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

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.

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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.)

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Under any sensible definition of "antiderivative", your function $f$ has the antiderivative $F(x)=|x|$. – TonyK Nov 17 '12 at 19:32
@TonyK: This is a fair point. By an "antiderivative" of $f$, I mean a differentiable function $F$ such that $F' = f$. The question asks if $f$ "has an antiderivative on domain $D$". By this, I assume he means on the entire domain $D$. With this interpretation, since $F(x) = |x|$ fails to be differentiable at $x = 0$, it is not an antiderivative for $f$ at $x = 0$ and thus not on all of $[-1, 1]$. – user48944 Nov 17 '12 at 21:29
THank you so much for a simple and clear example. – Jiddu Krishnamurti Nov 18 '12 at 1:46
@TonyK Well, that is not entirely true. $F'(0)$ doesn't exist, while he's setting $f(0)=1$. – Peter Tamaroff Nov 18 '12 at 2:13
@Peter: This is a sensible definition: $F$ is the antiderivative of $f$ if $F$ is continuous everywhere, differentiable almost everywhere, and $F' = f$ wherever $F'$ exists. – TonyK Nov 18 '12 at 10:47
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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.

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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.

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 The question was not about formal integrability. Your answer is irrelevant! – TonyK Nov 17 '12 at 19:28 @TonyK Why not supply a "relevant" answer? – Fly by Night Nov 17 '12 at 23:23 Is that supposed to be a defence of your answer? – TonyK Nov 18 '12 at 10:49

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

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.

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