$a(\omega) = E_0 \int \limits_{0}^{\infty}\left( e^{[i (\omega_0-\omega)-\gamma] t} + e^{[-i(\omega_0+\omega)-\gamma]t}\right) \mathrm{d}t$$
$$=E_0 \left( \frac{1}{i (\omega_0-\omega)-\gamma} + \frac{1}{-i(\omega_0+\omega)-\gamma} \right)\left( e^{[i (\omega_0-\omega)-\gamma] t} + e^{[-i(\omega_0+\omega)-\gamma]t}\right) \Bigm|^\infty_0$$
The integral vanishes for $t\to \infty$, as you can easily convince yourself by writing the exponential function as a product or by looking at the graph of the amplitude. Thus, we only have to evaluate at $t=0$ where it is of course $1$.
$$=E_0 \left( \frac{1}{i (\omega_0-\omega)-\gamma} + \frac{1}{-i(\omega_0+\omega)-\gamma} \right) \cdot (0-1)$$
a bit of algebra
$$=-E_0 \left( \frac{1}{i (\omega_0-\omega)-\gamma} - \frac{1}{i(\omega_0+\omega)+\gamma} \right) \cdot$$
There you go.