Proving the integral of an inverse function

Question

If $$\int F(x)\ dx=G(x),$$ show that $$\int F^{-1}(x)\ dx=xF^{-1}(x)-G(F^{-1}(x)).$$ All functions exist and are continuous.

I'm thinking about integration of parts, but I don't know what to do with that inverse.

Answer 1

Consider following change of variables: $x=F(z)$ $$ \int F^{-1}(x) {\rm d} x =\int F^{-1}(F(z)) {\rm d} F(z) =\int z {\rm d} F(z) $$

Now it's time for integration by parts: $$ \int z {\rm d} F(z) =zF(z) - \int F(z) {\rm d} z =F^{-1}(x) F(F^{-1}(x)) - G(z) =x F^{-1}(x) - G(F^{-1}(x)) $$

Answer 2

Let $H = F^{-1}$ Then $RHS = xH(x) - G(H(x))$. Differentiating this RHS with respect to $x$ we get: $$1H(x) + xH'(x) - G'(H(x))H'(x) = H(x) + xH'(x) - F(H(x))*H'(x) = H(x) + xH'(x) - xH'(x) = H(x) = F^{-1}(x)$$ We're done.