\right. is not exactly an invisible delimiter, because it inserts a small space, which is usually desirable, but no symbol and so can be used to pair a \left. Similarly, also \left. will insert a small space.
The space inserted has width \nulldelimiterspace; the default value is 1.2pt, so it's small but noticeable. In some cases one doesn't desire such a space and it's necessary to eliminate it. However it should be kept in mind that TeX processes a formula, inserting kerns and penalties only when it has scanned the "end-of-formula" marker ($ or $$, the latter usually masked by macros in LaTeX) and uses the values of the parameters which are valid at that moment.
How can one define a "really void" delimiter that matches a \left or \right without adding spaces? Here's how:
\newcommand{\zerodel}{.\kern-\nulldelimiterspace}
Now
\[ \left(\frac{a}{b}\right\zerodel \]
won't add the small space after the formula. One can think to better examples. Here's code that shows the differences (the parameter \nulldelimiterspace has been set to 10pt for better showing the result, the \bigg| is to make the delimiter grow and | indicates the end of the space taken by the "invisible delimiter"):
$\left(\bigg|\right.|$
$\left(\bigg|\right.|\nulldelimiterspace=0pt$
$\left(\bigg|\right\zerodel|$
$|\left.\bigg|\right)$
$|\left.\bigg|\right)\nulldelimiterspace=0pt$
$|\left\zerodel\bigg|\right)$

\[ \int_a^b f(x) = \left. F(x) \right |_{x=a}^{x=b} \]. – Kris Harper Nov 15 '11 at 14:44F(x)is of normal height. It turned out to be better just to use\Bigr|in that situation. – Ryan Reich Nov 15 '11 at 18:28