probability - CDF and PDF of semaphore waiting time

Imagine we have a semaphore that alternates every 40 seconds between green and red.

Waiting time is 0 when the semaphore is green, and when it is red it is the remaining time until it turns green.

I want to model the distribution of waiting times on this semaphore.

Starting with the CDF I have:

$$F(x) = \begin{cases} 0 && \text{if } x < 0\\ 0.5 && \text{if } x = 0 && \textit{half the time we don't need to wait}\\ 0.5 + \frac{0.5}{40} x && \text{if } x > 0 \text{ and } x<=40 && \textit{all waiting times ]0-40] are equally likely}\\ 1 && \text{if } x > 40 \end{cases}$$

Is the PDF of this distribution given by the following function?

$$PDF(x) = \begin{cases} 0 && \text{if } x < 0\\ 0.5 && \text{if } x = 0\\ \frac{0.5}{40} && \text{if } x > 0 \text{ and } x<=40\\ 0 && \text{if } x > 40 \end{cases}$$

And is the expected time waiting on this semaphore given by:

$$\begin{align*} \int_0^{40} x f(x) dx = \int_0^{40} x . \frac{0.5}{40} dx = 10 \end{align*}$$

Answer

Posting Henry's answer here for future reference.

This distribution's density is not defined at $x=0$, but instead we have a point probability $P(X=0) = 0.5$.

The expected value is calculated with a mix of discrete and continuous calculation:

$$\begin{align*} E[\text{waiting time}] &= \int_0^{40} x f(x) dx + 0 * P(X=0)\\ &= \frac{0.5}{40} . \frac{x^2}{2} \bigg|_0^{40} + 0 \\ &= \frac{0.5}{40}.\frac{40^2}{2}\\ &= 10 \end{align*}$$