I want to derive $G\in\mathcal{S}'(\mathbb{R})$ satisfying
$$\left(\Delta+\frac{\omega^{2}}{c^{2}}\right)G=\delta,\qquad(\star)$$ where $\Delta$ is the Laplace operator, $\omega\in\mathbb{R}$ and $c>0$ is a constant.
For all $\phi\in\mathcal{S}(\mathbb{R})$, taking the Fourier transform of $(\star)$ with respect to $x\in\mathbb{R}$ yields
$$ \begin{aligned} &\langle\left(\frac{\omega^2}{c^2}-\xi^2\right)\mathcal{F}G,\phi\rangle=1 \\ \iff&\langle G,\mathcal{F}\phi\rangle=\frac{1}{\left(\frac{\omega}{c}+i\varepsilon\right)^2-\xi^2}, \end{aligned}$$
where $\varepsilon>0$ is arbitrarily small to avoid a singularity when we invert the Fourier transform. In particular, applying an inverse Fourier transform gives
$$\begin{aligned} &\langle\mathcal{F}^{-1}G,\mathcal{F}\phi\rangle=\lim_{\varepsilon\downarrow 0}\frac{1}{2\pi}\int_{-\infty}^\infty\frac{e^{ix\cdot \xi}}{\left(\frac{\omega}{c}-i\varepsilon\right)^2-\xi^2}\,d\xi \\ \iff&\langle G,\mathcal{F}^{-1}\mathcal{F}\phi\rangle=\lim_{\varepsilon\downarrow 0}\frac{1}{4\pi}\int_{-\infty}^{\infty}\frac{e^{ix\cdot \xi}-e^{-ix\cdot \xi}}{\left(\frac{\omega}{c}+i\varepsilon\right)^2-\xi^2}\,d\xi \\ \iff&\langle G,\phi\rangle=\frac{ic}{2\omega}e^{i\omega|x|/c}, \end{aligned}$$
I have three question:
1) Why is the $\phi$ still there? I want to have $G(x)=...$
2) I don't fully understand the final step, when the limit is taken. Could someone perhaps explain that to me?
3) Can we prove this starting with $G\in\mathcal{D}'(\mathbb{R})$; and what information does one need to have to solve for $G\in\mathcal{S}'(\mathbb{R})$, since fundamental solutions are generally deemed to belong to the former vector space?