# 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: 743
 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: 743

## 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: 273 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: 743 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.
HW Helper
PF Gold
P: 2,655
 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.)
 Sci Advisor HW Helper P: 9,406 after all is said and done, i realize you may be primarily interested in question 3) above. Hence I refer you to this article: http://www.claymath.org/programs/out...s05/Conrad.pdf
 P: 273 Since you can always keep on taking the anti-derivative of an anti-derivative and so on and since functions continuously become more exotic, does that mean that there is an infinite number of new functions that we have yet to discover? If we keep taking the anti-derivative, will we ever loop back to something simple?
 Sci Advisor HW Helper P: 9,406 I don't think so. But it a little confusing to me, since we can get a hold on the situation by using abstract properties of functions rather than trying to write down specific examples. I.e. abstractly, antiderivatives are more special, and hence less exotic in some sense than the functions we integrate to get them. I.e. in some sense a piecewise linear function, which is the antiderivative of a step function, is more exotic because it is not piecewise constant, but less exotic because it is continuous. In general an antiderivative is Lipschitz continuous, in particular continuous, whereas integrable functions do not need to be continuous anywhere. So antiderivatives have better properties than integrable functions. But as far a specific functIONS GOES, it seems to me that taking antiderivatives is generally a way to get new functions from old. But I am not an expert on question 3). It's 1 and 2 that I have been thinking about.
Engineering
Thanks
P: 6,055
 Quote by tahayassen If we keep taking the anti-derivative, will we ever loop back to something simple?
Sometimes, yes. Maybe you haven't learned about derivatives of exponential and trig functions yet!
 Sci Advisor HW Helper P: 9,406 as i read conrad's paper, the existing theory only shows that many functions which are elementary in the usual sense of precalculus, do not have elementary antiderivatives. So my assumption that the situation continues as you add more new functions is entirely speculative.

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