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