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The fundamental theorem of calculus states that if $f:[a,b] \to \mathbb{R}$ is integrable and $F(x) = \int_{a}^{x} f(t)\ dt$, then $F'(x) = f(x)$ at every point $x$ at which $f$ is continuous. This means that if $f$ is integrable, $F'(x) = f(x)$ almost everywhere. If $f$ is continuous on $[a,b]$, then $F$ is an antiderivative of $f$, since $F'(x) = f(x)$ holds for all $x \in [a,b]$.

But what if $f$ has a discontinuity at some $x \in [a,b]$? In this case, it is not necessarily true that $F'(x) = f(x)$, and so we cannot necessarily conclude that $F$ is an antiderivative of $f$. Does this mean that there is no antiderivative of $f$? Is it possible for $f$ to have an antiderivative but the indefinite integral $F$ is not an antiderivative of $f$?

I know that if $f$ has a jump discontinuity, then $f$ can have no antiderivative (since the derivative of a function must satisfy the intermediate value property), but what if we have some other type of discontinuity?

TheProofIsTrivium
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5 Answers5

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Consider the map$$\begin{array}{rccc}F\colon&\Bbb R&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}x^2\sin\left(\frac1x\right)&\text{ if }x\ne0\\0&\text{ otherwise.}\end{cases}\end{array}$$Then $F$ is differentiable and$$(\forall x\in\Bbb R):F'(x)=\begin{cases}-\cos\left(\frac1x\right)+2x\sin\left(\frac1x\right)&\text{ if }x\ne0\\0&\text{ otherwise.}\end{cases}$$So, $F'$ is discontinuous at $0$. But $F$ is an antiderivative of $F'$.

José Carlos Santos
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  • I understand that $F'$ has an antiderivative and is discontinuous, but how do you know that $\left(\int_{a}^{x} F'(t)\ dt\right)'(0) \neq F'(0)$? – TheProofIsTrivium May 20 '20 at 22:43
  • What has that to do with your question? Wasn't your question to determine whether a discontinuous function can have an antiderivative? Besides, in my case we have $(\int_a^xF')'(0)=F'(0)$. – José Carlos Santos May 20 '20 at 22:47
  • My question was, can $f$ have an antiderivative even if the indefinite integral $F$ is not an antiderivative. In your example, the indefinite integral is an antiderivative, even though this is not guaranteed by the FTC because $F'$ is not continuous at $0$. – TheProofIsTrivium May 20 '20 at 22:49
  • Sorry, I realized the title and the body are not really asking the same question. Is there a good example of a function $f$ where $F = \int_{a}^{x} f(t)\ dt$ is not an antiderivative but there is some antiderivative? – TheProofIsTrivium May 20 '20 at 22:53
  • I can't tell you right now. Yes, I read the question from your title and I provided an answer to that question. – José Carlos Santos May 20 '20 at 22:59
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Assume $f:[a, b] \to\mathbb {R} $ is Riemann integrable on $[a, b] $ and possesses an anti-derivative $g:[a, b] \to\mathbb {R} $ on $[a, b] $ ie $g'(x) =f(x) \, \forall x\in[a, b] $. Then by Fundamental Theorem of Calculus we have $$g(x) =g(a) +\int_{a} ^{x} f(t) \, dt=g(a) +F(x) $$ It follows that $$F'(x) =g'(x) =f(x) $$ for all $x\in[a, b] $.

Therefore the indefinite integral $F:[a, b] \to\mathbb {R} $ defined by $$F(x) =\int_{a} ^{x} f(t) \, dt$$ also acts as an anti-derivative of $f$ on $[a, b] $.

Without symbols we can summarize as

If a Riemann integrable function possesses an anti-derivative over some closed interval then its indefinite integral also acts an anti-derivative over the same interval.

In short the answer to your question in bold is NO if we assume Riemann integrability of $f$. However not every Riemann integrable function possesses an anti-derivative. Continuity is only a sufficient for existence of anti-derivative and jump discontinuity is only a sufficient condition for non-existence of anti-derivative.

