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I am talking about indefinite integration of a function what ever its nature be as far as continuity and differentiabilty is Concerned. Can we integrate any function irrespective of the result be in elementary form or some special functions or infinite series. For example

$$\int e^{x^2}dx=\int 1+\frac{x^2}{1!}+\frac{x^4}{2!}+\cdots=x+\frac{x^3}{3}+\frac{x^5}{10}+\cdots$$

$$\int |x|dx=\int x dx$$ if $x \gt 0$ and $$\int |x|dx=\int -x dx$$ if $x \lt 0$

So is it True that any function can have Indefinite integral?

Umesh shankar
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    Every continuous function has an antiderivative. However, more general functions need not have antiderivatives, see here. – Wojowu Apr 11 '16 at 16:45

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No, of course not. If $f$ has an "indefinite integral" $F$, this means that $F' = f$, therefore your question is: "can any function be the derivative of some other function"? Well, it turns out that if $F$ is derivable, then $F'$ must posess the intermediate value property. Therefore, every function that does not have this property does not admit an indefinite integral. For instance, the function defined by $f(x) = \begin{cases} 1, x \in \Bbb Q \\ 0, x \notin \Bbb Q \end{cases}$ does not have Darboux's property. On the other hand, every continuous function has an indefinite integral.

This page has more of the information that you need. Notice that all the functions that do not admit an indefinite integral are somewhat pathological and artificial.

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Alex M.
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