Let $g(x) = e^x + ae^{-x}$. Since the function $f(x) = x(e^x + ae^{-x})$ is an even function, it implies that $g(x) = e^x + ae^{-x}$ is an odd function.
Since the domain of the function $f(x)$ is $\mathbb{R}$, we have $g(0) = 0$,
which means $g(0) = 1 + a = 0$, solving this gives $a = -1$.
Therefore, the answer is: $\boxed{-1}$.