Q:

What does Dirac delta function of a constant mean?

I have seen a formula that unit step function is the integration of Dirac delta function.
$$H(x) = \int_{-\infty}^{x} \delta(t)~\mathrm dt $$
In evaluating the integral if we take the integral as sum of infinite terms ,what does delta function of  a constant represent?

A:

The $\delta$-function is not actually a function - it's a distribution. The idea of finding an integral by using lots of thin rectangles doesn't work here. This is a more abstract form of integration. In fact, lots of integrals from Quantum Mechanics do not converge in the classical sense.
The $\delta$-function has the property that $\delta(x) = 0$ for all $x \neq 0$.
So, for example, $\delta(2) = 0$ and $\delta(-3.4)=0$.
The value of $\delta(0)$ is not well-defined, but we do know that
$$\int_{-\infty}^{\infty}\delta(x)~\mathrm dx = 1$$
Since $\delta(x) = 0$ for all $x \neq 0$, the values of $x$ away from zero contribute nothing to the integral:
$$\int_{-\varepsilon}^{\varepsilon} \delta(x)~\mathrm dx = 1$$
for any $\varepsilon > 0$, as small as you like!
If $S$ is some open subset of the real numbers then
$$\int_S \delta(x)~\mathrm dx \ \ = \ \ \left\{ \begin{array}{ccc} 1 & : & 0 \in S \\ 0 & : & 0 \notin S \end{array}\right.$$
In fact, you can make even stronger statements, e.g. $S$ doesn't need to be open, but you need to be careful how you word it.
In your example
$$H(x) := \int_{-\infty}^{x} \delta(\tau)~\mathrm d\tau$$
the set $S$ is the interval $(-\infty,x)$. If $x < 0$ then $0 \notin S$ and so $H(x) = 0$ for all $x < 0$. If $x>0$ then $0 \in S$ and so $H(x) = 1$ for all $x > 0$. What happens when $x=0$ depends on whom you speak to.

