# Why there's no chain rule for integrals of elementary functions which are expressible in terms of elementary functions?

The derivative of every elementary function is elementary; this is owing to the existence of the chain rule for differentiation.

On the other hand, the integral of an elementary function may turn out to be elementary or not elementary ($\text{e.g:}\int e^{-x^2}dx$). There's Risch algorithm, which for a given integral of an elementary function, tells you whether the integral is elementary or not, and if it's elementary, it finds the solution.

However I think it's still valid to ask, for integrals of elementary functions that are expressible in terms of elementary functions, why there's no chain rule for them?

• The integral version of the chain rule is substitution. Oct 3 '15 at 0:37
• @Chappers Substitution undoes the chain rule. There is no chain rule for integrals. I.e. there is no general rule for $\int (f\circ g)(x)dx$.
– user137731
Oct 3 '15 at 1:30
• @Bye_World Yes, in the same way that the product rule and integration by parts are related. Oct 3 '15 at 1:40
• I agree that integration by parts also undoes the product rule. I don't agree that the "integral version of the chain rule is substitution". I wish it were. Life would be so much better if we had a chain rule for integrals.
– user137731
Oct 3 '15 at 1:42
• More generally, one may wonder why symbolic differentiation is straightforward and always possible, while integration is arduous and exceptionally doable.
– user65203
May 30 '16 at 15:48

Differentiation is a function that satisfies linearity f(x + y) = f(x) + f(y) and f(ax) = af(x). It also satisfies the rule f(xy) = f(x)y + xf(y). Integration can be thought of as the inverse function much like division can be thought of as the inverse function to multiplication. However, just as division has only some of the same algebraic properties as multiplication but not all, it is not commutative, nor is it closed on a set with zero, integration does not have all the algebraic properties of differentiation. Indeed the reason for this is precisely because of division not having all the algebraic properties as multiplication. I say "precisely", making this a rigorous logical proof is I'm afraid beyond me.

• It'd be interesting if you pointed out the connection between division/multiplication and integration/differentiation. May 30 '16 at 17:14
• @OmarNagib It has to do with Taylor's Theorem and integrating term by term.
– GYL
May 30 '16 at 17:22

Let's take the elementary functions as defined by J. Liouville and J. F. Ritt (see e.g. Wikipedia: Elementary function and the references therein). With Liouville's theorem (Wikipedia: Liouville's theorem (differential algebra)) follows, the elementary functions are not closed regarding integration.

A chain rule for integration would have the form $$\int f(g(x))dx=F(x)+c$$, where $$F$$ would be in dependence of $$f$$ and $$g$$. Because of the conclusion above, $$F$$ is non-elementary for certain elementary $$f$$ and $$g$$. Therefore a chain rule for integration of all elementary functions cannot exist.

But there are chain rules for integration known which are applicable in some cases. Recently, such chain rules for integration were collected in:

Will, J.: Produktregel, Quotientenregel, Reziprokenregel, Kettenregel und Umkehrregel für die Integration

Will, J.: Product rule, quotient rule, reciprocal rule, chain rule and inverse rule for integration. May 2017