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  1. Suppose you have a function $f(x)$ defined as an integral, such that $$f(x)=\int g(x)dx$$ like say the natural logarithm defined as $$\ln(x)= \int_{1}^{x} \frac{1}{t}dt.$$



  2. Unfortunately, the inverse function theorem, which is $$f^{-1}(x)'= \frac{1}{f'(f^{-1}(x))}$$ can at best only be arranged into this form after integrating both sides: $$f^{-1}(x)+C = \int \frac{dx}{f'(f^{-1}(x))}$$ As you can see, the inverse of the original function appears on both sides of the equation, so this is not in any way helpful for finding the unknown explicit inverse of a particular function defined as an integral unless you somehow already know the answer to begin with.



  3. So, I am wondering if there is a way to manipulate intergation by parts to find the inverse of a function defined as an integral. Starting the statement of the formula, $$ \int u(x) v'(x)dx = u(x)v(x) - \int v(x) u'(x)dx$$ we can see the generalization of the integral of simply a single function is $$ \int g(x)dx = g(x)x - \int x g'(x)dx$$ However, $x$ can also be assumed as $x=g^{-1}(g(x))$ which, when substituted, would make the equation $$ \int g(x)dx = g(x)g^{-1}(g(x)) - \int g^{-1}(g(x)) g'(x)dx.$$



  4. Now for me, this is where I'm not completely sure, because if you look at the second term of the right side of the equation, that looks an awful lot like an instance of the chain rule for differentiation. It would seem the best arrangement of the equation is $$ \int g(x)dx = g(x)g^{-1}(g(x)) - \int g^{-1}(g(x)) g'(x)dx$$ $$ \int g(x)dx = g(x)x - \int g^{-1}(x)dx$$ For some reason, this doesn't seem completely correct to me, and I don't know how to take it further to accomplish the goal. Because, if you integrate $e^x$ for instance, the integral is still just $e^x$, and, $$ \int e^x dx \neq e^x*x- x(\ln(x)-1)$$ But, if I finished the integral, it still doesn't work, $$ \int g(x)dx = g(x)x - g^{-1}(x)-C$$ $$\int e^x \neq e^x x - \ln(x)$$
    Can this method be salvaged so that one can define the explicit inverse of a function defined as an integral?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Daniel Fischer Aug 6 '17 at 11:30
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Ok here is the main mistake in your argument. The equation $$\int g(x) \, dx=xg(x) - \int g^{-1}(x)\,dx\tag{1}$$ does not hold and the reason is that the following equation does not hold $$\int g^{-1}(g(x))g'(x)\,dx=\int g^{-1}(x)\,dx\tag{2}$$ That these equations don't hold can be easily checked by differentiating both sides of the equation. Perhaps you are trying to use substitution in integrals (which is more of a restatement of chain rule of derivatives) but not exactly using the new variable. The correct way to write the equation $(2)$ is as follows $$\int g^{-1}(g(x))g'(x)\,dx=\int g^{-1}(u)\,du\tag{3}$$ where $u=g(x) $.

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  • $\begingroup$ Okay, so then how could this new information be utilized to definitively say that the inverse can be explicitly defined in terms of algebraically manipulating the chain rule? If you do the u substitution, the inverse is not with respect to x. $\endgroup$ – DaneJoe Aug 6 '17 at 1:17
  • $\begingroup$ @DaneJoe: I think you have to stop at point 2. of your question and it is not possible to get hold of the inverse in an explicit manner. $\endgroup$ – Paramanand Singh Aug 6 '17 at 1:22
  • $\begingroup$ okay well that's part of what I'm asking. I might have gotten that step wrong, so I can't say whether the goal is feasible or not, but that's what I am trying to show. $\endgroup$ – DaneJoe Aug 6 '17 at 1:52
  • $\begingroup$ I think I will edit your example some if that's okay with you, and if not that's fine as well. But either way, I am still not clear on, and would like your interpretation of: that it is impossible to isolate $f^{-1}(x)$. I am not seeing with certainty that has been shown. $\endgroup$ – DaneJoe Aug 6 '17 at 8:52

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