Dirac Delta Function Integral results in Heaviside?

Question

I read somewhere that:

$$\int_{0}^{t} \delta(\tau-a) \ d\tau = \begin{equation} u(t-a)=% \begin{cases} 0 &\text{if $t < a$} \\ 1 &\text{if $t \geq a$}. \end{cases} \end{equation}$$

How is that so?

Answer 1

Raw explanation.

In a distributional sense, the Dirac Delta is the derivative of the Heaviside theta:

$$\frac{d}{dx} \theta(x) = \delta(x)$$

But as I said, it's a distributional derivative, so take that writing with gloves.

Hence, the distributional integral of the Dirac Delta is the Theta.

Distributional explanation

Take a test function $\phi(x)$. Let's act on it with the theta derivative:

$$\langle \frac{d}{dx}\theta(x), \phi(x)\rangle = -\langle \theta(x), \frac{d}{dx}\phi(x)\rangle$$

$$ = -\int_{-\infty}^{+\infty}\theta(x)\phi'(x)\ \text{d}x = -\int_0^{+\infty}\phi'(x)$$

Due to the choice of the test function we get that at the infinity it does vanish, hence:

$$ = - \phi(\infty) + \phi(0) = \phi(0)$$

So we can finally deduce that

$$\langle \frac{d}{dx}\theta(x), \phi(x)\rangle = \langle\delta(x), \phi(x)\rangle = \phi(0)$$

Which is the meaning of

$$\int\delta(x)\phi(x)\ \text{d}x = \phi(0)$$

Now you just have to adapt it on your case.

Answer 2

In a, somewhat, more visual sense: You know, that heavyside defined his fuction to be 1 after a certrain point and zero before. Similiarly, the Dirac-Distribution ist very, very small (infinitly small) in width, but so incredibly high, that the area under the function is $1$. If you now using the "Integration in Area under a graph"-method to get to the result, you come from $-\infty$ and see nothing but zero. But at the point, where the dirac-impulse is located, you get an area. But this area is so small, that you don't really have some kind of slope. You just instantly hop to $1$ with your area. And since it's all zero afterwards, you will stay at that result. Hency you got the Heavyside-function.

Additionally: You can approximate a dirac by various functions. A Guassian distribution, I waas told about some $\tan$ function. They all have a parameter, that you can tune, to get a sharper impulse. (Using Gaussian Distribution: The variance) You can (often) get an integral of that function (or an approximation) that also depends on that parameter. As you tune that parameter to make the distribution sharper, you will also see, that the integral will go towards the heavyside function.

These methods are not rigorous but help to grasp the concept.

Answer 3

The Dirac delta has the property $\delta(\phi)=\phi(0)$ for test functions $\phi\in\mathscr{D}$. It is very common to write $\delta(\phi)$ as $\int_{-\infty}^\infty \delta(x)\phi(x)\,dx$. This doesn't really make sense in terms of Lebesgue integration, since there is no measurable function $\delta(x)\phi(x)$ to integrate, but people do it a lot.

Once you do that, you note that $\delta(x)$ is nonzero only at $x=0$ or, in more orthodox lingo, that the support of $\delta$ is $\{0\}$; as a result, you can write the integral in many, many other ways, basically as any $\int_u^v \delta(x)\phi(x)\,dx$ for any $u<0$, $v>0$. You can also note that $\int_u^v \delta(x)\phi(x)\,dx = 0$ if $uu>0$, for the same support reason and independent of $\phi$.

You can also note that $\int_u^v f(x-a)\,dx=\int_{u-a}^{v-a} f(x)\,dx$ for any $f$, a general property of integrals on the real line. Finally, although the function identically equal to $1$ is not a test function in $\mathscr{D}$ because its support isn't compact, when you use a finite integration interval you can take $\phi=1$.

Put it all together and I think you get what you want.