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In the way the derivative can be defined as a limit, specifically $$f'(x):=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$ or any of the other possible variants, is there a way to define the antiderivative, as in indefinite integral?

The handful of sources I've looked over (Wikipedia and MathWorld, to name a few) all refer to the antiderivative simply as a "nonunique inverse operator" (I'm paraphrasing). I can't say I'm completely satisfied with this notion. Is the following the best we can do?

If $F(x)$ is a function that satisfies $\dfrac{d}{dx}F(x)=f(x)$, then $F(x)$ is called an antiderivative of $f(x)$. (Forgive the lack of formality.)

Is it possible to come up with a sort of "inverse-limit" to describe antiderivatives?

user170231
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    an anti derivative of $f(x)$ is $\int_a^x f(t) , dt.$ – abel Jun 08 '15 at 23:10
  • The definition you give is more/less what I typically see. However, it should say that $F(x)$ is called an antiderivative of $f(x)$ since it is not unique (the implies unique). – TravisJ Jun 08 '15 at 23:14

2 Answers2

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Since the antiderivative is so far from unique, we can't give anything like a pointwise formula for it as it looks like you want without choosing one. Probably the best idea is the definite integral, $$F(x)=\int_0^x f(x)dx=\lim_{n\to \infty} \frac{x}{n}\left(\sum_{k=0}^{n-1} f(kx/n)\right)$$ Here I've used the left Riemann sum, which is of course not the only way to define an integral, but it works whenever the integral exists. It's possible to write a less arbitrary definition, but uglier.

Kevin Carlson
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  • This Riemann sum only works if $f$ is continuous! – Zardo Jun 08 '15 at 23:14
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    @Zardo Well, no, it computes the integral of the function that's $1$ at $0$ and $0$ everywhere else perfectly well. – Kevin Carlson Jun 08 '15 at 23:19
  • That's true. Continuity is only a sufficient condition. Piecewise continuity is sufficient as well. But (as far as I know) there are functions where this sum does not converge to the actual integral. – Zardo Jun 08 '15 at 23:26
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    I don't think that's actually true. If the limit over shrinking partitions exists, i.e. the function is Riemann integrable, then it hits the right integral no matter what path you take through partitions. I admit I haven't written down a proof using the Riemann integral in four or five years, but Pedro agrees with me here: http://math.stackexchange.com/questions/353452/necessary-and-sufficient-conditions-for-riemann-integrability. – Kevin Carlson Jun 08 '15 at 23:40
  • It might be preferable to say "whenever the limit exists" rather than "whenever the integral exists", to clarify that you don't mean all functions that can be integrated by any method? – Harry Johnston Jun 09 '15 at 02:23
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For all regulated functions you can define a sequence of step functions that converge to the function you wish to integrate. Then compute the Riemann sum over the partition.

RowanS
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