I'm trying to wrap my head around the following.
I understand the steps that lead from $$x(y)=A \cos(2 \pi f_0 t)$$ to $$X(y)=A \frac{\delta(f-f_0)+\delta(f+f_0)}{2}$$
$f_0$ is the carrier frequency.
The problem I have is that I would expect the integration of cosine from -$\infty$ to $\infty$ to "blow".
Through Euler, we have:
$$A\cos(2\pi f_0t)=\frac{A}{2}\left[e^{j2\pi f_0t}+e^{-j2\pi f_0t}\right]$$
Ok, now there is an important property of Dirac's Delta function regarding its integration:
$$\int_{-\infty}^{+\infty}\delta(x\pm x_0)f(x)dx=f(\mp x_0)$$
Then, by inspection of the Euler form of the cosine, we have:
$$e^{j2\pi f_0t}=\int_{-\infty}^{+\infty}\delta(f-f_0)e^{j2\pi ft}df$$
and
$$e^{-j2\pi f_0t}=\int_{-\infty}^{+\infty}\delta(f+f_0)e^{j2\pi ft}df$$
Which are, clearly, the inverse Fourier Transforms of $\delta(f-f_0)$ and $\delta(f+f_0)$. So:
$$\frac{A}{2}\left[e^{j2\pi f_0t}+e^{-j2\pi f_0t}\right]\stackrel{\mathcal{F}}\longleftrightarrow \frac{A}{2}\left[\delta(f-f_0)+\delta(f+f_0)\right]$$
This is also the procedure used for $A\sin(2\pi f_0t)$, giving:
$$\frac{A}{2j}\left[e^{j2\pi f_0t}-e^{-j2\pi f_0t}\right] \stackrel{\mathcal{F}}\longleftrightarrow \frac{A}{2j}\left[\delta(f-f_0)-\delta(f+f_0)\right]$$