Non integrable function

Question

There are some functions whose indefinite integration is not possible. What is the reason behind this?

If you mean "no elementary antiderivative", see https://math.stackexchange.com/questions/155/how-can-you-prove-that-a-function-has-no-closed-form-integral, https://math.stackexchange.com/questions/265780/how-to-determine-with-certainty-that-a-function-has-no-elementary-antiderivative, https://math.stackexchange.com/questions/560679/how-to-tell-if-an-integral-can-be-integrated-has-an-elementary-anti-derivative — Hans Lundmark, Jul 23 '17 at 16:04

score 1 · Answer 1 · answered Jul 23 '17 at 15:40

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There are several possible reasons. For instance, Darboux's theorem says that if $f$ is differentiable, then $f'$ has the mean value property. Therefore, a function that lacks that property cannot have an indefinite integral. That's the case, for instance, of the function$$f(x)=\begin{cases}1&\text{ if }x\geqslant 0\\0&\text{ otherwise.}\end{cases}$$

answered Jul 23 '17 at 15:40

José Carlos Santos

427,504

what about $$x[x\ge 0]+C=\int f(x),\mathrm dx$$ where I've used the Iverson bracket? – Masacroso Jul 23 '17 at 15:53
@Masacroso I have no idea about what you mean by that. – José Carlos Santos Jul 23 '17 at 15:56
what I wrote is equivalent to the ramp function more a constant. It represents the indefinite integral of $f$. – Masacroso Jul 23 '17 at 15:57
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@Masacroso That function is not differentiable at $0$. – José Carlos Santos Jul 23 '17 at 15:58
I know, but it represent the indefinite integration of $f$. – Masacroso Jul 23 '17 at 15:59
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@Masacroso No it doesn't. An indefinite integral of the function $f$ that I mentioned would have to be a function $F$ such that $F'=f$. Your function doesn't have this property. – José Carlos Santos Jul 23 '17 at 16:01
well, it depends on the definition of indefinite integral then. You are assuming that indefinite integral is equivalent to primitive, but I dont think its necessary. – Masacroso Jul 23 '17 at 16:04
@Masacroso Then tell us what's your definition. – José Carlos Santos Jul 23 '17 at 16:06
a definition of indefinite integral could be a function such that when taking values it represent any definite integral of $f$, that is, if $F$ is an indefinite integral of $f$ then $\int_a^b f(x),\mathrm dx=F(b)-F(a)$ – Masacroso Jul 23 '17 at 16:08
@Masacroso Yes, it could be. But I must assume that the OP is using the standard definition of indefinite integral. – José Carlos Santos Jul 23 '17 at 16:11

Non integrable function

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