This is something that came to me randomly:

We know that a complex-differentiable function is infinitely-differentiable, and that elementary complex functions follow the same formal rules for differentiation as elementary real functions.

Does this mean that whenever we have a function defined via elementary functions on R whose derivative is not differentiable everywhere on R, we can conclude that this function does not have an antiderivative that can be expressed in terms of elementary functions?

For example: the function $x^3 sin(\frac{1}{x^2})$ extended by zero at the origin. This function is differentiable everywhere on R, but its derivative is not continuous (thus not differentiable) at the origin.

Thus if $x^3 sin(\frac{1}{x^2})$ had an antiderivative expressible in terms of elementary functions, we would obtain a contradiction with the fact that differentiable complex functions are smooth.

Here are my questions:

1) Is my reasoning sound?

2) Is this discussed somewhere in more generality, and has it been formalized in some way?

3) Does anyone have any examples where this observation leads to interesting conclusions?

Thank you!


1 Answer 1


I would say that your reasoning is not sound, or at least that is lacking something.

Looking at your example, even though $$ f(x) = \begin{cases} x^3 \sin \frac1{x^2}, & x \neq 0 \\ 0, & x = 0 \end{cases} $$ is differentiable at $x=0$ as a function $\mathbb{R} \to \mathbb{R}$, the corresponding complex function $$ f(z) = \begin{cases} z^3 \sin \frac1{z^2}, & z \neq 0 \\ 0, & z = 0 \end{cases} $$ has an essential singularity at $z=0$, so it cannot have (complex, holomorphic) anti-derivative on $\mathbb{C}$. It does have a holomorphic anti-derivative on $\mathbb{C} \setminus \{ 0 \}$, but that's more of a coincidence. The residue at $z=0$ happens to be $0$. If you take $x^5 \sin\frac1{x^2}$ instead, the corresponding complex function doesn't even have an anti-derivative on the punctured plane.

Furthermore, starting with $$ g(x) = 4x^3\sin \frac1x - x^2\cos \frac1x $$ the same "argument" would show that $g$ doesn't have an elementary anti-derivative, since $$ g'(x) = \begin{cases} 12x^2\sin\frac1x - 6x\cos\frac1x - \sin\frac1x, & x \neq 0 \\ 0, & x= 0 \end{cases} $$ is discontinuous at $x=0$. But this is false, since $g$ is the derivative of $x^4\sin\frac1x$.

  • $\begingroup$ Ah thank you very much! This cleared up a lot of things for me. $\endgroup$
    – Brevity
    Jan 14, 2016 at 17:40

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