Consider two functions g and u over the interval $[0,\infty)$ defined by: $$ \begin{array}{clll} g(t)&=e^{-t},\;0\leq t\\[2mm] u(t)&=\left\{ \begin{array}{ccc} 1&;&0\leq t\\ 0&;&\text{otherwise}\\ \end{array} \right. \end{array} $$
Let $y(t)$ be the function defined by: $$ y(t)=\int_0^t g(t-\tau )u(\tau) d\tau $$
(a) Find the Laplace Transform $Y(s)$ for $y(t)$.
(b) Find $y(t)$
What I have tried:
(a) $y(t)=\int g(t-\tau) d\tau= \int e^{-(t-\tau)} d\tau = e^{\tau-t}+C$ therefore
$Y(s)=\int e^{-st} e^{\tau-t}d\tau=e^{-t(s+1)+\tau}+C$
Is my starting point correct?
Also, to find complete $y(t)$ do I have to integrate $e^{\tau-t}$ for $0\leq t$ and $t\leq 1$ separately?
Remember the definition of convolution, $(f*g)(t)=\displaystyle\int_0^t f(\tau)g(t-\tau)d\tau$, then $\mathcal L\{f*g\}=\mathcal L\{f\}\cdot\mathcal L\{g\}$.
Hence we have \begin{align*} Y(s)&=\mathcal L\left\{\int_0^t g(t-\tau )u(\tau) d\tau\right\}\\[2mm] &=\mathcal L\{g(t)\}\cdot \mathcal L\{u(t)\}\\[2mm] &=\mathcal L\{e^{-t}\}\cdot \mathcal L\{u(t)\}\\[2mm] &=\dfrac{1}{s+1}\cdot \dfrac{1}{s}\\[2mm] &=\dfrac{1}{s(s+1)} \end{align*} Now using simple fractions we have \begin{align*} Y(s)&=\dfrac{1}{s}-\frac{1}{s+1}\\[2mm] \mathcal L^{-1}\{Y(s)\}&=\mathcal L^{-1}\left\{\dfrac{1}{s}-\frac{1}{s+1}\right\}\\[2mm] y(t)&=\mathcal L^{-1}\left\{\dfrac{1}{s}\right\}-\mathcal L^{-1}\left\{\frac{1}{s+1}\right\}\\[2mm] y(t)&=u(t)-e^{-t} \end{align*}