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EDIT

In physics, most of us learned that \begin{align} \int_{a^-}^{a^+} f(x)\delta (x-a)dx&=f(a)\\ \frac{dH(x)}{dx}&=\delta(x). \end{align}

So, would it be natural to have the below? \begin{align} \int_0^t\delta(x+a)dx=H(t+a)-H(a) \end{align} However, it shows in Wolfram this weird complexity \begin{align} \int_0^t\delta(x+a)dx=(2H(t)-1)H(-a-tH(-t))H(a+tH(t)) \end{align}

It surprised me that my understanding of Heaviside function $H(x)$ was heavily overestimated.

MathArt
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    "We all learned that" ... personally, I learned that $\langle \delta_a,\varphi\rangle = \varphi(a)$ for $\varphi$ continuous ... but perhaps you come from physics? Perhaps you could write "I learned that" instead? – LL 3.14 Aug 20 '23 at 09:33
  • @LL 3.14 "You come from physics" isn't very kind... You should know that even mathematicians use this identity which can be considered if you want as an "abuse of notation" ... – Jean Marie Aug 20 '23 at 09:37
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    Yes, of course, I know that. I didn't want to offend by saying he was coming from physics, which is not bad from my opinion, just to make him aware that there are always a lot of point of views. And no, if you see the Dirac delta as a Radon measure, then it can be thought of as acting on continuous functions ... – LL 3.14 Aug 20 '23 at 09:40
  • Notice Wolfram's formula becomes a lot simpler if you assume $t>0$. (There's still a thing or two left to explain.) – aschepler Aug 20 '23 at 20:15

1 Answers1

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No worry. It is a different expression for the same final result.

I have programmed your "weird function" defined by :

$$W(t)=\underbrace{(2H(t)-1)}_{sign(t)}.H(-a-t.H(-t))H(a+\underbrace{t.H(t)}_{ramp}) \tag{1}$$

(Matlab code below)

It gives the same result as $H(t+a)-H(a)$ in the two different cases $a>0$ and $a \le 0$.

It is not the first time that I see equivalent expressions written very differently using Heaviside function.

A known example is the "characteristic function" of interval $[a,b]$ (value $1$ for $a<t<b$, $0$ elsewhere) which can be written using $H$ function in (at least) two ways :

$$\underbrace{H(t-a)-H(t-b)}_{additive \ way}=\underbrace{H(t-a)H(b-t)}_{multiplicative \ way}$$

Please note that the first factor in (1) is the sign function and that $t*H(t)$ defines the "ramp function" which is a primitive function of $H$.

t=-5:0.01:5;
H=@(x)((sign(x)+1)/2); % Heaviside
a=-2; % 
W=(2*H(t)-1).*H(-a-t.*H(-t)).*H(a+t.*H(t));
plot(t,W)
Jean Marie
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