= Solution
Use the <Fourier transform> convention
$$
\widetilde f(\omega)=\int_{-\infty}^{\infty}f(t)e^{-i\omega t}\,dt,\qquad f(t)=\frac1{2\pi}\int_{-\infty}^{\infty}\widetilde f(\omega)e^{i\omega t}\,d\omega.
$$
Since $\mathcal F[e^{i\Omega t}]=2\pi\delta(\omega-\Omega)$ in <distribution theory>, the carrier has transform
$$
\boxed{\widetilde c(\omega)=\pi[\delta(\omega-\Omega)+\delta(\omega+\Omega)].}
$$
The <modulation property of the Fourier transform> translates the message spectrum to the two carrier frequencies. Therefore
$$
\boxed{\widetilde\phi(\omega)=\pi[\delta(\omega-\Omega)+\delta(\omega+\Omega)]+\frac12[\widetilde s(\omega-\Omega)+\widetilde s(\omega+\Omega)].}
$$
The first term represents the unmodulated <carrier wave>; the other terms are the <amplitude modulation sidebands>.
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