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I'm having lectures on complex analysis, there is this theorem:

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I got a bit curious about the following: Having an antiderivative depends also on our choice of elementary functions, I am aware that we can enlarge (or shorten) this set (at least this is what was told to me in elementary calculus lectures, the professor gave a demonstration that$f(x) =\frac{1}{x}$ does not have an antiderivative if the set of elementary functions is the set of polynomial functions. This seemed to imply that the choice of this set is arbitrary).

So what happens to this theorem when we enlarge (or shorten) the set of elementary functions? Does it only work for the the standard set of elementary functions?

Red Banana
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  • What does "having an antiderivative" mean here? It seems to me that what this means to you is different from what I recall being standard. To me, having an antiderivative means that there exists a function that once differentiated yields the starting function. This is completely independent of elementary functions. Moreover, your definition of elementary function is also different from what is standard. – Git Gud Jul 30 '21 at 00:52
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    An antiderivative here just means any antiderivative, not necessarily a closed-form antiderivative, so the set of elementary functions doesn't make a difference. – Jair Taylor Jul 30 '21 at 00:53
  • @GitGud "Having an antiderivative" means having a function that can be written in terms of elementary functions. – Red Banana Jul 30 '21 at 00:54
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    @JairTaylor is right. "Having an antiderivative" in theorem 14.2 has nothing to do with elementary functions. – GEdgar Jul 30 '21 at 00:54
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    @RedBanana No, that's not the meaning of ”having an antiderivative” – jjagmath Jul 30 '21 at 00:56
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    It would be truly amazing if something as simple as (ii) or (iii) would characterize having an elementary antiderivative! – GEdgar Jul 30 '21 at 00:59
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    Perhaps what is confusing you is that over $\mathbb{R}$, every continuous function $f(x)$ has an antiderivative, namely $\int_a^x f(x) , dx$ for some $a$ in the interval (but not every function has an elementary antiderivative.) However, this is not the case over $\mathbb{C}$, where having any antiderivative is a very special property. – Jair Taylor Jul 30 '21 at 00:59
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    @JairTaylor Ah, I thought that "having an antiderivative" meant "having an antiderivative in terms of elementary functions". – Red Banana Jul 30 '21 at 01:03
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    Your elementary calculus professor seems to have been misleading you in several ways: there is a generally agreed definition of the term elementary function - it's not a term that is up for grabs. Many differentiable functions (i.e., functions with an antiderivative) are not elementary. The notion of whether a function has an antiderivative is not dependent on any putative alternative definition of "elementary function". – Rob Arthan Jul 30 '21 at 01:04
  • @RobArthan Yeah, it looks like this is another of those "calculus white lies". – Red Banana Jul 30 '21 at 01:07
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    @RedBanana: commiserations! Good on you for questioning what you are being taught. – Rob Arthan Jul 30 '21 at 01:08
  • @RobArthan I don't think that part is problematic. It's pretty clear what OP means by enlarging the set of elementary functions. Call them pseudo-elementary or something if you must. – Jair Taylor Jul 30 '21 at 01:12
  • @RobArthan You wrote ”Many differentiable functions (i.e., functions with an antiderivative)” but ”differentiable function” is not the same that ”function with an antiderivative” – jjagmath Jul 30 '21 at 01:19
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    @JairTaylor: I think redefining standard mathematical terminology in an ad hoc way is very bad practice. In this case, the notion of elementary function has absolutely nothing to do with the theorem that has led to the OP's confusion. It is the OP's elementary calculus professor who is to blame for giving a wrong understanding of the concept of antiderivative. – Rob Arthan Jul 30 '21 at 01:19
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    @jagmath: apologies, that was an error on my part. Please read my comment as saying "There are many functions with an antiderivative that are not elementary functions." – Rob Arthan Jul 30 '21 at 01:22

1 Answers1

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This statement is wrong: ”Having an antiderivative depends also on our choice of elementary functions”.

A function $f$ may or may not have an antiderivative. If it have an antiderivative $F$, that's independent of what kind of functions you are willing to accept as elementary.

We say, for example, that $f(x) = e^{x^2}$ doesn't have an elementary antiderivative not because it doesn't have an antiderivative, but because its antiderivative is not elementary.

jjagmath
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  • Isn't your second phrase equivalent to the phrase you're quoting? – Red Banana Jul 30 '21 at 00:57
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    No. Take the following more concrete examples: Any continuous function $f:\Bbb R \to \Bbb R$ have an antiderivative. Doesn't matter what kind of function it is, it can be elementary or not, but it's an antiderivative. On the other hand, there exists functions that are very discontinuous and don't have an antiderivative (elementary or not). – jjagmath Jul 30 '21 at 01:09