"Routine methods" here mean "methods like integration by parts and integration by substitution" (as the original asker referred to) (too long for the title).
The original question on zhihu.com was asked in another way: "Is there an elementary function, whose indefinite integral cannot be computed using only routine methods like integration by parts and integration by substitution, yet the result of which is unexpectedly still elementary?"
I think there are no such functions-if we already know that the antiderivative of an elementary function $f(x)$, $F(x)$, is also elementary, then we can always compute $\int f(x)\mathrm{d}x=F(x)+C$ with "routine methods". My proof is as below:
When we compute $F'(x)=f(x)$, we only need:
(1)the derivatives of constant, power, exponential, logarithmic, trigonometric and inverse trigonometric functions;
(2)the sum(difference) rule ($(u\pm v)'=u'\pm v'$);
(3)the product rule ($(uv)'=u'v+uv'$);
(4)the chain rule ($[u(v(x))]'=u'(v(x))v'(x)$).
At the same time, we can do inverse operations for all above, by using:
(1)the antiderivatives of constant, power, exponential, logarithmic, trigonometric and inverse trigonometric functions;
(2)the sum(difference) rule ($\int(u\pm v)\mathrm{d}x=\int u\mathrm{d}x\pm\int v\mathrm{d}x$);
(3)integration by parts ($\int u\mathrm{d}v=uv-\int v\mathrm{d}u$);
(4)integration by substitution ($\int uv'\mathrm{d}x=\int u\mathrm{d}v$).
So, we can compute $\int f(x)\mathrm{d}x=F(x)+C$ by inverting the computation of $F'(x)=f(x)$, only using "routine methods".
I wonder whether I am right.
[Note]
Unlike similer questions which discuss about primitive functions of elementary functions, in this question we know that the antiderivative is elementary from the beginning.
- $\begingroup$ What do you mean by calling a function "primitive"? $\endgroup$Gerry Myerson– Gerry Myerson2025-07-06 01:47:21 +00:00Commented Jul 6 at 1:47
- 1$\begingroup$ It would be worth your while to read up on the Risch algorithm, en.wikipedia.org/wiki/Risch_algorithm would get you started. $\endgroup$Gerry Myerson– Gerry Myerson2025-07-06 01:50:30 +00:00Commented Jul 6 at 1:50
- 1$\begingroup$ Also, you might find van der Poorten, A., Tran, X. Quasi-Elliptic Integrals and Periodic Continued Fractions. Mh Math 131, 155–169 (2000). doi.org/10.1007/s006050070018 of interest. $\endgroup$Gerry Myerson– Gerry Myerson2025-07-06 01:59:09 +00:00Commented Jul 6 at 1:59
- 1$\begingroup$ @GerryMyerson It should be "elementary". Edited. Thank you. $\endgroup$JC Q– JC Q2025-07-06 06:18:33 +00:00Commented Jul 6 at 6:18
1 Answer
The answer is yes, but with a caveat. If a function is known to have an elementary antiderivative, then it is itself necessarily elementary and there is a finite sequence of applications of your mentioned rules to compute it.
The yes part: Say $F'(x)=f(x)$ with $F$ elementary. Then, by applying the known differentiation rules, there is a sequence of equalities $F'(x)=\ldots=\tilde{f}(x)$ with $\tilde{f}$ an elementary function (which we know to be equal to $f$). In principle, if only you were clever enough to notice each equality in the reverse order $f(x) = \tilde{f}(x) = \ldots = F'(x)$ then you would have computed the antiderivative.
Example: $$f(x)=\cos(\cos(x))\frac{1-\cos^2(x)}{\sin(x)} \stackrel{(1)}{=} (\sin)'(\cos(x))(-\sin(x)) \stackrel{(2)}{=} (\sin)'(\cos(x))(\cos)'(x) = (\sin \circ \cos)'(x).$$
Note: at the end of the day, a computation boils down to a sequence of equalities.
Note: this has nothing to do with integrals, just algebraic operations, known algebraic relations between elementary functions and the differential rules.
The problem of course is that knowing $F$ has an elementary representation guarantees that it is possible to compute it with purely algebraic rules, but it doesn't tell us how. To find this way, you always encounter two problems:
How to decide algorithmically whether two different representations of elementary functions are equal? For example step 1 in the computation above or: $e^{\cos^2(x)}e^{\log(x)}e^{\sin^2(x)}/x = e$.
How to find algorithmically a sequence of relations given by rules of differentiation? For example step 2 in the computation above.
Problem 1) is unclear to me how difficult it is, whereas Problem 2) is certainly much harder and to my knowledge there is no such algorithm for the whole class of elementary functions. Welcome to Computer Algebra System Engineering.
- $\begingroup$ I understand what you mean by "this has nothing to do with integrals, just algebraic operations, known algebraic relations between elementary functions and the differential rules". I also agree that to be "clever enough" is in a way all we need. I find other questions discussing problem 1. As for problem 2, I remember that it is open. Further studies can be carried out. Thank you for your contribution. $\endgroup$JC Q– JC Q2025-07-06 06:43:20 +00:00Commented Jul 6 at 6:43