Deriving the Hemholtz equation from the wave equation

Question

I am trying to derive the Hemholtz equation in electromagnetism. In the Lorentz gauge $\partial_\mu A^\mu = 0$ the inhomogeneous maxwell equations read: \begin{equation}\label{maxwell inh} \square A^\mu = \mu_0 J^\mu \iff \bigg(\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\nabla^2\bigg)A^\mu = \mu_0 J^\mu \end{equation} Let us take a temporal Fourier transform: \begin{align} A_\mu(\textbf{r},t) = \int_{-\infty}^\infty \frac{d\omega}{2\pi} \tilde{A}_\mu(\textbf{r},\omega) e^{-i\omega t}, \ J_\mu(\textbf{r},t) = \int_{-\infty}^\infty \frac{d\omega}{2\pi} \tilde{J}_\mu(\textbf{r},\omega) e^{-i\omega t} \end{align} and substitute into the wave equation: \begin{align} &\int_{-\infty}^\infty\frac{d\omega}{2\pi} \bigg(\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\nabla^2\bigg)\tilde{A}_\mu(\textbf{r},\omega) e^{-i\omega t} =\int_{-\infty}^\infty \frac{d\omega}{2\pi} \mu_0 \tilde{J}_\mu(\textbf{r},\omega) e^{-i\omega t}\\ &\implies \bigg(\nabla^2 + \frac{\omega^2}{c^2}\bigg)\tilde{A}_\mu = -\mu_0 \tilde{J}_\mu \end{align} The last equation is known as the Hemholtz equation. However, I'm not sure how to justify the last step since the integral over $\mathbb{R}$ of a non-zero quantity can still be zero i.e. an odd function.