I don't know what is meant by a "single formula" but if, for instance, $f(x,a)$ is continuous, then $f(x,a) dx$ is a continuous 1-form, so has a well-defined path integral along any reasonable path. The integral will be path dependent if $f(x,a)$ depends at all on $a$, but let's take a path that moves first in the $a$ direction, then in the $x$ direction, i.e., set $F(x,a) = \int_{x_0}^x f(x,a)dx$ for some reasonable choice of basepoint $x_0$. Then if $f$ is differentiable in both variables, by Stokes' theorem $\frac{\partial F}{\partial a}(x,a) = \int_{x_0}^x \frac{\partial f}{\partial a}(x,a)dx$, which is still continuous.

Concretely, for the example $\int x^a dx$, take $x_0 = 1$. Then for $a \ne 1$, we have $\int_1^x x^a dx = \frac{x^{a+1}-1}{a+1}$, which approaches $\log x$ as $a \to 1$. (This is the compound interest limit.)

I don't know what is meant by a "single formula" but if, for instance, $f(x,a)$ is continuous, then $f(x,a) dx$ is a continuous 1-form, so has a well-defined path integral along any reasonable path. The integral will be path dependent if $f(x,a)$ depends at all on $a$, but let's take a path that moves first in the $a$ direction, then in the $x$ direction, i.e., set $F(x,a) = \int_{x_0}^x f(x,a)dx$ for some reasonable choice of basepoint $x_0$. Then if $f$ is differentiable in both variables, by Stokes' theorem $\frac{\partial F}{\partial a}(x,a) = \int_{x_0}^x \frac{\partial f}{\partial a}(x,a)dx$, which is still continuous.

Concretely, for the example $\int x^a dx$, take $x_0 = 1$. Then for $a \ne -1$, we have $\int_1^x x^a dx = \frac{x^{a+1}-1}{a+1}$, which approaches $\log x$ as $a \to -1$. (This is the compound interest limit.)