I am learning a bit about distributions and came across the following...
In "Theory of Distributions, a nontechnical introduction" by Richards and Youn there is a formula with no explanation in the introduction. It states that as a distribution the following makes sense: $$\sum_{n=-\infty}^\infty \delta''(x-2\pi n)=\frac{-1}{\pi}\sum_{n=1}^\infty n^2\cos(nx) $$
Could someone please explain why the singularities don't pose an issue- how does one make sene of this summation? i am most curious about the singularity when the $n$ on the LHS=0. is this a known cosine representation of $\delta''$? thanks
The equality means that for all $\phi \in C_c^{\infty}(\mathbb{R})$, $$\sum_{n = -\infty}^{\infty}\int_{\mathbb{R}} \delta''(x - 2\pi n)\phi(x)\,dx = -\frac{1}{\pi}\sum_{n = 1}^{\infty}\int_{\mathbb{R}} n^2\cos(nx)\phi(x)\,dx.$$ Applying the definition of differentiation (integration by parts) and $\delta$ yields $$\sum_{n = -\infty}^{\infty}\phi''(2\pi n) = -\frac{1}{\pi}\sum_{n = 1}^{\infty}n^2\int_{\mathbb{R}}\cos(nx)\phi(x)\,dx.$$ The left hand side makes sense since only finitely many terms of the sum are nonzero due to $\phi$ having compact support. On the right hand side, we notice $\int_{\mathbb{R}}\cos(nx)\phi(x)\,dx$ is essentially $\text{Re } \hat{\phi}(n)$, and since $\phi$ is smooth and compactly supported, $\hat{\phi}$ is a Schwartz function, so the sum on the right is absolutely convergent.