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Give an example of Riemann Integrable function (discontinuous) which has a primitive. Justify

Please help me. I need an example of such a discontinuous function.

Is $f(x)=\cos{1/x}$ a right choice

user1942348
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  • A primitive of $f$ is a function $F$ where $F' = f$. – RRL Mar 05 '18 at 01:16
  • @RRL Yes. its true. Would you provide me an example of a discontinuous function like that – user1942348 Mar 05 '18 at 01:19
  • Please edit the question to be clear about what you are looking for. Obviously $f(x) = 2x$ is Riemann integrable with a primitive $F(x) = x^2$. Seems like you want something else. – RRL Mar 05 '18 at 01:36
  • @RRL Edited above – user1942348 Mar 05 '18 at 01:38
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    Your example $\cos(1/x)$ is almost correct. The function $f(x) =\cos(1/x),f(0)=0$ is discontinuous at $0$ and Riemann integrable in any closed interval and has a primitive $F$ given by $F(x) =\int_{0}^{x}f(t),dt$ as we have $F'(x) =f(x) $ everywhere. – Paramanand Singh Mar 05 '18 at 04:29
  • @ParamanandSingh What will be the primitive then – user1942348 Mar 05 '18 at 09:42
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    As mentioned in previous comment the primitive is the function $\int_{0}^{x}\cos(1/t),dt$. – Paramanand Singh Mar 05 '18 at 10:35
  • @ParamanandSingh Understand, but how to integrate that? – user1942348 Mar 05 '18 at 12:01
  • There is no need to integrate that. We have a function $F$ which is primitive of $f$. That's all that matters. – Paramanand Singh Mar 05 '18 at 12:27
  • @ParamanandSingh: What would be the best way to show that $\int_0^x \cos(1/t) \mathrm{d}t$ is the primative? The fundamental theorem of calculus doesn't quite cover it. – Theo Bendit Mar 07 '18 at 13:05
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    @TheoBendit: If $F(x) =\int_{0}^{x}f(t),dt$ where $f(x) =\cos(1/x),f(0)=0$ then by FTC $F'(x) =f(x) $ if $x\neq 0$. It requires some effort to show that $F'(0)=0=f(0)$ and happens via an indirect use of FTC. See https://math.stackexchange.com/a/1784625/72031 – Paramanand Singh Mar 07 '18 at 13:51

1 Answers1

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Consider $$f(x) = \begin{cases} x^2 \sin(x^{-1}) & \text{if } x \neq 0 \\ 0 & \text{if }x = 0 \end{cases}.$$ Then $$f'(x) = \begin{cases} 2x \sin(x^{-1}) - \cos(x^{-1}) & \text{if } x \neq 0 \\ 0 & \text{if }x = 0 \end{cases}.$$ (The derivative at $0$ follows from the derivative squeeze theorem, as $-x^2 \le f(x) \le x^2$.) Note that $f'$ is discontinuous at $0$, and has a primitive $f$.

All that remains is showing that $f'$ is Riemann Integrable. Note that we may express

$$f'(x) = \begin{cases} 2x \sin(x^{-1}) & \text{if } x \neq 0 \\ 0 & \text{if }x = 0 \end{cases} - \begin{cases} \cos(x^{-1}) & \text{if } x \neq 0 \\ 0 & \text{if }x = 0 \end{cases}.$$

The former terms is continuous by the squeeze theorem, and hence is integrable over any compact interval. The latter term can be proven Riemann Integrable by modifying my answer here.

Theo Bendit
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  • Very interesting. Thanks. How to check the Riemann Integration counterpart – user1942348 Mar 05 '18 at 02:13
  • $f′(x)$ is R-integrable as it has only one point of discontinuity at $x=0$. Right? – user1942348 Mar 05 '18 at 02:30
  • That's certainly one way to look at it (it's also important to note that $f'$ is bounded, locally). I also linked to that answer because it almost directly deals with $\cos(x^{-1})$. The only difference is that it's compressed by a factor of $\pi$, and potentially you want a symmetric interval like $[-1, 1]$, instead of something like $[0, 1]$. – Theo Bendit Mar 05 '18 at 02:37
  • Thanks. So the conclusion is "Discontinuities in $f(x)$ can allow the existence of its Primitive $F(x)$". But it contradints with "no discontinuous function has a primitive" – user1942348 Mar 05 '18 at 02:40
  • Yes. Mine and RRL's answer contradict each other. – Theo Bendit Mar 05 '18 at 02:42
  • Please see: https://books.google.co.in/books?id=7I-RCwAAQBAJ&pg=PA122&lpg=PA122&dq=discontinuous+function+that+has+a+primitive,&source=bl&ots=0UI7XLml-S&sig=o-4lJKzu-tQLnfT1jawDrmOreMI&hl=en&sa=X&ved=0ahUKEwjr_7ekiNTZAhVKp48KHTOEC1cQ6AEIdjAI#v=onepage&q=presence%20of%20jump&f=false – user1942348 Mar 05 '18 at 02:45
  • It says that presence of jump discontinuity in f(x) prevents the existance of primitive – user1942348 Mar 05 '18 at 02:46
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    That's true. A jump discontinuity in a derivative will violate Darboux's theorem, which RRL refers to. If you take a small enough interval around the jump discontinuity, the function to the left will be close enough to the left limit, and similarly to the right as well, that there will be a gap, locally, in the range of the function. This violates the intermediate value property. However, not every discontinuity is a jump discontinuity! – Theo Bendit Mar 05 '18 at 02:58
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    @TheoBendit: I think you are correct. Nice answer. – RRL Mar 05 '18 at 03:29