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See edit
It is known that the anti derivative of $\sin(x^2)$ is not an elementary function, and one can represent it using a power series by term-by-term integration of its Taylor series.

However, is there any way to show that it's antiderivative is transcendental and cannot be represented by an elementary function?

My thoughts are to assume that there is an elementary function $f$ such that $f'=\sin(x^2)$, and show a contradiction. However, are there any special properties that only elementary or only transcendental functions have (so I can show that $f$ does not satisfy one or satisfies one)?

Edit:
As people pointed out in comment, the opposite of transcendental is algebraic, (and therefore $\int \sin(x^2)$ is certainly transcendental)... Maybe ignore transcendental and imagine it to mean "non elementary".

user12986714
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  • What is the definition of a transcendental function? – herb steinberg Apr 29 '20 at 02:37
  • @herb Transcendental functions are those that are not elementary (maybe google "elementary functions") – user12986714 Apr 29 '20 at 03:04
  • @herb Or maybe google "transcendental functions" and it will lead you to a Wikipedia article :-) – user12986714 Apr 29 '20 at 03:23
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    I don't think one can prove this just by following their nose and using standard real analysis. This sort of question is much deeper than one would think and leads to a whole area of math called "Differential Galois Theory". – Ivo Terek Apr 29 '20 at 03:24
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    (It essentially deals with differential fields instead of fields, and Picard-Vessiot extensions instead of Galois extensions. This should give enough keywords to look further into the subject if you want to.) – Ivo Terek Apr 29 '20 at 03:25
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    Transcendental means not algebraic. Anyway the function in question is neither transcendental nor elementary. For the first part note that $\sin x^2$ is not algebraic and hence its anti-derivative is also not algebraic. – Paramanand Singh Apr 29 '20 at 03:33
  • I thought that the problem here is that elementary functions include trig and exp/log ones. Otherwise it will be very easy to show that $\int \sin(x^2)$ cannot be represented by finite polynomials. Maybe I'd better give up as I am just self learning "Introductory to Real Analysis" (and calculus of variations in parallel)... – user12986714 Apr 29 '20 at 03:35
  • @Paramanand "Transcendental means not algebraic", which I thought meant that a function is either transcendental or algebraic (law of the excluded middle). So I wonder why it is neither transcendental nor algebraic? Also that isn't $\sin(x^2)$ a composition of$\sin(x) $ and $x^2$? Thank you. – user12986714 Apr 29 '20 at 03:41
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    $\sin(x^2)$ is elementary, any antiderivative is not. – Ivo Terek Apr 29 '20 at 03:59
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    Yes either a function is algebraic or transcendental. In your case both $\sin x^2$ and its anti-derivative are transcendental. Out of all transcendental functions there is a special category called elementary functions consisting of functions obtained by a finite number of arithmetic operations and compositions on algebraic, trigonometric and exponential functions and their inverses. $\sin x^2$ is an elementary function. – Paramanand Singh Apr 29 '20 at 05:45
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    One can prove using the techniques first championed by Liouville (and now its abstraction called differential galois theory) that anti-derivative of $\sin x^2$ is non-elementary. – Paramanand Singh Apr 29 '20 at 05:47
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    Also contrary to what you may think proving a function to be non-elementary is an algebraic problem and not an analytical one. – Paramanand Singh Apr 29 '20 at 05:49
  • It looks like the definition of transcendental function is not so obvious. – herb steinberg Apr 30 '20 at 00:22

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