Finding the Fourier series of $\delta(x)$ on $ (-\pi, \pi)$ (dirac delta)

Question

So $\delta(x)$ is even, so it will be a cosine series. I get to $$a_n = \int_{-\pi}^{\pi}\delta(x)cos(nx)dx $$ for $a_0$ I get $\frac{1}{\pi}$. Now there's the following property of the delta function: $$ \int_{-\infty}^{\infty}\delta(x-a)f(x)dx = f(a)$$ Can I apply that property, where $a=0$, and there's an $n$ in $\cos(nx)$?

Answer 1

Hint. We have that $$a_n = \int_{-\pi}^{\pi}\delta(x)\cos(nx)dx=\int_{-\infty}^{+\infty}\delta(x)f(x)dx$$ where $f(x)=\cos(nx)\cdot \mathbf{1}_{[-\pi,\pi]}(x)$ where $\mathbf{1}_{[-\pi,\pi]}$ is the characteristic function of the interval $[-\pi,\pi]$.

Answer 2

When you apply $a = 0$ it looks like you just get $a_n = 1$.

The resulting series would just be $1+\Sigma_{n=1}^\infty \cos{nx}$

I graphed a few partial sums on Mathematica and they look pretty interesting.

Here's 5 terms:

Here's 10 terms:

Here's 20 terms:

The graphs do appear to be converging on an impulse at the origin, with periodic impulses at intervals of $2\pi$.

I should note that the series itself appears to diverge, which means all those wiggly bits at to the left and right of the impulse (where the original function is zero) grow without bound as more terms are added on.