# Why do un-integrable functions exist?

By un-integrable I mean functions whose antiderivative can not be expressed in terms of elementary functions.

I recently learnt that any differentiable function can be expanded using the Taylor expansion, which essentially provides a polynomial representation. And polynomial funcitons are very easy to integrate. Then, why is it that there are functions whose antiderivative can not be expressed in terms of elementary functions?

• "...any differentiable function can be expanded using the Taylor expansion...": not true, at least in $\mathbb R$. "...which essentially provides a polynomial representation": absolutely not! A polynomial has a finite number of terms, which makes it completely different from an infinite series. – TonyK Jun 23 '15 at 21:27
• Look at the thread How can you prove that a function has no closed form integral? – Zev Chonoles Jun 24 '15 at 1:46

Two reasons:

1) Not all integrable functions are differentiable, not all differentiable functions are smooth (i.e., have derivatives of all orders, which is necessary for the series to exist), and not all smooth functions converge to their Taylor series. The canonical example for the latter is $f(x) = e^{-1/x^2}$ (with $f(0) = 0$), which is smooth with $f^{(n)}(0) = 0$ for all $n$ and thus has a Taylor series around $0$ that's exactly $0$.

2) An arbitrary Taylor series is not a polynomial or even an elementary function. Take $\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}\int_0^x dt\; e^{-t^2}$, for example, or $\Gamma(x) = \int_0^\infty dt\;t^{x-1}e^{-t}$.

Because a Taylor expansion of a function is in general not finite and an infinite sum of polynomials can not always be expressed as a finite combination of elementary functions.

• How are these integrals evaluated then? (Other than numerically) – Aritra Das Jun 23 '15 at 20:49
• @AritraDas, most interesting integrals (read: integrals computed for real applications) are all done numerically (because there is often not a "nice" function to be integrated). – TravisJ Jun 23 '15 at 20:54
• @AritraDas The non elementary functions can be expressed in form of integrals or sums of elementary functions, so if an integral, evaluated by some rule leads to such a sum or integral you have the elementary function. – Lykos Jun 23 '15 at 20:55
• @AritraDas finding By deriving functions rules for integration can be found and often integrals can be transformed using these rules to find the non elementary integrals but there is no general way of solving every integral, sadly. – Lykos Jun 23 '15 at 21:10

Why do un-integrable functions exist ?

Look at it this way: The inverse of an elementary function is not always elementary, so why should its anti-derivative be such ? So, if you accept the fact that $f(x)=x+\sin x$ is elementary, but $f^{-1}(x)$ is not, you should also accept the fact that, even though $g(x)=e^{-x^2}$ is elementary, its antiderivative is not.

This is sort of a pedantic answer, but there is the Risch algorithm for finding elementary anti-derivatives for elementary functions, or determining that no elementary anti-derivative exists. So if you study the Risch algorithm and its proof of correctness closely enough, you should be able to begin to see what characteristics guarantee that an elementary function will have an elementary anti-derivative formula, or that it doesn't. And in particular, that it's possible that no elementary anti-derivative formula exists for some given elementary functions.

The definition of Taylor expansion can be loosely defined as "The best polynomial approximation of $f(x)$ near $a$", which means that the approximation gets worse and worse as the neighborhood of $a$ gets larger. Therefore the Taylor expansion is not a sufficient approximation with all values of $x$. More than that, sometimes the Taylor expansion of a function is infinite, and infinite sum of polynomials is not an elementary function.

• Okay, but will this method of integration, over some span, ever give a wrong result? – Aritra Das Jun 23 '15 at 20:48
• Say that the Taylor expansion of the integrand in $a$ and $b$ doesn't have a closed form. Then you must choose the degree of approximation so that you can evaluate a finite Taylor expansion, which is therefore "wrong result" as it is a numerical result, not analytical. – Eemil Wallin Jun 23 '15 at 20:54
• @AritraDas, not every integrable function has a Taylor series representation. Only analytic functions have Taylor series representation. – TravisJ Jun 23 '15 at 20:55
• This answer is misleading. The Taylor expansion is always infinite if f isn't a polynomial. – Benjamin Lindqvist Jun 26 '15 at 23:12