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Concept
Integral of inverse functions
Summary
In mathematics, integrals of inverse functions can be computed by means of a formula that expresses the antiderivatives of the inverse of a continuous and invertible function , in terms of and an antiderivative of . This formula was published in 1905 by Charles-Ange Laisant.
Let and be two intervals of .
Assume that is a continuous and invertible function. It follows from the intermediate value theorem that is strictly monotone. Consequently, maps intervals to intervals, so is an open map and thus a homeomorphism. Since and the inverse function are continuous, they have antiderivatives by the fundamental theorem of calculus.
Laisant proved that if is an antiderivative of , then the antiderivatives of are:
where is an arbitrary real number. Note that it is not assumed that is differentiable.
In his 1905 article, Laisant gave three proofs. First, under the additional hypothesis that is differentiable, one may differentiate the above formula, which completes the proof immediately. His second proof was geometric. If and , the theorem can be written:
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 is just assumed to be strictly monotone (but not necessarily continuous, let alone differentiable). In this case, both and are Riemann integrable and the identity follows from a bijection between lower/upper Darboux sums of and upper/lower Darboux sums of . The antiderivative version of the theorem then follows from the fundamental theorem of calculus in the case when is also assumed to be continuous. Laisant's third proof uses the additional hypothesis that is differentiable. Beginning with , one multiplies by and integrates both sides. The right-hand side is calculated using integration by parts to be , and the formula follows.
Nevertheless, it can be shown that this theorem holds even if or is not differentiable: it suffices, for example, to use the Stieltjes integral in the previous argument.
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