Does this integral have a proper closed form solution? The integral is given by:
$\int_0^L\delta(x-v)H(x-d)dx$
$\delta$ is the dirac delta function and $H(x-d)$ is the heaviside function (equal to 0 for $x\leq d$ and equal to 1 for $x>d$).
Both $v$ and $d$ belong to the intercal $0\leq x\leq L$
Does this integral have a closed form solution?
 A: The comments are trying to make this more complicated than needed so I have no option but to remove the answer or continue pointless discussions about things that are peculiar and really not what this question was about.
I hope commentators are now thrilled.
I cannot delete the answer so I let someone else make it as complicated as he wants.
A: In general
$$\int_{-\infty}^{+\infty} \delta(x) f(x) dx = f(0)$$
or
$$\int_{-\infty}^{+\infty} \delta(x-v) f(x) dx = f(v)$$
So in general you would have $H(v-d)$ being $1$ if $v>d$. Elsewhere it is $0$.
A small problem is what to do if $v=d$, for example $v=d=0$ since nothing is nicely defined, there has to be some additional step that justifies the accepted value $0$ if we can justify it.
Probably, it would be advisable then to treat the integral as an improper integral and use this form of Riemann–Stieltjes integral that is shifting $f(x)\delta(x)dx$ towards $f(x)dH(x)$ (We understand here $H(x)$ as a cumulative distribution.)
$$\lim_{\epsilon \to 0} \int_{\epsilon}^{L}H(x)dH(x)$$
We do it this way since we need to make $H(x)$ a continuous function.
This gives the result $0$ as $H(x)$ is simply a constant either for $L$ or any $\epsilon$. The same goes for any $v=d$.
