I have the following distribution: $$\Bbb R\times\Bbb R^s\ni(t,x)\mapsto\Delta(t,x)= c \Theta(t) \int_{\Bbb R^s} e^{ikx} \frac{\sin(t\Omega(k))}{\Omega(k)} d^sk.$$ Here $\Omega(k)=\sqrt{m^2+|k|^2}$ and $\Theta$ is the Heaviside function. I would like to show that it lies in $\mathcal S'(\Bbb R^{1+s})$. The distribution contains a spatial Fourier transform, so I thought I could fix $t\in\Bbb R$, and discuss the distribution $u_t(k):=\Theta(t) \sin(t\Omega(k))/\Omega(k)$ in the $k$ variable, as follows: define the brackets $\langle \xi\rangle :=\sqrt[+]{1+|\xi|^2}$, so that e.g. $\Omega(k) = m\langle k/m \rangle$, and observe that $$ \begin{split} |\langle u_t(k),\phi\rangle| &= c\Theta(t)\left|\int_{\Bbb R^s} \frac{\sin(t\Omega(k))}{m\langle k/m \rangle} \phi(k) d^sk\right| \\ &\leq c m^s \int_{\Bbb R^s} \frac{d^s\xi}{\langle\xi\rangle^{1+s}} \cdot \sup_{k\in\Bbb R^s} \left\langle \frac k m \right\rangle^s |\phi(k)| \\ &\leq \gamma \sum_{|\alpha|\leq s} \sup_{k\in\Bbb R^s} |k^\alpha \phi(k)|, \end{split} $$ where $\gamma>0$ is an appropriate constant and we have used the fact that $\langle\xi\rangle^{-p}$ is integrable over $\Bbb R^s$ as soon as $p>s$. Thus $u_t(k)$ is tempered and Fourier transformable. On the other hand, the distribution $v_k(t) = \Theta( t) \sin(t\Omega(k))/\Omega(k)$ in the variable $t$ is a tempered distribution, because $\mathcal F_t\Theta(\omega) = i(\omega + i0^+)^{-1}$ lies in $\mathcal S'(\Bbb R)$, and $\mathcal F_t\sin(\Omega(k)\,\cdot\,)(\omega) = i\pi(\delta_{\Omega(k)} - \delta_{-\Omega(k)})$ is compactly supported, and thus the convolution of the transforms is tempered, and its inverse transform, namely $v_k(t)$, must be tempered as well.
However, I am not sure whether this allows me to conclude that the overall distribution $\Delta$ is tempered. For instance, I am uncertain whether I may exchange the partial Fourier transforms, seeing as the definition contains a sinc function, which is famously not Lebesgue integrable.