Why is $\int e^{-t}u(t) dt = (1-e^{-t})u(t) + Constant$?

Question

How do you solve $\int e^{-t}u(t) dt $? In which u(t) is the unit step function. $\int e^{-t}u(t) dt = (1-e^{-t})u(t) + Constant$ But what are the intermediate steps?

Unit step

1. u(t) = \begin{cases} 0 & \mbox{for } t < 0 \\ \\ 1 & \mbox{for } t > 0 \end{cases}

Integration by parts:

$f=-e^{-t}$

$g=u(t)$

$\int f'g = fg - \int fg'$

$\int e^{-t}u(t) = -e^{t}u(t) - \int -e^{-t}\delta(t) = -e^t u(t) + u(t) = (1-e^t ) u(t)$

Answer 1

Let $\varphi\in\mathcal{C}^\infty_c\left(\mathbb{R};\mathbb{R}\right)$ a test function; we have \begin{align} \int_\mathbb{R}\left(1-\mathrm{e}^{-t}\right)u(t)\varphi'(t)\mathrm{d}t & = \int_0^{+\infty}\left(1-\mathrm{e}^{-t}\right)\varphi'(t)\mathrm{d}t\\ & = \left[\left(1-\mathrm{e}^{-t}\right)\varphi(t)\right]_0^{+\infty}-\int_0^{+\infty}\mathrm{e}^{-t}\varphi(t)\mathrm{d}t\\ & = -\int_\mathbb{R}\mathrm{e}^{-t}u(t)\varphi(t)\mathrm{d}t\\ \end{align} because $\varphi$ has compact support. So the derivative in the sense of distributions of $t\mapsto \left(1-\mathrm{e}^{-t}\right)u(t)$ is given by $t\mapsto\mathrm{e}^{-t}\varphi(t)u(t)$.

Answer 2

\begin{align*} \int e^{-t} u(t) dt &= \int_{-\infty}^t e^{-s} u(s) ds +C\\ &= \begin{cases} C & t \leq 0\\ \int_0^te^{-s} ds +C & t >0\\ \end{cases} \\ &= \begin{cases} C & t \leq 0\\ \big(1-e^{-t}\big) +C & t >0\\ \end{cases} \\ &= \big(1-e^{-t}\big) u(t) + C. \end{align*}