When I have an even function $f(x)=f(-x)$ we can write
$$ \int_{-\infty}^{\infty} f(x) = 2 \int^{\infty}_0 f(x)$$ Now if we allow $f(x)$ to be a distribution, specifically to contain delta functions, the above result no longer seems to hold.
It overcounts the contribution at 0. This normally doesn't matter for a function since it has measure zero but if $f(x)$ contains Dirac Delta function at $x=0$ the result is now different. So when I use this trick I guess I have to be careful with end points where $f(x)$ is not a function.
Should I split the integral into $\int_{-\infty}^{-\epsilon}f(x) + \int_{-\epsilon}^{\epsilon}f(x) + \int_{\epsilon}^\infty f(x)$. Then the first and last term will be equal by evenness of $f(x)$.
Does this subtlety only arise when integrating over a distribution that is not a function at the point of symmetry (and thus at the end-point when we try to split the integral)?