However, I don't really see how to exponentiate the infinitesimal form
to actually get to the finite form

In addition to the detailed answer of @LubošMotl, you may notice that conformal transformations transform light cones in light cones.

This means the following. Starting from a light cone :

$$(x'-a')^2 = x'^2-2x'.a' + a'^2=0\tag{1}$$

you must find that it is also a light cone for $x$, that is, it exist $a$ such as :

$$(x-a)^2 = x^2-2x.a + a^2=0\tag{2}$$

To do that, replace simply $x'$ by its value in function of $x$, by :

$$x'^{\mu} = \frac{x^{\mu} - b^{\mu}x^2}{1-2x\cdot b + b^2 x^2} \tag{3}$$

and, with some algebra, you will find that it is possible to exhibit $a(a',b)$ ($a$ as a function of $a'$ and $b$) , which satisties ($2$).:

$$a = \dfrac{a' + a'^2b}{1-2a'.b+a'^2b^2}\tag{4}$$

This means that the transformation $(3)$ is a conformal transformation whose infinitesimal expression is clearly:
$$x'^{\mu} = x^{\mu} + 2(x \cdot b)x^{\mu} - b^{\mu}x^2 \tag{5}$$
(because only quadratic terms in x are allowed in infinitesimal conformal transformations)

Any other expression of the global special conformal transformation $(3)$, having infinitesimal expression $(5)$, will not transform light cones in light cones.

This post imported from StackExchange Physics at 2014-06-27 11:33 (UCT), posted by SE-user Trimok