The derivative of the Heaviside function $\theta(x – a)$ is normally taken to be the delta function $\delta(x – a)$.
This question has two parts, the first is whether a constant coefficient is introduced when differentiating the Heaviside function. According to Wolframalpha (I know it is not the best reference site but unable to find information elsewhere),
$$\frac{\partial}{\partial x}\theta(x/a – 1) = \frac{1}{a}\delta(x/a – 1),$$
whereas,
$$\frac{\partial}{\partial x}\theta(x-a) = \delta(x – a),$$
although these two Heaviside functions are essentially the same. Wolframalpha: first equation second equation.
The second part is whether is is correct to say that,
$$\theta(x – a) = \theta(x/a – 1),$$
which the above seems to imply is not true but I would normally freely switch between the two. I guess a similar question would be if,
$$\delta(x – a) = \delta(x/a – 1).$$
This could be a mistake by me in either formulating the problem or not really a proper mathematical question (I am a physicist and so skirt the boundaries of mathematical rigour).
Many thanks
Best Answer
You want to know $\frac{d}{dx} \theta(x/a-1)$. You write the definition:
$$\int_{-\infty}^\infty \frac{d}{dx} \left ( \theta(x/a-1) \right ) f(x) dx = -\int_{-\infty}^\infty \theta(x/a-1) f'(x) dx = -\int_a^\infty f'(x) dx = f(a)$$
Thus $\frac{d}{dx} \theta(x/a-1) = \delta(x-a)$ in the sense of distribution theory. But this is the same as $\frac{1}{a} \delta(x/a-1)$, because of the definition of $\delta(x/a-1)$.