What is the integral of $\int_{-\infty}^{+\infty} H(t)\delta(t)dt$ ($H(t)$ Heaviside step, $\delta(t)$ Dirac delta)? I was trying to figure out what is the integral of 
$$\int_{-\infty}^{+\infty} H(t)\delta(t)dt,$$ 
where $H(t)$ is the Heaviside step and $\delta(t)$ is the Dirac delta.
A first approach:  We observe that $\delta(t) = H'(t)$, and hence:
$$\int_{-\infty}^x H(t)\delta(t)dt = \int_{-\infty}^{+\infty} H(t)H'(t)dt = \left.\frac{1}{2}H^2(t)\right|_{t=-\infty}^{t = +\infty} = \frac{1}{2}(1-0) = \frac{1}{2}.$$ 
A second approach: Since $\int_{-\infty}^{+\infty} f(t)\delta(t)dt = f(0)$, then:
$$\int_{-\infty}^x H(t)\delta(t)dt = H(0) = 1.$$
What's wrong?
 A: Let $a\in (0,1)$ and $\delta_n(x)$ be the regularized Dirac Delta given by
$$\delta_n(x)=\begin{cases}n&,x\in[-\frac an,\frac{1-a}n]\\\\0&,\text{elsewhere}\tag1\end{cases}$$
For all smooth functions with compact support $\phi$, we have
$$\lim_{n\to \infty}\int_{-\infty}^\infty \phi(x)\delta_n(x)\,dx=\phi(0)$$

Now, let's analyze the integral of $\delta_n(x)H(x)$.  Proceeding, we have
$$\begin{align}
\lim_{n\to\infty}\int_{-\infty}^\infty H(x)\delta_n(x)\,dx&=\lim_{n\to\infty}\int_{0}^{(1-a)/n}n\,dx
\\\\&=1-a\tag2
\end{align}$$
Inasmuch as the value the integral in $(2)$ depends on the regularization of the Dirac Delta, $\delta(x)$, we assert that the distribution $\langle H,\phi\rangle$ fails to exist.

It is of interest to note that the regularization $\delta_n(x)$ as given in $(1)$ is consistent with defining the Heaviside function as 
$$H(x)=\begin{cases}1&,x>0\\\\
a&,x=0\\\\0&,x<0\end{cases}$$ 
And naively evaluating $\langle H,\delta\rangle$ as $H(0)$ would give $\int_{-\infty}^\infty H(x)\delta(x)\,dx=a$, which doesn't agree with the result in $(2)$ unless $a=1/2$.
