9
$\begingroup$

When we differentiate a function $f(x)$, there are conditions under which the derivative would not exist and cannot become differentiable. However, I have tried looking online for any conditions for integration and I haven't found anything.

Are there any cases where $F(x)$ does not exist from $\int f(x)dx$? In other words, what makes a function non-integrable?

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ If $f$ has more than countably many discontinuities, it depends what kind of integral you are using. $\endgroup$ – D_S Jul 28 '16 at 15:25
  • 1
    $\begingroup$ Usually one is interested in existence of definite integrals. But one interesting fact is a theorem of Darboux, that any derivative has the Intermediate Value Property. So if $f$ does not have that property it cannot have an everywhere antiderivative. $\endgroup$ – André Nicolas Jul 28 '16 at 15:28
  • 1
    $\begingroup$ countably many, not countless. lol $\endgroup$ – D_S Jul 28 '16 at 15:28
  • 2
    $\begingroup$ Oops. Yes, you're correct. I misread that haha. $\endgroup$ – KingDuken Jul 28 '16 at 15:28
  • 5
    $\begingroup$ Note also that calculus students also often misunderstand and think that "integrable" means "has an antiderivative expressible as an elementary function". This is not the case, but this is indeed an interesting issue, which is quite complicated (but more or less solved). $\endgroup$ – Ian Jul 28 '16 at 15:35
8
$\begingroup$

A bounded function $f:[a,b]\subset\mathbb{R}\to\mathbb{R}$ is integrable if it is continuous. Actually, $f$ only needs to be almost continuous, meaning it can be discontinuous at countably-many points and still be integrable.

|cite|improve this answer
$\endgroup$
  • 10
    $\begingroup$ The necessary and sufficient condition is that it is discontinuous at a set of points which has Lebesgue measure zero. (This is a theorem due to Lebesgue, considerably more recent than Riemann's work.) Countable sets are in this class, but there are others. $\endgroup$ – Ian Jul 28 '16 at 15:31
  • $\begingroup$ @Ian Is that also true for Riemann integrals? $\endgroup$ – Rhymoid Jul 28 '16 at 21:45
  • 3
    $\begingroup$ @Rhymoid It's a theorem due to Lebesgue which uses the Lebesgue measure but is actually about the Riemann integral. There are plenty of nowhere continuous Lebesgue integrable functions. $\endgroup$ – Ian Jul 28 '16 at 21:47
6
$\begingroup$

An interesting result due to Darboux is that any derivative has the Intermediate Value Property. As a consequence, if $f$ does not have the Intermediate Value Property, then $f$ cannot have an everywhere defined antiderivative $F$.

Remark: Your question asked about the indefinite integral. However, the notion of primary interest is the definite integral. For most mathematical purposes, the other answers are the useful ones.

|cite|improve this answer
$\endgroup$
4
$\begingroup$

The function $f:[0;1]\to \mathbb R$ defined by $$f(x) = \begin{cases}0 \mathrm{\;if\;}x\notin\mathbb Q\\1 \mathrm{\;if\;}x\in\mathbb Q\end{cases}$$ is not Riemann-integrable : see here.

|cite|improve this answer
$\endgroup$
2
$\begingroup$

You're asking when $$\int\limits_E f(x)dx$$ exists, where $E$ is some subset of the real line. This depends on your definition of integral. For example, take $f(x) = x$. One way to interpret $$\int_{-\infty}^{\infty} xdx$$ is as $$\lim\limits_{a \to \infty} \int_{-a}^a xdx$$ and this is clearly $0$. But there are other ways to interpret this integral and have it not converge. For example, if you let the positive bound to go infinity faster than the negative bound: $$\lim\limits_{a \to \infty} \int_{-a}^{2a} xdx$$ Actually, you can modify how the positive and negative bounds go to infinity and make that integral come out to whatever you want.

On the other hand, the Lebesgue integral $\int\limits_{(-\infty,\infty)} f(x)$ is defined quite differently and is not open to interpretation. It agrees with the Riemann integral on bounded intervals on most nice functions, but by definition $\int\limits_{(-\infty,\infty)} f$ exists if and only if when you only integrate on parts where $f$ is positive, and when you integrate on the parts where $f$ is negative, each of those converges. So $$\int\limits_{(-\infty,\infty)} xdx$$ is not defined as a Lebesgue integral.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.