Prove that $\int f^{−1}(x) dx = yf(y) −\int f(y) dy $

Question

Prove that $\int f^{−1}(x) dx = yf(y) −\int f(y) dy $.

Note: I came across this formula in this very interesting (at least to me) survey article on the Lambert W function:

http://www.apmaths.uwo.ca/~djeffrey/Offprints/W-adv-cm.pdf

Answer 1

Let $x=f(y)$. Then $$\int f^{-1}(x)dx=\int f^{-1}(f(y))f^{\prime}(y)dy=\int yf^{\prime}(y)dy=yf(y)-\int f(y)dy$$ On integration by parts.

Answer 2

Similar to the other answer, I think it goes like this.

By integration of parts, one gets:

$∫ f^{-1}(x) dx = x \cdot f^{-1}(x) - ∫ \frac{x}{f'(f^{-1}(x))}dx$

Making the substitution $x=f(y)$, just for the RHS:

$∫ f^{-1}(x) dx = f(y) \cdot y - ∫ \frac{f(y)}{f'(y)}dx$

However: $\frac{dx}{dy}=f'(y)$, so one gets:

$∫ f^{-1}(x) dx = f(y) \cdot y - ∫ f(y) \frac{dy}{dx} dx$

And the result follows:

$∫ f^{-1}(x) dx = f(y) \cdot y - ∫ f(y) dy$

Notice that if you make the substitution on both sides, then you would yield the identity :

$∫ y \cdot f'(y) dy = f(y) \cdot y - ∫f(y) dy$

Which follows by direct integration by parts for $y$ and $f(y)$.