Paramanand Singh
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  • Aren’t you assuming continuity of $f$ at $x$ here? My question is about what happens when this is not true. – TheProofIsTrivium May 21 '20 at 03:38
  • @yonatano: no I am not assuming the continuity of $f$ at any point. All I assume is that $f$ is Riemann integrable and it possesses an anti-derivative $g$ on the interval under consideration so that $g'=f$. Since $F(x) =g(x) - g(a) $ we have $F'(x) =g'(x) =f(x) $. – Paramanand Singh May 21 '20 at 05:25
  • @yonatano: You will now appreciate the fact that there are two parts of the Fundamental Theorem of Calculus. First part deals with the derivative of indefinite integral. The second part deals with anti-derivatives and their role in evaluation of the integral. It is the second part which is at work here and says that if an anti-derivative exists it must differ from indefinite integral by a constant and therefore they possess the same derivative (the original function which was being integrated). – Paramanand Singh May 21 '20 at 05:34
  • I don't see why it follows from $g(x) = F(x) + g(a)$ that $F'(x) = g'(x) = f(x)$. How do you know that $F$ is differentiable at $x$? – TheProofIsTrivium May 21 '20 at 05:38
  • @yonatano: because $g(x) $ is differentiable at $x$ with derivative $f(x) $ so $g(x) - g(a) $ is also differentiable at $x$ with same derivative $f(x) $. – Paramanand Singh May 21 '20 at 05:39
  • @yonatano: You are trying to think too deep. If $\sin x$ is differentiable with derivative $\cos x$ then $\sin x - 10$ is also differentiable with same derivative $\cos x$. – Paramanand Singh May 21 '20 at 05:42
  • @yonatano: the fact that $f$ possesses an anti-derivative is a very severe condition on $f$ and that controls lot of its behavior. – Paramanand Singh May 21 '20 at 05:42
  • Thank you, this makes sense now. You're saying that if $f$ is Riemann integrable it does not necessarily have an antiderivative, but if it does have an antiderivative then also the indefinite integral must be an antiderivative. Are there necessary and sufficient conditions for a function $f$ to have an antiderivative? Does this change if we use the Lebesgue integral instead of Riemann? – TheProofIsTrivium May 21 '20 at 05:50
  • @yonatano: Sorry, I am not well versed in Lebesgue integral. – Paramanand Singh May 21 '20 at 06:03
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If there is a function $g$ which is differentiable at every point with $g'=f$ and if $\int_a^{x} f(t)dt$ exists for all $x$ then it is necessary that $g$ is absolutely continuous and $g(x)=g(a)+\int_0^{x} g'(t)dt=g(a)+\int_0^{x} f(t)dt$. This is proved in Rudin's RCA. Hence $\int_0^{x} f(t)dt$ is an anti-derivative.

Kavi Rama Murthy
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One can prove that if $f:J \rightarrow \Bbb R $, $J$ is an interval, is locally Riemann integrable (i.e. integrable in every compact interval of $J$) and has an antiderivative then the function $$ h(x)=\int^x_af(t)dt,\; a\in J $$ is an antiderivative of $f(x)$. That is because if $f(x)$ is Riemann integrable and have an antiderivative $g(x)$ then $$ \int^b_af(t)dt=g(b)-g(a) $$ A proof can be find in Bartle's book.

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As I understand the question: If $F'=f$ on $[a,b],$ is it true that $F(x)=F(a) + \int_a^x f$ for $x\in [a,b]?$ Answer: No if we stay in the realm of the Riemann integral. One reason is that $f$ need not even be Riemann integrable (RI). A very nice example shows that $F$ can be differentiable everywhere and strictly increasing, and yet $f(x)=F'(x)=0$ for $x$ in a dense subset of $[a,b].$ Such an $f$ cannot be RI, so we can't get even get off the ground. See https://en.wikipedia.org/wiki/Pompeiu_derivative.

What if $F'=f$ on $[a,b]$ and $f$ is RI? Then the answer is yes. This is a well known result, following quickly from the MVT.

You might ask: Suppose $F'=f$ on $[a,b]$ and $f\in L^1[a,b].$ Do we have $F(x)=F(a) + \int_a^x f$ for $x\in [a,b]?$ Yes. You can find a proof of this in Rudin's Real and Complex Analysis.

Beyond RI and LI you could look at improper integrals I suppose, but I'm out of ammo at this point.

zhw.
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