# Functions that are easy to antidifferentiate but whose inverses are hard to antidifferentiate

Some functions have antiderivatives that are harder to compute than the antiderivatives of their inverse, i.e. $$\ln(x)$$ is such a function. Another is $$\arctan(x)$$. So at the beginning of an integral calculus course, before any antidifferentiation techniques are learned, you can motivate the switch to an integral $$dy$$ by showing that you can compute the area under $$\ln(x)$$ between $$x=1$$ and $$x=e$$ by a switch to an integral $$dy$$ of the function $$e^y$$. (This is better motivation for the change of variables than I usually see calculus books give, which is that you can sometimes get away with one integral $$dy$$ instead of a sum of two or three integrals $$dx$$.)

Taking this to the extreme, I'd like a list of elementary functions that are reasonably easy to antidifferentiate, and whose inverses are still elementary but are either impossible to integrate with the standard techniques of a first year of calculus or have provably non-elementary antiderivatives.

• Hints for integration and simplicity: "A heuristic program that solves symbolic integration problems in freshman calculus; 1955", and Richard J. Fateman, Theodore H. Einwohner: A Proposal for Automated Integral Tables; 1992: 3 How do we compute integrals?
– IV_
Jan 31, 2019 at 19:09
• Further hints for integration and simplicity: Geddes/Czapor/Labahn: Algorithms for Computer Algebra. Springer, page 473 ff., and Heck: Introduction to Maple. Springer, page 226 ff., and Terelius: Symbolic Integration, 2009.
– IV_
Jan 31, 2019 at 20:27
• According to Ritt's theorem (my comment above), the elementary invertible elementary functions are repeated compositions. Because this multicomposition is bijective, all component functions of this multicomposition are bijective, that means their inverses exist. For integrating this multicompositions, one could use a "chain rule of integration". The complexity of the resulting forms of the integrands decide if the integration of the function or of its inverse is more complicated.
– IV_
Jan 31, 2019 at 21:37
• Feb 2, 2019 at 22:28

Until now, I don't have a complete answer. The single parts of my answer are only hints that should be considered.

1). Complexity of Antidifferentiation

There are some approaches for complexity measures of antidifferentiation or integration. But I don't know if there are approaches we need here.

Antiderivative means i.a. integration in closed form. Let's concentrate here on the elementary functions in closed form.

The elementary functions are analytic. Each elementary function can be splitted into bijective elementary functions by restricting its domain (that were the partial inverses).

2.) Non-Existence of Non-elementary Integrals

This section treats elementary functions that have a non-elementary integral.

Let $$f$$ be an elementary function, $$f^{-1}$$ the elementary inverse of $$f$$, and $$c$$ the integration constant.

The Integral of inverse functions gives

$$\int f(x)dx=xf(x)-\int f^{-1}(f(x))df(x)+c.$$

Because $$f$$ is elementary, $$xf(x)$$ is elementary. If $$\int f(x)dx$$ is (non)elementary, $$\int f^{-1}(f(x))df(x)$$ is therefore (non)elementary too. Marchisotto and Zakeri (see the reference below) give therefore the following Integrals of inverse functions theorem.

"If $$f$$ and $$f^{-1}$$ are elementary functions over some closed interval, then $$\int f(x)dx$$ is elementary if and only if $$\int f^{-1}(x)dx$$ is elementary."

That means, there are no elementary functions that are reasonably easy to antidifferentiate, and whose inverses are still elementary but have provably non-elementary antiderivatives.

That means, each elementary invertible elementary function with an elementary antiderivative has an inverse with an elementary antiderivative.

Marchisotto, E. A.; Zakeri, G.-A.: An invitation to integration in finite terms. College Math. J. 25 (1994) (4) 295-308

3.) The Structure of Elementary Invertible Elementary Functions

J. F. Ritt proved in [Ritt 1925] that if a bijective elementary function $$f$$ has an elementary inverse, then $$f=f_n\circ\ ...\ \circ f_1$$, where each $$f_i$$ is either an algebraic function of one variable or else $$\exp$$ or $$\ln$$.
R. H. Risch proved Ritt's result in [Risch 1979].

That means: If a function $$f$$ and its inverse are elementary, $$f$$ is composited by $$\exp$$, $$\ln$$ and/or unary univalued algebraic functions. There exists a representation of $$f$$ as elementary function that doesn't contain multiary algebraic functions.

All the functions and inverse functions that are relevant for us must have this structure.

If the set of allowed basic functions in the compositions is the set {transcendental elementary standard functions, unary univalued algebraic functions}, the composition representation of $$f$$ may contain also multiary univalued algebraic functions with arguments that are algebraically dependent allowed basic functions of algebraically dependent functions (algebraic dependence over $$\mathbb{C}$$ is meant).

4.) The Structure of Elementary Integrable Elementary Functions

How can the structure mentioned in section 3 be combined with the structure from Liouville's theorem for integration in finite terms?

Until now, I cannot describe the elementary functions that are elementary invertible and elementary integrable. And I cannot describe the elementary functions that are elementary invertible and not elementary integrable.

