Is “integrability” equivalent to “having antiderivative” ???

Question

I am wondering if "a function $f(x)$ is integrable on a domain $D$" this proposition is equivalent to "$f(x)$ has antiderivative on domain $D$". If it is not the case, give me a counter example. Thank you.

Answer 1

There is an even simpler example. Take, e.g.,

$$ f(x) = \begin{cases} 1 & x \geq 0 \\ -1 & x < 0 \end{cases} $$

This function is Riemann-integrable on $[-1, 1]$, but it cannot be the derivative of any function because derivatives cannot have jump discontinuities. (See, e.g., this question.)

Answer 2

Not really. The function

$$f(x)=\begin{cases} 0\text{ if } x\in\Bbb R\setminus \Bbb Q\\\frac 1 q \text{ if } \frac p q \in\Bbb Q\end{cases}$$

is defined on $\Bbb R$, has period one, and it is Riemann integrable over $[0,1]$ with $$\int_0^1 f=0$$ but the function has no antiderivative at all (it is not differentiable either). It is known as Thomae's function.

Answer 3

It's important to specify what you mean by "having an anti-derivative". Let's just concentrate on elementary functions. One example is $f(x) = e^{-x^2}.$ There is no elementary function with the property that its derivative is $e^{-x^2}.$ However,

$$\int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi} \, . $$

The reason I mentioned semantics is because, at some point, someone defined a special function called the error function, and denoted by $\text{erf}$, with the property that

$$\int e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2}\text{erf}(x) + C \, . $$

In some sense, you could just define a new function with the property that its derivative is the function whose anti-derivative you seek. You need to think about continuity though.

Answer 4

If you can handle a little more advanced material, let me provide you with a reference. Let $ I $ be a closed interval. As shown above by Peter, there exists a bounded function on $ I $ that is Riemann-integrable on $ I $ but does not have an antiderivative on $ I $. On the other hand, if you refer to

http://books.google.com/books?id=fXfEG-F2zJUC&pg=PA34&lpg=PA34&dq=derivatives+are+continuous+on+a+dense+set&source=bl&ots=-_8l2Zj5T1&sig=XnFiJniYei69Fbi0IhhzTdQ9Evs&hl=en&sa=X&ei=JMOnULTeDpKI9ASvyoHwCg&ved=0CDgQ6AEwAg#v=onepage&q=derivatives%20are%20continuous%20on%20a%20dense%20set&f=false,

you will see that there also exists a function on $ I $ that has an antiderivative on $ I $ but is not Riemann-integrable on $ I $.

Let me explain further. As mentioned in the reference, if $ A $ is a dense $ G_{\delta} $-subset of $ I $ (by definition, a $ G_{\delta} $-subset is the intersection of countably many open subsets), then there exists a function $ f $ on $ I $ that is (i) the derivative of a function on $ I $, (ii) continuous at all points in $ A $, and (iii) discontinuous at all points in $ I \setminus A $.

Now, Lebesgue's theorem on the necessary and sufficient condition for Riemann-integrability states that a bounded function on $ I $ is Riemann-integrable on $ I $ if and only if it is continuous almost everywhere on $ I $, i.e., the set of discontinuities of the function has measure $ 0 $. Produce a dense $ G_{\delta} $-subset $ A $ of $ I $ that has measure $ 0 $ (take the set of Liouville numbers contained in $ I $ for example). Then there is a function $ f $ on $ I $ that has an antiderivative and is discontinuous on $ I \setminus A $; by Lebesgue's theorem, $ f $ cannot be Riemann-integrable on $ I $.

Conclusion Riemann-integrability does not imply the existence of an antiderivative, and the existence of an antiderivative does not imply Riemann-integrability.