This is a very fundamental question. I am missing something very important here.
I stumbled upon this issue while solving a integration step in a problem as follows:
I am having a discontinuous function as $F_X(x)$ given by
$$F_X(x)=\begin{cases}
1 &, 0\leq x< 3 \\\\
\exp(-\mu x)&, 3\leq x<\infty
\end{cases}$$
I could integrate $F_X(x)$ easily by a piecewise integration.
However if I could represent this function as $\displaystyle1-(1-\exp(-\mu x))H(x-3)$ then proceed for integration as $\displaystyle \int_0^{\infty} [1-\{1-\exp(-\mu x)\}H(x-3)] dx$ , I couldn't solve it.
Further a more fundamental question to ask is:
I want to integrate rectangular function decomposed as Heaviside functions as follows.
$$\int_0^{\infty} \left(H(t-2)-H(t-3)\right) \,dt$$
It is quite straight forward that the answer is 1, however I am unable separate the integral using the linearity of integration and find the answer. Obviously I am missing some fundamentals here.
Please any advice will be helpful, is there some silly mistake I am making in the whole question.
Best Answer
It's unclear as to the specific comment in the OP
But, note that we have for $L>3$
$$\begin{align} \int_0^L (H(t-2)-H(t-3))\,dt&=\left(\int_0^L H(t-2)\,dt\right)-\left(\int_0^L H(t-3)\,dt \right)\\\\ &=(L-2)-(L-3)\\\\ &=1 \end{align}$$
Now letting $L\to \infty$, we obtain the expected result.
Similarly, for $L>3$, we have
$$\begin{align} \int_0^L \left(1-(1-e^{-\mu x})H(x-3)\right)\,dx&=\int_0^L (1)\,dx- \int_0^L (1-e^{-\mu x})H(x-3)\,dx\\\\ &=L-\int_3^L(1-e^{-\mu x})\,dx\\\\ &=L-(L-3)+\int_3^L e^{-\mu x}\,dx\\\\ &=3+\frac{e^{-3\mu}-e^{-L\mu}}{\mu} \end{align}$$
Letting $L\to \infty$ we find that
$$\int_0^\infty \left(1-(1-e^{-\mu x})H(x-3)\right)\,dx=3+\frac{e^{-3\mu}}{\mu}$$