I have the following function
\begin{equation} P(x) = \frac{1}{L} + \frac{2}{L} \sum_{n = 1}^{\infty} \cos\left( \frac{n \pi x}{L} \right) \end{equation}
and I try to show that it represents a Gaussian. I'm trying to do this via the Euler-MacLaurin formula. My approach was to convert the sum to the infinite sum:
\begin{equation} P(x) = \frac{1}{L} \left\{ 1 + 2\sum_{n = 0}^{\infty} \cos\left( \frac{n \pi x}{L} \right) \right \}= \frac{1}{L} \sum_{n = - \infty}^{\infty} \cos \left ( \frac{n \pi x}{L} \right) \end{equation}
Now, we can apply the Euler-MacLaurin formula to obtain:
\begin{equation} LP(x) = \sum_{n = -\infty}^{\infty} \cos \left ( \frac{n \pi x}{L} \right) = \int_{-\infty}^{\infty} \cos \left ( \frac{n \pi x}{L} \right) dx + \frac{\cos(\infty) + \cos(-\infty)}{2} + \sum_{k = 1}^{p/2} \frac{B_{2k}}{(2k)!} \left ( f^{(2k - 1)}(\infty) - f^{(2k - 1)}(-\infty)\right) + R_p \end{equation}
where $f^{(\xi)}(x)$ is the $\xi$-th derivative of $\cos \left( \frac{n \pi x}{L} \right )$, $B_k$ are the corresponding Bernoulli numbers, and $R_p$ is an error term. So, here I get stuck. What are the values of sine and cosine at infinity and negative infinity? Any suggestions on how to proceed from here? Once again, the goal is to show that $P(x)$ represents a Gaussian. Thanks!
That is a Dirac comb, that is the periodic repetition of unit pulses, whose width tends to zero and so to a Dirac Delta.