But I can give some examples of the latter:
List of Functions Without Antiderivatives
Yadav D. K.: Six Conjectures on Indefinite Nonintegrable Functions or Nonelementary Functions. 2016, conjectures 4 and 5

5.) Complexity of Antidfifferentiation in Dependence of the Integrand's Class of Functions

Which classes of functions are easier to antidifferentiate?
The elementary standard functions? The rational functions? The explicit algebraic functions? The (explicit and implicit) algebraic functions?

And which of the algebraic functions? Antiderivatives of algebraic functions can be i.a. purely algebraic or purely logarithmic.

6.) Antiderivatives of the Elementary Standard Functions

Decide which antidifferentiation is easier.

$$f(x)\rightarrow \int f(x)dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f^{-1}(x)\rightarrow \int f^{-1}(x)dx$$

$$x^n\rightarrow \frac{x^{n+1}}{n+1}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x^{\frac{1}{n}}\rightarrow \frac{n{x}^{\frac{n+1}{n}}}{n+1}$$

$$a^x\rightarrow \frac{a^x}{\ln(a)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \log_a(x)\rightarrow \frac{x(\ln(x)-1)}{\ln(a)}$$

$$e^x\rightarrow e^x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ln(x)\rightarrow x\ (\ln(x)-1)$$

$$\sin(x)\rightarrow -\cos(x)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \arcsin(x)\rightarrow x\ \arcsin(x)+\sqrt{1-x^2}$$

$$\cos(x)\rightarrow \sin(x)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \arccos(x)\rightarrow x\ \arccos(x)-\sqrt{1-x^2}$$

$$\tan(x)\rightarrow -\ln(\cos(x))\ \ \ \ \ \ \ \ \ \ \ \arctan(x)\rightarrow x\ \arctan(x)-\frac{1}{2}\ln(1+x^2)$$

$$\cot(x)\rightarrow \ln(\sin(x))\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ arccot(x)\rightarrow \frac{1}{2}x\pi-x\ \arctan(x)+\frac{1}{2}\ln(1+x^2)$$

$$\sec(x)\rightarrow \ln\left(\frac{1+\sin(x)}{\cos(x)}\right)\ \ \ \ \ \ \ \ \ \ \ \ \ arcsec(x)\rightarrow x\ arcsec(x)-\ln\left(x\left(1+\sqrt{{\frac{-1+x^2}{x^2}}}\right)\right)$$

$$\csc(x)\rightarrow -\ln\left(\frac{1+\cos(x)}{\sin(x)}\right)\ \ \ \ \ \ \ \ \ \ arccsc(x)\rightarrow x\ arccsc(x)+\ln\left(x\left(1+\sqrt{\frac{-1+x^2}{x^2}}\right)\right)$$

$$\sinh(x)\rightarrow \cosh(x)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ arcsinh(x)\rightarrow x\ arcsinh(x)-\sqrt{1+x^2}$$

$$\cosh(x)\rightarrow \sinh(x)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ arccosh(x)\rightarrow x\ arccosh(x)-\sqrt{x-1}\sqrt{x+1}$$

$$\tanh(x)\rightarrow \ln(\cosh(x))\ \ \ \ \ \ \ \ \ \ \ \ arctanh(x)\rightarrow x\ arctanh(x)+\frac{1}{2}\ln(-(x-1)(x+1))$$

$$\coth(x)\rightarrow \ln(\sinh(x))\ \ \ \ \ \ \ \ \ \ \ \ \ arccoth(x)\rightarrow x\ arccoth(x)+\frac{1}{2}\ln((x-1)(x+1))$$

$$sech(x)\rightarrow \arctan(\sinh(x))\ \ \ \ \ \ arcsech(x)\rightarrow x\ arcsech(x)-\arctan\left(\sqrt{-\frac{x-1}{x}}\sqrt{\frac{x+1}{x}}\right)$$

$$csch(x)\rightarrow \ln(\tanh(\frac{x}{2}))\ \ \ \ \ \ \ \ \ \ \ \ arccsch(x)\rightarrow x\ arccsch(x)+\ln\left(x\left(1+\sqrt{\frac{1+x^2}{x^2}}\right)\right)$$

7.) Complexity of the Integrands from Integration Rules

The common integration rules beside Change of Variable are Integration by Substitution and Partial Integration. The complexity of the resulting integrands depends on the standard functions in the integrand. See the following reference for the form of the integrands.
Produktregel, Quotientenregel, Reziprokenregel, Kettenregel und Umkehrregel für die Integration

Here are techniques for Integration By Parts:
Alcantara, E.: Integrals of composite functions through tabular integration by parts. Asia Pacific Higher Educ. Res. J. 2 (2015) (1)
Alcantara, E.: On the Derivation of Some Reduction Formula through Tabular Integration by Parts. Asia Pacific J. Multidisciplinary Res 3 (2015) (1) 80-84
Mardeli Jandja, M.; Lutfi, M.: The Five Columns Rule in Solving Definite Integration by Parts Through Transformation of Integral Limits. J. Phys.: Conf. Ser. 1028 (2018) 012109

I think this probably answer your question: if $$F$$ is an antiderivative of $$f$$, then $$f^{-1}$$ has the antiderivative $$x f^{-1}(x)-F(f^{-1}(x))$$