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?
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 $H \delta$ 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 $a=1/2$.