Let's consider the function: $$ \int_{K}^{+\infty}f(x)dx$$ with $f(x) = \delta(x-K)$ if $x=K$ where $\delta$ is the Dirac delta function (cf https://en.wikipedia.org/wiki/Dirac_delta_function#Definitions)
and $f(x) = -x^{-3}$ for $x>K$
Does the value of $f$ at $x=K$ has an impact on the value of the integral? My intuition says yes, but I don't really know how to calculate the value with the dirac bound.
If
$$f(x) := \left\{\begin{array}{cc} -x^{-3} &: x\not = K\\ \delta(x-K) &: x = K \end{array} \right.$$
then $f(x) = \delta(x-K) + g(x)$, where
$$g(x) := \left\{\begin{array}{cc} -x^{-3} &: x\not = K\\ 0 &: x = K \end{array} \right.$$
Since we know how to integrate $g(x)$, by linearity of integration it's enough to consider how to integrate $\delta(x-K)$ when one of the bounds of integration is $K$.
One property of $\delta(x-K)$ which can be taken to be the defining property is that
$$\int \delta(x-K)\, dx = H(x-K),$$ where
$$H(x) = \left\{\begin{array}{cc} 0 &: x<0\\ 1 &: x>0\end{array} \right.$$ Various conventions can be given to $H(0)$ (see e.g. here), typically $H(0) = 1/2$, but these are only conventions since $H$ is discontinuous there.
The value of the integral
$$\int_K^\infty \delta(x-K) \, dx$$ depends on this convention. If you choose $a$ to be the value of the integral, then by necessity you will get
$$\int_{-\infty}^K \delta(x-K) \, dx = 1-a.$$
This assumes you are doing Riemann integration. The (Lebesgue) notation of integrating over a set removes this ambiguity.