Inmathematics,integrals of inverse functions can be computed by means of a formula that expresses theantiderivatives of theinverse${\displaystyle f^{-1}}$ of acontinuous and invertible function${\displaystyle f}$, in terms of${\displaystyle f^{-1}}$ and an antiderivative of${\displaystyle f}$. This formula was published in 1905 byCharles-Ange Laisant.^[1]

Let${\displaystyle I_1}$ and${\displaystyle I_2}$ be twointervals of${\displaystyle \mathbb{R}}$. Assume that${\displaystyle f:I_1\to I_2}$ is a continuous and invertible function. It follows from theintermediate value theorem that${\displaystyle f}$ isstrictly 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 thefundamental 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${\displaystyle f^{-1}}$ isdifferentiable.

In his 1905 article, Laisant gave three proofs.

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

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 aproof 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/upperDarboux sums of${\displaystyle f}$ and upper/lower Darboux sums of${\displaystyle f^{-1}}$.^[2][3] 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.

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.

One may also think as follows when${\displaystyle f}$ is differentiable. As${\displaystyle f}$ is continuous at any${\displaystyle x}$,${\displaystyle F:={\int }_0^{x}f}$ 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. 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,${\displaystyle G(y):=C+{\int }_0^y{f}^{-1}}$ 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${\displaystyle {\int }_{I_2}{f}^{-1}}$.

Nevertheless, it can be shown that this theorem holds even if${\displaystyle f}$ or${\displaystyle f^{-1}}$ is not differentiable:^[3][4] 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${\displaystyle f^{-1}}$ isabsolutely continuous.^[4]

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${\displaystyle f^{-1}(y)}$.^{[citation needed]} 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 themean value theorem to${\displaystyle F}$ between${\displaystyle x}$ and${\displaystyle x+h}$, taking into account that${\displaystyle f}$ is monotonic.

1. Assume that${\displaystyle f(x)=\exp (x)}$, hence${\displaystyle f^{-1}(y)=\ln (y)}$. The formula above gives immediately$${\displaystyle \int \ln (y)dy=y\ln (y)-\exp (\ln (y))+C=y\ln (y)-y+C.}$$
2. Similarly, with${\displaystyle f(x)=\cos (x)}$ and${\displaystyle f^{-1}(y)=\arccos (y)}$,$${\displaystyle \int \arccos (y)dy=y\arccos (y)-\sin (\arccos (y))+C.}$$
3. With${\displaystyle f(x)=\tan (x)}$ and${\displaystyle f^{-1}(y)=\arctan (y)}$,$${\displaystyle \int \arctan (y)dy=y\arctan (y)+\ln \left|\cos (\arctan (y))\right|+C.}$$

Apparently, this theorem of integration was discovered for the first time in 1905 byCharles-Ange Laisant,^[1] who "could hardly believe that this theorem is new", and hoped its use would henceforth spread out among students and teachers. This result was published independently in 1912 by an Italian engineer, Alberto Caprilli, in an opuscule entitled "Nuove formole d'integrazione".^[5] It was rediscovered in 1955 by Parker,^[6] and by a number of mathematicians following him.^[7] Nevertheless, they all assume thatf orf⁻¹ isdifferentiable. The general version of thetheorem, free from this additional assumption, was proposed by Michael Spivak in 1965, as an exercise in theCalculus,^[2] and a fairly complete proof following the same lines was published by Eric Key in 1994.^[3]This proof relies on the very definition of theDarboux integral, and consists in showing that the upperDarboux sums of the functionf are in 1-1 correspondence with the lower Darboux sums off⁻¹. In 2013, Michael Bensimhoun, estimating that the general theorem was still insufficiently known, gave two other proofs:^[4] The second proof, based on theStieltjes integral and on its formulae ofintegration by parts and ofhomeomorphicchange of variables, is the most suitable to establish more complex formulae.

The above theorem generalizes in the obvious way to holomorphic functions:Let${\displaystyle U}$ and${\displaystyle V}$ be two open and simply connected sets of${\displaystyle \mathbb{C}}$, and assume that${\displaystyle f:U\to V}$ is abiholomorphism. Then${\displaystyle f}$ and${\displaystyle f^{-1}}$ have antiderivatives, and if${\displaystyle F}$ is an antiderivative of${\displaystyle f}$, the general antiderivative of${\displaystyle f^{-1}}$ is

${\displaystyle G(z)=z{f}^{-1}(z)-F\circ f^{-1}(z)+C.}$

Because all holomorphic functions are differentiable, the proof is immediate by complex differentiation.

• Integration by parts
• Legendre transformation
• Young's inequality for products

• Calculus
• Theorems in mathematical analysis
• Theorems in calculus

• Articles with short description
• Short description is different from Wikidata
• Pages using sidebar with the child parameter
• All articles with unsourced statements
• Articles with unsourced statements from April 2020