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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

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

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  • \$\begingroup\$ 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?) \$\endgroup\$ – The Photon Oct 27 '14 at 16:46
  • \$\begingroup\$ isnt limit of u(t) as t goes to infinity 1? \$\endgroup\$ – Fawaz Oct 27 '14 at 16:48
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    \$\begingroup\$ @Fawaz, no, u(t) is simply 1, t>=0, by definition. No need to invoke limits \$\endgroup\$ – Scott Seidman Oct 27 '14 at 17:53
  • \$\begingroup\$ @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. \$\endgroup\$ – The Photon Oct 27 '14 at 18:11
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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

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

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There's no contradiction.

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

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  • \$\begingroup\$ Is this a proper limit problem? The integral of delta(t) is u(t), so the integral is u(infinity)-u(-infinity) = 1-0 =1. I haven't seen this handled as limits. \$\endgroup\$ – Scott Seidman Oct 27 '14 at 17:56

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