# 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?

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 $$\int_0^{\infty} e^{-\tau} \text{sign}(t-\tau) d \tau = \begin{cases}1-2e^{-t} & \text{if } t \geq 0\\ -1 & \text{if } t \leq 0 \end{cases}$$
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 $(f*g)(t) = \max(1-2e^{-t},-1)$.