Derive the indefinite integral of a Heaviside Function

Question

How can I prove that

$$\int H(x-a)dx=(x-a)H(x-a)+constant$$

where $H(x-a)$ represents a jumped Heaviside function

Thanks in advanced.

Answer 1

Consider Ramp function \begin{align} R(x) &= \dfrac{x+|x|}{2} \\ &= \begin{cases}x&x\geq0,\\0&x<0\end{cases} \\ &= x{\bf H}(x) \end{align} It's easy to see for $x\neq a$, $\dfrac{d}{dx}R(x-a)={\bf H}(x-a)$ then $$\int{\bf H}(x-a)dx=R(x-a)+C=(x-a){\bf H}(x-a)+C$$

Answer 2

The Heaviside function is a piecewise step: $\mathcal H(x-a)=\begin{cases}1 &:& x-a\geqslant 0\\ 0 & :& x-a<0\end{cases}$

So consider the two cases, $x\geqslant a$ and $x<a$ for the directed integral $$\int \mathcal H(x-a)\mathrm d x = \mathscr c+ \begin{cases}\phantom+\int_a^x \mathcal H(s-a)\,\mathrm d s &:& x-a\geqslant 0\\ -\int_x^a \mathcal H(s-a)\,\mathrm d s & :& x-a<0\end{cases}$$

Answer 3

The function $H$ is piecewise constant, so that it is the derivative of a piecewise linear function. Furthermore, the slopes are $0$ then $1$.

We have

$$I(x):=\begin{cases}x\le a\to C_-,\\x\ge a\to C_++x.\end{cases}$$

To ensure continuity at $a$, $C_-=C_++a$ and

$$I(x):=\begin{cases}x\le a\to C,\\x\ge a\to C+x-a.\end{cases}$$

This is equivalent to

$$I(x)=C+(x-a)H(x-a).$$