Integral of inverse functions

Statement of the theorem

Let ${\displaystyle I_1}$ and ${\displaystyle I_2}$ be two intervals of ${\displaystyle \mathbb{R}}$. Assume that ${\displaystyle f:I_1\to I_2}$ is a continuous and invertible function. It follows from the intermediate value theorem that ${\displaystyle f}$ is strictly monotone. Consequently, ${\displaystyle f}$ maps intervals to intervals, so is an open map and thus a homeomorphism. Since ${\displaystyle f}$ and the inverse function ${\displaystyle f^{-1}:I_2\to I_1}$ are continuous, they have antiderivatives by the fundamental theorem of calculus.

Laisant proved that if ${\displaystyle F}$ is an antiderivative of ${\displaystyle f}$, then the antiderivatives of ${\displaystyle f^{-1}}$ are: $${\displaystyle \int f^{-1}(y)dy=y{f}^{-1}(y)-F\circ f^{-1}(y)+C,}$$ where ${\displaystyle C}$ is an arbitrary real number. Note that it is not assumed that ${\displaystyle f^{-1}}$ is differentiable.

In his 1905 article, Laisant gave three proofs.

First proof

First, under the additional hypothesis that ${\displaystyle f^{-1}}$ is differentiable, one may differentiate the above formula, which completes the proof immediately.

Second proof

His second proof was geometric. If ${\displaystyle f(a)=c}$ and ${\displaystyle f(b)=d}$, the theorem can be written: $${\displaystyle {\int }_c^d{f}^{-1}(y)dy+{\int }_a^{b}f(x)dx=bd-ac.}$$ The figure on the right is a proof without words of this formula. Laisant does not discuss the hypotheses necessary to make this proof rigorous, but this can be proved if ${\displaystyle f}$ is just assumed to be strictly monotone (but not necessarily continuous, let alone differentiable). In this case, both ${\displaystyle f}$ and ${\displaystyle f^{-1}}$ are Riemann integrable and the identity follows from a bijection between lower/upper Darboux sums of ${\displaystyle f}$ and upper/lower Darboux sums of ${\displaystyle f^{-1}}$. The antiderivative version of the theorem then follows from the fundamental theorem of calculus in the case when ${\displaystyle f}$ is also assumed to be continuous.

Third proof

Laisant's third proof uses the additional hypothesis that ${\displaystyle f}$ is differentiable. Beginning with ${\displaystyle f^{-1}(f(x))=x}$, one multiplies by ${\displaystyle f^{\prime }(x)}$ and integrates both sides. The right-hand side is calculated using integration by parts to be $xf(x)-\int f(x)dx$, and the formula follows.

Details

One may also think as follows when ${\displaystyle f}$ is differentiable. As ${\displaystyle f}$ is continuous at any ${\displaystyle x}$, $F:={\int }_0^{x}f(t)dt$ is differentiable at all ${\displaystyle x}$ by the fundamental theorem of calculus. Since ${\displaystyle f}$ is invertible, its derivative would vanish in at most countably many points. Sort these points by ${\displaystyle ...<t_{-1}<t_0<t_1<...}$. Since ${\displaystyle g(y):=y{f}^{-1}(y)-F\circ f^{-1}(y)+C}$ is a composition of differentiable functions on each interval ${\displaystyle (t_i,t_{i+1})}$, chain rule could be applied ${\displaystyle g^{\prime }(y)=f^{-1}(y)+y/f^{\prime }(y)-f\circ f^{-1}(y).1/f^{\prime }(y)+0=f^{-1}(y)}$ to see ${\displaystyle {g|}_{(t_i,t_{i+1})}}$ is an antiderivative for ${\displaystyle {f|}_{(t_i,t_{i+1})}}$. We claim ${\displaystyle g}$ is also differentiable on each of ${\displaystyle t_i}$ and does not go unbounded if ${\displaystyle I_2}$ is compact. In such a case ${\displaystyle f^{-1}}$ is continuous and bounded. By continuity and the fundamental theorem of calculus, $G(y):=C+{\int }_0^y{f}^{-1}(t)dt$ where ${\displaystyle C}$ is a constant, is a differentiable extension of ${\displaystyle g}$. But ${\displaystyle g}$ is continuous as it's the composition of continuous functions. So is ${\displaystyle G}$ by differentiability. Therefore, ${\displaystyle G=g}$. One can now use the fundamental theorem of calculus to compute ${\int }_{I_2}{f}^{-1}(t)dt$.

Nevertheless, it can be shown that this theorem holds even if ${\displaystyle f}$ or ${\displaystyle f^{-1}}$ is not differentiable: it suffices, for example, to use the Stieltjes integral in the previous argument. On the other hand, even though general monotonic functions are differentiable almost everywhere, the proof of the general formula does not follow, unless ${\displaystyle f^{-1}}$ is absolutely continuous.

It is also possible to check that for every ${\displaystyle y}$ in ${\displaystyle I_2}$, the derivative of the function ${\displaystyle y\mapsto y{f}^{-1}(y)-F(f^{-1}(y))}$ is equal to ${\displaystyle f^{-1}(y)}$. In other words: $${\displaystyle \mathrm{\forall }x\in I_1\underset{h\to 0}{lim}\frac{(x+h)f(x+h)-xf(x)-\left(F(x+h)-F(x)\right)}{f(x+h)-f(x)}=x.}$$ To this end, it suffices to apply the mean value theorem to ${\displaystyle F}$ between ${\displaystyle x}$ and ${\displaystyle x+h}$, taking into account that ${\displaystyle f}$ is monotonic.