Special integrals

There are special integrals such as the logarithmic integral and exponential integrals. I want to know if there are primitives for such integrals. If not, why not?

A simple starting point (as indicated by Qiaochu) is Liouville's theorem (or 'principle') based on differential algebra and was extended with the Risch algorithm.

This last link should clarify some of the ideas used :

1. the only new term appearing during an integration (i.e. that was not in the integrand) is a linear combination of logarithms (because logarithms alone may disappear during differentiation...)
2. exponentials $e^f$ had to be in the integrand first (since differentiation doesn't make them disappear) and will reappear as $h\,e^f$ (of course subtle points exist like considering $\sqrt{x}=e^{\,\large{\ln(x)/2}}\cdots$)
3. differentiation of an algebraic function $\theta$ (i.e. there is a polynomial $P(\theta)=0$) will give a rational function $\dfrac {d(\theta)}{e(\theta)}$ with $\,d$ and $e\,$ two polynomials.

These 3 ideas will provide logarithmic, exponential and algebraic extensions to the differential algebra (starting for example with the field of rational functions over $\mathbb{Q}$) that will give all the elementary functions.

An excellent tutorial about this is "Symbolic Integration" from Manuel Bronstein.

Geddes, Czapor and Labahn's book "Algorithms for Computer Algebra" is very clear too.

Now let's use these ideas to study $\;\displaystyle\int\frac {e^x}x\,dx$.

From a more precise version of $2.$ a primitive must be of type $\ I(x)=h(x)\,e^x\;$ with $h(x)$ a rational function. Let's suppose this and differentiate $\,I(x)$ : $$(h'(x)+h(x))\,e^x=\frac {e^x}x$$ so that we need : $$h'(x)+h(x)=\frac 1x$$ We supposed $h$ rational so that it may be decomposed in simple elements but $h'(x)$ can't give $\dfrac 1x$ so that $\dfrac 1x$ must be part of $h(x)$. In this case $h'(x)$ will create a term $-\dfrac 1{x^2}$ that must be compensated by a $\dfrac 1{x^2}$ term inside $h(x)$ that will generate a $-\dfrac 2{x^3}$ term... This process clearly doesn't end !

The same method could be used for the sine integral : $\;\displaystyle\int \frac {\sin(x)}x\,dx\,$ simply by writing $\ \sin(x)=\dfrac{e^{ix}-e^{-ix}}{2i}$.
(this method was presented by Matthew P Wiener in an old post at sci.math : recommended reading too !)

Concerning the logarithmic integral we have $\ \operatorname{li}(x)=\operatorname{Ei}(\ln(x))\ $ so that the non-elementary proof for the one should apply for the other as well.