Lets say we have a function $f:\mathbb{R} \to \mathbb{R}$. How is it possible to tell that we cannot find the anti-derivative of the function? Is there any specific test for it?
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asked Sep 14, 2016 at 10:33
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2$\begingroup$ Do you mean anti-derivative or anti-derivative in terms of elementary functions? $\endgroup$– mvwSep 14, 2016 at 10:38
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$\begingroup$ I think that you want this. $\endgroup$– Ivo TerekSep 14, 2016 at 10:42
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1$\begingroup$ As a rule of thumb, functions have no closed-form antiderivative. :-) $\endgroup$– user65203Sep 14, 2016 at 10:49
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$\begingroup$ If you mean in terms of elementary functions, see here: math.stackexchange.com/questions/265780/… $\endgroup$– Hans LundmarkSep 14, 2016 at 12:47
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If you mean anti-derivatives in general, there seems to be no simple test.
However mathematicians have proven that certain classes of functions have an anti-derivative. E.g. a continous function has an anti-derivative. So all tests that prove a function to be continous would apply.
For the more weird cases see here.
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$\begingroup$ how about exp(x^2) ? that can be proved to be not integrable without the error function, which is defined in terms of the integral of exp(x^2) - if a function is continuous and continuously differentiable, then it is always possible to integrate its Taylor Series expansion - then call that function $\endgroup$– CatoSep 14, 2016 at 10:53
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$\begingroup$ You mean $e^{-x^2}$. Has an antiderivative. That it is defined in terms of an integral expression or a special function is just a matter of convenience. $\endgroup$– mvwSep 14, 2016 at 10:55
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$\begingroup$ what is the antiderivative of $e^{-x^2}$ or $e^{x^2}$? Since they are surely continuous? $\endgroup$– CatoSep 14, 2016 at 11:09
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$\begingroup$ @AndrewDeighton $f(x)=\int_0^x e^{-t^2}\,dt$ has the property that $f'(x)=e^{-x^2}$. See math.stackexchange.com/questions/523824/… $\endgroup$– egregSep 14, 2016 at 11:33