Fourier transform of $\cos ^ 2$

Question

I have to find the Fourier transform of $ [\cos (2 \pi f_0 t)] ^{2}$ using particularly product theorem. I know that, from this theorem, $$ F x(t)= F [\cos (2 \pi f_0 t)] \ast F [\cos (2 \pi f_0 t)] $$ where $\ast$ is the convolution. I found that $ F [\cos (2 \pi f_0 t)] = \frac{1}{2} \delta ( f-f_0 ) + \frac{1}{2} \delta ( f + f_0 ) $. So now $ F [\cos (2 \pi f_0 t)]^{2} = [ \frac{1}{2} \delta ( f-f_0 ) + \frac{1}{2} \delta ( f + f_0 )] \ast [ \frac{1}{2} \delta ( f-f_0 ) + \frac{1}{2} \delta ( f + f_0 )] $ but I have problems solving this convolution.

Answer 1

Define$$h(f):=F[\cos(2\pi f_0t)](f)=\tfrac12[\delta(f-f_0)+\delta(f+f_0)]$$so$$\begin{align}\int_{\Bbb R}h(g)h(f-g)dg&=\tfrac14\int_{\Bbb R}[\delta(g-f_0)+\delta(g+f_0)][\delta(f-g-f_0)+\delta(f-g+f_0)]dg\\&=\tfrac14[\delta(f-2f_0)+\delta(f)+\delta(f+2f_0)+\delta(f)],\end{align}$$since $\int_{\Bbb R}\delta(g-c)u(f,\,g)dg=u(f,\,c)$.