Let's say I have an arbitrary elementary function $f : \mathbb{R} \to \mathbb{R}$, is there an upper bound on the number of functions that satisfies the following? $$\frac{d}{dx}\left(\int f(x) \ dx\right) = f(x) \ \ \ \ \ \ \ \ \ \ \ \ \ \text(1)$$

For example if I have $\int e^{x}\ dx$, there seems to be only one function that satisfies $(1)$, that being $e^{x}$. However if I have $\int \frac{1}{\sqrt{4-x^2}} \ dx$, there are at least two functions that satisfy this, those being $\sin^{-1}(\frac{x}{2})$ and $-\cos^{-1}(\frac{x}{2})$.

So let $S$, be the set of all functions satisfying $(1)$ above :

$$S = \left\{f : \frac{d}{dx}\left(\int f(x) \ dx\right) = f(x)\right\}$$

What is $|S|$?

  • Is |S| finite or infinite?
  • Is there any way to determine if $|S| = 1$ (i.e. there is only one possible anti-derivative for a function?
  • Does $|S|$ vary for different classes of elementary functions (e.g. polynomial, rational, trigonometric, exponential, or more generally, transcendental or algebraic)

I ask this particular question as for some functions such as $\int x^2 dx$, it seems like there can only ever be one function satisfying $(1)$, but I've had the same feeling about other functions, but using another technique of integration for those functions often yields more than one valid function (anti-derivative) satisfying $(1)$, and I have no way of telling the difference.

Is there a rigorous answer/definition in line with what I'm asking? If it goes into topics in Analysis, I'm all ears as currently it seems a bit hand-wavy to me, as if someone was saying: 'Sometimes you find more than one anti-derivative, sometimes you don't, eh what you gonna do about it?'

  • 1
    $\begingroup$ In (1), should one of the $f$s be a $g$? $\endgroup$ – Noah Schweber Aug 4 '16 at 17:18

It's unclear what question you're asking - I believe you are over-using "$f$." If you are asking how many antiderivatives a given function has, consider the following:

Suppose $G'=f$, and $c$ is a constant. Show that $(G+c)'=f$.

This shows that any function which has an antiderivative, has infinitely many (in fact, continuum many). In particular, $e^x$ is not the only antiderivative of $e^x$ - there's also things like $e^x+17$.

In the other direction, suppose $G'=H'=f$. Then what can you say about $(G-H)'$? What does that tell you about the function $G-H$ (under reasonable niceness assumptions on $G$ and $H$ - in particular, you want to assume here that the domain of $G$ and $H$ is all of $\mathbb{R}$, or at least is connected)?

Note that this isn't always obvious - for example, $\sin^{-1}({x\over 2})$ and $-\cos^{-1}({x\over 2})$ appear to not differ by a constant, but in fact they do.

  • $\begingroup$ Yes that is exactly what I'm asking, how many anti-derivatives a given function has. With your $G-H$ example, you're trying to show that anti-derivatives will only ever differ by a constant, regardless of how different the anti-derivatives appear algebraically, am I correct in saying that? $\endgroup$ – Perturbative Aug 4 '16 at 17:42
  • $\begingroup$ @Perturbative (You should edit your question then to make that clear.) Yes - although you need an assumption on the domain in order to get this. Consider the function $G(x)=0$ if $x<0$, and $1$ if $x>0$, on the domain $\mathbb{R}\setminus\{0\}$. $G$ is differentiable on this domain - $G'$ is the constant function $f: x\mapsto 0$. Another function on this domain with the same derivative is $H(x)=0$ for all $x$. However, $G-H$ is not a constant. This is made possible by the "split" in the domain; note that $G-H$ is "locally constant." $\endgroup$ – Noah Schweber Aug 4 '16 at 19:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.