Convolution with sign function I am having some trouble calculating the convolution $ (f*g)(t) $ between these two functions:
$$ f(t)=e^{-t}1(t) $$
where $1(t)$ is the unit step function, and
$$ g(t)=\mathrm{sgn}(t) $$
Using the definition of the convolution I get:
$$ (f*g)(t)=\int_{-\infty}^\infty e^{-\tau}1(\tau)\mathrm{sgn}(t-\tau)\,d\tau=\int_{0}^\infty e^{-\tau}\mathrm{sgn}(t-\tau)\,d\tau  $$
This is where I get stuck. Can someone point me in the right direction?
 A: Split the integral as follows.
$$I = \int_0^{\infty} e^{-\tau} \text{sign}(t-\tau) d \tau = \int_0^{t} e^{-\tau} \text{sign}(t-\tau) d \tau + \int_t^{\infty} e^{-\tau} \text{sign}(t-\tau) d \tau $$
Now recall the definition of $\text{sign}(t-\tau)$.
$$\text{sign}(t-\tau) = \begin{cases} -1 & \text{if }t-\tau < 0 \text{ i.e. }\tau > t\\1 & \text{if } t -\tau > 0 \text{ i.e. }\tau < t\end{cases}$$
Hence, if $t>0$, we have
$$I = \int_0^{t} e^{-\tau} d \tau - \int_t^{\infty} e^{-\tau} d \tau = (1-e^{-t}) - (e^{-t}) = 1-2e^{-t}$$
If $t < 0$, we have
$$I = -\int_0^{\infty} e^{-\tau} d \tau = -1$$
Hence, we have
$$\int_0^{\infty} e^{-\tau} \text{sign}(t-\tau) d \tau = \begin{cases}1-2e^{-t} & \text{if } t \geq 0\\ -1 & \text{if } t \leq 0 \end{cases}$$
A: If $t\ge 0$, we have 
\begin{eqnarray}
(f*g)(t) &=& \int_{-\infty}^\infty e^{-x}1_{[0,\infty)}(x) \operatorname{sgn}(t-x) dx\\
&=&  \int_{0}^\infty e^{-x} \operatorname{sgn}(t-x) dx \\
&=& \int_{0}^t e^{-x} dx - \int_{t}^\infty e^{-x} dx \\
&=& 1-e^{-t} - e^{-t} \\
&=& 1-2 e^{-t}
\end{eqnarray}
If $t<0$, we have
\begin{eqnarray}
(f*g)(t) &=& \int_{-\infty}^\infty e^{-x}1_{[0,\infty)}(x) \operatorname{sgn}(t-x) dx\\
&=&  \int_{0}^\infty e^{-x} \operatorname{sgn}(t-x) dx \\
&=& -\int_{0}^\infty e^{-x}  dx \\
&=& -1
\end{eqnarray}
Hence $(f*g)(t) = \max(1-2e^{-t},-1)$.
