How to prove the two formulas are equal in the sense of distribution

Question

$1+2\sum_{n=1}^\infty \cos2n\pi x=\sum_{k=-\infty}^\infty \delta(x-k)$.

I couldn't have an idea to prove it, maybe we can discuss how to get it clearly.

Answer 1

I would integrate it and show that both sides are equal. The Dirac Delta is defined by its action under the integral, so it should be enough.

But I am physicists, so maybe it is more subtle.

Answer 2

We have that \begin{align*} 1+2\sum_{n=1}^\infty \cos 2n\pi x=\sum_{k=-\infty}^\infty \exp(2k\pi i x), \end{align*} and the right side is the Dirichlet kernel which means that \begin{align*} &\int_{-\infty}^\infty \sum_{k=-\infty}^\infty \exp(2k\pi i x) \phi(x)dx\\ &=\lim_{N\to\infty}\int_{-\infty}^\infty 2\pi D_N(2\pi x)\phi(x)dx\\ &=\lim_{N\to\infty}\left[\cdots+\int_{-\frac{3}{2}}^{-\frac{1}{2}}2\pi D_N(2\pi x)\phi(x)dx+\int_{-\frac{1}{2}}^{\frac{1}{2}}2\pi D_N(2\pi x)\phi(x)dx+\int_{\frac{1}{2}}^{\frac{3}{2}}2\pi D_N(2\pi x)\phi(x)dx+\cdots\right]\\ &=\lim_{N\to\infty}\left[\cdots+\int_{-3\pi}^{-\pi} D_N(t)\phi(\frac{t}{2\pi})dt+\int_{-\pi}^{\pi} D_N(t)\phi(\frac{t}{2\pi})dt+\int_{\pi}^{3\pi} D_N(t)\phi(\frac{t}{2\pi})dt+\cdots\right], \end{align*} and by the periodicity($2\pi$) of Dirichlet kernel and Fourier convergence theorem we have \begin{align*} &\lim_{N\to\infty}\left[\cdots+\int_{-3\pi}^{-\pi} D_N(t)\phi(\frac{t}{2\pi})dt+\int_{-\pi}^{\pi} D_N(t)\phi(\frac{t}{2\pi})dt+\int_{\pi}^{3\pi} D_N(t)\phi(\frac{t}{2\pi})dt+\cdots\right]\\ =&\lim_{N\to\infty}\left[\cdots+\int_{-\pi}^{\pi} D_N(\tau)\phi(\frac{\tau-2\pi}{2\pi})d\tau+\int_{-\pi}^{\pi} D_N(\tau)\phi(\frac{\tau}{2\pi})d\tau+\int_{-\pi}^{\pi} D_N(\tau)\phi(\frac{\tau+2\pi}{2\pi})d\tau+\cdots\right]\\ =& \sum\limits_{k=-\infty}^\infty \phi(k), \end{align*} Therefore in the sense of distribution we know that \begin{align*} 1+2\sum\limits_{n=-\infty}^\infty \cos 2n\pi x=\sum\limits_{k=-\infty}^\infty \delta(x-k) \end{align*}