In quantum theory it is common to use what is known as Wick rotation, which says for a function defined over $\mathbb{R}^n$, for at least one of the variables, make the substitution $t \mapsto it.$ So for example, we may have $$f(x_1, x_2, \ldots, x_n) \mapsto f(ix_1, x_2, \ldots, x_n).$$ It is said that such a transformation is an analytic continuation.
How is this an analytic continuation if they no longer share a common open set in their respective domains? In the example above, the only set they share is $\{0, x_2,, \ldots, x_n : x_2, \ldots, x_n \in \mathbb{R} \}$ but this is not open.
Let's recall that an analytic continuation consists in extending the domain of definition of an analytic function (and keeping analytic properties at the same time).
In quantum mechanics, we usually work with wavefunctions $f : \mathbb{R}^n \rightarrow \mathbb{C}$, which are extended to $\tilde{f} : \mathbb{C}\times\mathbb{R}^{n-1} \rightarrow \mathbb{C}$ in a first step and then are Wick-rotated in a second step, i.e. $\tilde{f}(z_1,x_2,\ldots,x_n) \mapsto \tilde{f}(iz_1,x_2,\ldots,x_n)$. Wick's rotation isn't an analytic continuation itself but assumes such one.
In conclusion, $f$ is analytically continuated before the transformation due to Wick's rotation and the common open subset among the domains of definition is simply given by $D_1 \cap D_2 = \mathbb{C}\times\mathbb{R}^{n-1} \cap \mathbb{R}^n = \mathbb{R}^n$.