# Why do some functions not have Anti derivatives??

by Hysteria X
Tags: anti, derivatives, functions
 P: 21 as the title says why are some functions like ## √cotx##(root cotx) not integratable??
P: 708
 Quote by Hysteria X as the title says why are some functions like ## √cotx##(root cotx) not integratable??
Are you sure ? : http://www.wolframalpha.com/input/?i...8cot%28x%29%29
Can you give us a true example of what you call "not integrable" ?
 P: 635 Are you referring to functions such as sin(x^2) or log(log(x))? They do have antiderivatives, it's just impossible to express them finitely in terms of elementary functions.
P: 708

## Why do some functions not have Anti derivatives??

Hi Hysteria X !
The remark of Whovian is pertinent. If it is that what you mean, I suggest to have a look at a funny example, on page 2 of the paper "Sophomore's Dream Function" :
http://www.scribd.com/JJacquelin/documents
 P: 271 I believe the only functions that are impossible to integrate are ones with an infinite amount of discontinuities.
 P: 635 Such as the Dirichelt Function? Yep, "most" functions (being most functions people regularly encounter) are integrable. "Most" functions (in that we take a random number from f's codomain to be f(x), where x is an arbitrary number in f's domain, for all x in f's domain) are going to be non-integrable, however.
 P: 708 Yes, the Dirichelet function is a good example (considering the Riemann definition of integration). Of course, there are an infinity of functions of the Dirichelet kind. But, we have also to take account of the kind of integration considered : The Dirichelet function is Lebesgue's integrable.
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P: 2,561
 Quote by tahayassen I believe the only functions that are impossible to integrate are ones with an infinite amount of discontinuities.
If we are speaking of the Riemann integral, it can in fact handle infinitely many discontinuities, as long as the "infinity" is not too large. To be precise, a function is Riemann integrable if and only if it is bounded and its set of discontinuities has Lebesgue measure zero. All countably infinite sets have measure zero, and some uncountably infinite ones do, too, for example the Cantor set.

The Dirichlet function (i.e. the indicator function of the rational numbers) is discontinuous everywhere, so it fails the test.

On the other hand, the function (also, confusingly, sometimes called the Dirichlet function) defined by
$$f(x) = \begin{cases}0 & \textrm{ if }x \textrm{ is irrational} \\ 1/q & \textrm{ if }x \textrm{ is rational and }x = p/q \textrm{ in lowest terms}\end{cases}$$
is continuous at every irrational x, and discontinuous at every rational x, so its set of discontinuities is countable. Therefore this function is Riemann integrable. (Its integral is zero.)