Area under Dirac Delta function

What is the difference of the area calculation for the Dirac delta function when using different limits of integration?

$$\int_{-\infty}^{\infty}\delta(x)dx = 1$$

but

$$\int_{-\infty}^{t}\delta(x)dx = u(t)$$

• What is the limit of u(t) as t goes to infinity? (Alternatively, what is the value of u(t) for any t > 0? and is +infinity > 0?) Oct 27 '14 at 16:46
• isnt limit of u(t) as t goes to infinity 1? Oct 27 '14 at 16:48
• @Fawaz, no, u(t) is simply 1, t>=0, by definition. No need to invoke limits Oct 27 '14 at 17:53
• @ScottSeidman, "infinity" isn't a real number, it is only something we can approach as a limit. (e.g. whenever we talk about "infinity" we are actually using a shorthand to talk about a limit) However, I do admit I'm only doing "engineering math" here. If OP wants a mathematician's answer, they should go to math.SE. Oct 27 '14 at 18:11

When integrating to only $t$ there are two cases: if $t < 0$ then the integral is $0$, if $t \geq 0$ then the integral is $1$: $$\int_{-\infty}^{t}\delta(x)dx = \begin{cases} 0\text{, }t < 0 \\ 1\text{, }t \geq 0\end{cases}$$

But this is just another way of writing the unit step function $u(t)$ so

$$\int_{-\infty}^{t}\delta(x)dx = u(t)$$

Since $$\lim_{t \to \infty} u(t) = 1$$ then it is also true that

$$\int_{-\infty}^{\infty}\delta(x)dx = 1$$

$$\begin{matrix} \int_{-\infty}^\infty\ \delta(t) \mathrm{d}t &=& \lim_{\tau\to\infty}\int_{-\infty}^\tau\ \delta(t) \mathrm{d}t\\ &=& \lim_{\tau\to\infty}u(\tau)\\ &=& 1 \end{matrix}$$