I am aware this is not a perfectly rigorous question, but bear in mind I am a physicist (though I would be open to a rigorous discussion). Suppose I want to integrate a function $f$ together with the delta function derivative kernel $δ$'($x$), but instead of integrating from minus to plus infinity, I want to do it from $0$ to $\infty$.
I am realizing this is not entirely rigorous, and probably the right way to discuss it is working with the distribution u$δ$', where u is the Heaviside distribution. But then again I believe the product of the distributions is ill defined. So how do I do it, assuming f is a test function that vanishes close to $x=0$?
The Dirac Delta is not a function and the linear functional that is often represented by an integral is not an integral.
If $f$ is a suitable test function (i.e., $f\in C_C^\infty$), then we have
$$\langle \delta, f\rangle=f(0)$$
However, $fH$ is not a suitable test function since it is not continuous at the origin. And both $\langle \delta,fH\rangle$ and $\langle \delta H,f\rangle$ are ill-defined.
To see this, let $\delta_n$ be the sequence of functions defined for any number $0<a<1$, as
$$\delta_n=\begin{cases}n&,x\in (-a/n,(1-a)/n)\\\\0&,\text{elsewhere} \end{cases}$$
Then, for any suitiable test function $f$ we have clearly that
$$\lim_{n\to\infty}\int_{-\infty}^\infty \delta_n(x) f(x)\,dx=f(0)$$
We call $\delta_n$ a regularization of the Dirac Delta. Now, let's apply this regularization to the function $fH$, which is equivalent to applying $\delta_n H$ to the test function $f$. Proceeding, we have
$$\lim_{n\to\infty}\int_{-\infty}^\infty \delta_n(x)f(x)H(x)dx=\lim_{n\to\infty}\int_0^{(1-a)/n}nf(x)dx=(1-a)f(0)$$
Inasmuch as this result depends on the value of $a$, the limit is not uniquely defined and $\delta H$ is, therefore, not a distribution.