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Note on duplicates
This question is not a duplicate; it asks when the inverse of the derivative equals the function, mine asks when the inverse of the derivative equals the integral.

The question
Let $f$ be any function. Define $F(x)$ to be the integral of $f(x)$, and $(f')^{-1}$ to be the inverse function of the derivative of $f$. When is $$F(x)=(f')^{-1}?$$ I don't know much about how to solve this equation. Any help would be appreciated.

  • $(f')^{-1}$ is not the reciprocal of $f'$.
  • I could do this, but I don't know where to begin from. Any hint of where to start from would also be helpful.

My thoughts and work

  • Does such a function even exist? Why/why not?
  • From the initial equation, we get $$F^{-1}(x)=f'(x)$$ Now we use this formula, to get $$f=xF^{-1}(x)-G\circ F^{-1}(x)+C$$ Where $G$ is the integral of $F$. So, we get $$f=xf'(x)-G\circ f'(x)+C$$ Now what can be done?

Motivation
I have seen many functional/differential equations that are just simple combinations of integrals, derivatives, like this which interested me in solving differential equations, that look simple but have hard solutions. This is not from any website/book/article, just my thought.

russian bot
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  • Not any function is integrable. Furthermore, the inverse operator of $\frac{\mathrm{d}}{\mathrm{d}x}$ does not exist as this operator is not injective (it kernel is $\mathbb{R}$ or $\mathbb{C}$ depending on the underlying field). – Nicolas Dec 20 '20 at 15:23
  • @Nicolas so it does not exist? – russian bot Dec 20 '20 at 15:24
  • The fundamental theorem of calculus tells you that if, say, $f$ is continuous on $[a,b]\subset\mathbb{R}$ and $f'$ exists on $[a,b]\setminus Z$ where $Z$ is countable, then the anti-derivative of the derivative gives you back the function $f$ MODULO constants (modulo $\mathbb{R}$ here). – Nicolas Dec 20 '20 at 15:27
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    Well, if $(f')^{-1}=F(x)$ then $F^{-1}(x)=f'(x)$ – Raffaele Dec 20 '20 at 15:46
  • Try $f(x):=cx^\alpha$ for $x>0$, and see whether you can fulfill your conditions with suitable choices of $c$ and $\alpha$. – Christian Blatter Dec 20 '20 at 16:54
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    @GregMartin: read carefully. The question you suggested asks when is the inverse of the derivative equals the function, I am asking when the inverse of the derivative equals the integral. – russian bot Dec 21 '20 at 05:38
  • Make an Ansatz $F(x)=ax^b$, find $f’=F’’$ and use $F(f’(x))=x$ to solve for $a$ and $b$ – Hagen von Eitzen Dec 21 '20 at 10:19
  • This question is not a duplicate of the quoted question. – Christian Blatter Dec 22 '20 at 16:18
  • @ChristianBlatter yes, I mentioned it at the beginning of the question, but they still closed it. – russian bot Dec 23 '20 at 03:35
  • The derivative of $F(x)$ is $f(x)$; the derivative of $(f')^{-1}(x)$, by the Inverse function Theorem, is $1/(f''(f'(x))$. So taking derivatives of your original equation, you get $f(x)f''(f'(x)) = 1$, don't you? – Arturo Magidin Dec 28 '20 at 22:16

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