Actually, even though you've written $\dot{e}_x$, what I assume you've meant to write is
$$\mathring{e}_x\text{.}$$
Let $T_x$ be the time-to-death random variable for $(x)$. Then, a very well-known formula is
$$\mathring{e}_x = \mathbb{E}[T_x] = \dfrac{\omega - x}{2}\text{.}$$
Then it follows that
$$\mathring{e}_{x+1} = \dfrac{\omega-(x+1)}{2} = \dfrac{\omega-x}{2}-\dfrac{1}{2}=\mathring{e}_x-\dfrac{1}{2}\text{.}$$
This really only works well for the uniform distribution. When people generally refer to the "recursive" formula for $\mathring{e}_x$, they mean
$$\mathring{e}_x = \mathring{e}_{x:\overline{n}|}+{}_{n}p_{x}\cdot\mathring{e}_{x+n}\text{.}$$
If $\mathring{e}_{x+1}$ is given, $$\mathring{e}_x = \mathring{e}_{x:\overline{1}|}+p_{x} \cdot \mathring{e}_{x+1}\text{.}$$
Now recall that the PDF of $T_x$ is given by
$$f_{T}(t) = \dfrac{1}{\omega - x}\text{, } t \in [0, \omega - x]\text{.}$$
Then, another formula you should have memorized by now is
$${}_{t}p_{x} = \dfrac{\omega - x - t}{\omega - x}$$
from which you can find $p_x$ easily, and
$$\mathring{e}_{x:\overline{1}|} = \int_{0}^{1}{}_{t}p_{x}\text{ d}t\text{.}$$