Consider $f(x) = \exp(-x^2)$. It is a known fact that it does not have a primitive function that can be written from elementary functions. However, since $f$ is continuous, one can apply the Fundamental Theorem of Calculus and a primitive function can be derived, namely, $$ F(x) = \int_a^x f(t)dt $$
My question: the FTC is for compact intervals $[a,b]$, and the variable $x$ verifies $x\in [a,b]$. Can we define the primitive $F$ in all $\mathbb{R}$?
Let $f(x)=e^{-x^2}$. Since $f(x)$ is integrable all over $\mathbb{R}$, $$\int_{-\infty}^{a}e^{-x^2}\,dx\leq\int_{-\infty}^{\infty}e^{-x^{2}}\,dx=\sqrt{\pi}$$ is well defined the function $$F(x)=\int_{-\infty}^{x}e^{-t^{2}}\,dt$$ Notice that $$\lim_{x\rightarrow +\infty}F(x)=\sqrt\pi$$ The primitive of the Gaussian function has a lot of uses, mainly in statistic, as one defines the error function as the following $$\text{erf}(x)=\frac{2}{\sqrt\pi}\int_{0}^{x}e^{-t^{2}}\,dt$$