Is there a theorem for the integral of an inverse of a function?

[MWresearch]

Say I have f(x) and then I have Σ(f(x))dx (I'm using Σ in place of an integral since there's no character for an integral that fits on one line), is there some formula relating Σ(f(x))dx to f^-1(x)? Sort of like the opposite of the derivative rule for inverse functions?

[ajb]

The closest that comes to mind is the sort of thing that can be found here: https://en.wikipedia.org/wiki/Integral_of_inverse_functions and http://mathworld.wolfram.com/InverseFunctionIntegration.html

Edited July 16, 2015 by ajb

[MWresearch]

Well thanks for that then.

[ajb]

Well thanks for that then.

Is that close to what you need?

[MWresearch]

Not really but if its the only thing I guess I'll take it. I was thinking more in terms of logic that the "proof without words" theorem would yield something nice like "the inverse of the integral of f-1(x) - the integral of f(x)." But, if that that complicated theorem is the best it gets then that's the best it gets.

[ajb]

I don't know what result you are looking for. However, you can play about with expressions and see if anything closer to what you need comes up.

[MWresearch]

The result I'm looking for is what I tried to say before. To understand it, I will clear up what I mean and say "inverse of the inverse." If I took f(x), and then took the inverse of that function, I would get inverse of f(x) or f-1(x). Now, if I take the inverse of the inverse, or the inverse of f-1(x), I get f(x).

With that concept in mind, it seems intuitive that there would be a formula which shows that the inverse of the integral of f-1(x) is the integral of f(x).

I know that's not actually right, because when I test it with f(x)=x^2 and sqrt(x) the theorem doesn't work.

[4h710r5]

The closest that comes to mind is the sort of thing that can be found here: https://en.wikipedia.org/wiki/Integral_of_inverse_functions and http://mathworld.wolfram.com/InverseFunctionIntegration.html

This might be useful for reduction formulae (writing an integral $I_n$ in terms of $I_{n \pm k}$).