Why do un-integrable functions exist?

Solution 1:

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 $\Gamma(x) = \int_0^\infty dt\;t^{x-1}e^{-t}$.

Solution 2:

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.

Solution 3:

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.

Solution 4:

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.