Prove $xf(x) - \int_{0}^{x} f(t) \,dt = \int_{f(0)}^{f(x)} g(t) \,dt, g(x) = f^{-1}(x), \forall f(x)$?

Question

I have proved it for all functions of the form $f(x) = x^{a}, \forall a, x,$ but I am not sure if it is true for all functions.

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Proposition

$$xf(x) - \int_{0}^{x} f(t) \,dt = \int_{f(0)}^{f(x)} g(t) \,dt, g(x) = f^{-1}(x), \forall f(x)$$

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Proof

Let $f(x) = x^{a}, a \neq -1$, then:

$f(x)x = x^{a + 1}$,

$F(x) = \frac{1}{a + 1} x^{a + 1}$,

$F(0) = 0$, and

$f(x)x - \int_{0}^{x} f(t) \,dt = \color{lime}{\frac{a}{a + 1} x^{a + 1}}$.

Since $f(x) = x^{a}, g(x) = x^{\frac{1}{a}}$,

$G(x) = \frac{a}{a + 1} x^{\frac{a + 1}{a}}$,

$G(f(x)) = \frac{a}{a + 1} x^{a + 1}$,

$G(f(0)) = 0$, and

$G(f(x)) - G(f(0)) = \color{lime}{\frac{a}{a + 1} x^{a + 1}}$.

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Let $f(x) = x^{a}, a = -1$, then:

$f(x)x = 1$,

$F(x) = \ln|x|$,

$F(0) = -\infty$, and

$f(x)x - \int_{0}^{x} f(t) \,dt = \color{lime}{\infty}$.

Since $f(x) = x^{a}, g(x) = x^{a}$,

$G(x) = \ln|x|$,

$G(f(x)) = -\ln|x|$,

$G(f(0)) = -\infty$, and

$G(f(x)) - G(f(0)) = \color{lime}{\infty}$.

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I believe that there may be a few exceptions to this, but I am not sure.

I am not sure as to whether or not this formula has been derived / used before.

If you would like to see how I derived it, please, ask me!

Answer 1

You can use the Fundamental Theorem of Calculus (Leibniz's Rule) to show that the derivatives of the LHS and RHS wrt $x$ are identical ($=xf'(x)$).

Which means that LHS = RHS $+ c$.

All you need to do now is to show $c=0$ and that's trivial by taking $x=0$.

Answer 2

Yes, it's true for all invertible functions. Think of the integrals in terms of areas under graphs; notice that the areas from the two integrals combine to form an $x$ by $f(x)$ rectangle. (This seems similar to integration by parts; maybe there's a proof using it)