# Is there a function whose antiderivative can be found but whose derivative cannot?

Does a function, $f(x)$, exist such that $\int f(x) dx$ can be found but $f' (x)$ cannot be found in terms of elementary functions.

For example, if $f(x)=e^{x^2}$, then the derivative is easily calculated by using the chain rule. However, there does not exist an anti-derivative in terms of elementary functions.

Does a function exist with the opposite property?

If the antiderivative $F$ of $f$ is elementary, then so is $f' = F''$ (for any reasonable definition of “elementary function”). Thus, no such example can be found.

EDIT

1. What I assume here is that for your favorite definition of “elementary function”, the following is true: Every elementary function is differentiable and the derivative is again an elementary function.

This is indeed fulfilled (on the respective domains) if you take as your elementary functions all functions which can be obtained from $\exp, \ln, \sin, \cos$ and polynomials by taking sums/quotients/products and compositions of these functions. This is a consequence of the chain rule.

It is not fulfilled, however, if you also want to include roots, since e.g. $x \mapsto \sqrt{x}$ is not differentiable at the origin. But note that it is true if you only consider the roots as functions on $(0,\infty)$ instead of $[0,\infty)$.

2. I assume that if your function $f$ has a continuous version (with respect to equality a.e.), you identify it with its continuous version.

As noted in the answer of @RossMillikan, the Dirichlet function $f = 1_\Bbb{Q}$ is (Lebesgue)-integrable with “antiderivative” $x \mapsto 0$, but not differentiable. But note that $f = 0$ almost everywhere, which is elementary and has an elementary derivative.

Finally, if $F(x) = \int_a^x f(t) \, dt$ is elementary, then (by Lebesgue’s differentiation theorem) you have $f(x) = F'(x)$ almost everywhere. Hence, if $F$ is elementary (as outlined in point 1), then $F'$ is elementary and hence continuous, so that we get $f = F'$ almost everywhere. Since we agreed to identify $f$ with its continuous version, we get $f = F'$ everywhere, so that $f$ is differentiable with $f' = F''$ elementary, as claimed above.