Prove that $\sum\limits _{n=-\infty}^{n=\infty}\cos\left(2\pi nt\right)=\sum\limits _{n=-\infty}^{n=\infty}\delta\left(t-n\right)$

Question

I've tried using Fourier transforms on both but didn't quite get anything useful. I'd really appreciate some help.

Answer 1

The thing on the right is a periodic distribution with period $1$. So we can expand it into a Fourier series in terms of $\cos 2\pi n t$ and $\sin 2\pi n t$. The coefficients of sines are all $0$, because $$b_n=2\int_{-1/2}^{1/2} \delta(t)\sin 2\pi n t\,dt = 2\sin (2\pi n \cdot 0) =0$$ The cosine coefficients are $$a_n=2\int_{-1/2}^{1/2} \delta(t)\cos 2\pi n t\,dt = 2\cos (2\pi n \cdot 0) =2$$ Hence, the expansion is $$1+2\sum_{n=1}^\infty \cos 2\pi n t \tag{1}$$ which is the same as $\sum_{n=-\infty}^\infty \cos 2\pi n t$ because cosines are even. The series (1) converges, in the sense of distributions, to the distribution $\sum_{n\in\mathbb Z}\delta(t-n)$.