We define the Fourier transformation as $$ \int_{\mathbb{R}} e^{-iwt}dt=\hat{f}(w) $$ If we apply the Fourier transformation on $$ \int_{\Omega}\int_0^{\infty}f(x,t)dtdx $$ by definition is $$ \int_{\mathbb{R}} e^{-iwt}(\int_{\Omega}\int_0^{\infty}f(x,t)dtdx)dt $$ which if we switch the order of integration becomes $$ \int_{\Omega} \int_{\mathbb{R}}e^{-iwt}( \int_0^{\infty}f(x,t)dt)dt dx $$ I am thinking of pulling the exponential into the most inner integration which gives $$ \int_{\Omega} \int_0^{\infty}\hat{f}(w,x)dtdx $$ but this does not make sense given the inner integration goes to infinity.
Or if we take $$ F(x) = \int_0^{\infty}f(x,t)dt $$ then the Fourier transform of a 'constant' goes to a delta function(transforming on t only not considering x) then the whole thing reduces to $$ \int_{\Omega} \delta(x-F(x))dx $$
Your notation is ambiguous. If I understand you well, you want to determine the Fourier transform of the quantity $\int_\Omega\mathrm{d}x \int_0^\infty\mathrm{d}t \,f(x,t)$. Yet it is not a function, because $x$ and $t$ are the variables of integration and don't have any meaning outside the two integrals. In consequence, it is just a constant and its Fourier transform is given by a Dirac delta, more precisely : $$ \mathfrak{F}\left[\int_\Omega\mathrm{d}x \int_0^\infty\mathrm{d}t \,f(x,t)\right](\omega) = \int_\mathbb{R}\mathrm{d}\tau \,e^{-i\omega\tau} \int_\Omega\mathrm{d}x \int_0^\infty\mathrm{d}t \,f(x,t) = 2\pi\delta(\omega) \int_\Omega\mathrm{d}x \int_0^\infty\mathrm{d}t \,f(x,t) $$