TITLE: Dirac Delta function
QUESTION [9 upvotes]: I know that $\int_{-\epsilon}^\infty f(x)\delta(x)dx=f(0)$ but what about $\int_0^\infty f(x)\delta(x)dx$? I suppose we have to do this by definition since the lower limit is bang on $0$?

REPLY [4 votes]: You are something of a victim of ambiguity in notation.  The definition of the integral
$$
\int_A f \; da
$$
is taken over the open or closed (or neither) object $A$.  As Carl M notes, we have that
$$
\int_{\{0\}} f(x) \delta(x) \; dx = f(0)
$$
by construction of the delta functional, while
$$
\int_{(0,\infty)} f(x) \delta(x) \; dx = \int_{(-\infty,0)} f(x) \delta(x) \; dx = 0.
$$
More generally,
$$
\int_A f(x) \delta(x) \; dx = \begin{cases}f(0)\qquad&0 \in A\\0\qquad&\text{otherwise.}\end{cases}
$$
When someone writes
$$
\int_0^\infty g(x) \; dx
$$
they are not being clear what the domain of integration really is.