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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".

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    $\begingroup$ 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". $\endgroup$
    – Ivo Terek
    Apr 29, 2020 at 3:24
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    $\begingroup$ (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.) $\endgroup$
    – Ivo Terek
    Apr 29, 2020 at 3:25
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    $\begingroup$ 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. $\endgroup$
    – Paramanand Singh
    Apr 29, 2020 at 3:33
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    $\begingroup$ $\sin(x^2)$ is elementary, any antiderivative is not. $\endgroup$
    – Ivo Terek
    Apr 29, 2020 at 3:59
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    $\begingroup$ Also contrary to what you may think proving a function to be non-elementary is an algebraic problem and not an analytical one. $\endgroup$
    – Paramanand Singh
    Apr 29, 2020 at 5:49

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