Computing a messy convolution Consider the functions
$$
x(t) = u(t - \frac{1}{2}) - u(t - \frac{3}{2})
$$
and
$$
h(t) = tu(t)
$$
where $u(t) = 1$ if $t \geq 0$ and $u(t) = 0$ if $t < 0$.
I'm trying to compute
$$
(x*h)(t) = \int\limits_{-\infty}^{\infty} x(t)h(\tau - t) \, d\tau.
$$
However, the problem gets nasty very quickly and ends up as the sum of terrible looking expressions. Could anyone give me an idea on how to simplify this problem, if there is a way?
 A: An easy (and pretty standard) way of computing convolutions is by taking the Laplace transform of the functions, multiplying them (convolution transforms to multiplication in Laplace transformed space), and taking the inverse Laplace transform. As for this question $$\eqalign{
  & X(s) = {{{e^{ - {s \over 2}}} - {e^{ - {{3s} \over 2}}}} \over s}  \cr 
  & H(s) = {1 \over {{s^2}}} \cr} $$ so that $$X(s)H(s) = {{{e^{ - {s \over 2}}} - {e^{ - {{3s} \over 2}}}} \over {{s^3}}}$$ Now it is easy to see that $$x(t)*h(t) = {1 \over 2}\left( {{{\left( {t - {1 \over 2}} \right)}^2}\mathscr{U}(t - {1 \over 2}) - {{\left( {t - {3 \over 2}} \right)}^2}\mathscr{U}(t - {3 \over 2})} \right).$$
A: This convolution can be seen as a sliding average in the window $[t-\frac32,t-\frac12]$, over a linear ramp function.
As long as $t-\frac12\le0$, the average is $0$.
When $t-\frac32\ge0$, gives the value of the ramp at the window center, i.e. $t-1$.
In between, you get a piece of a parabola that ensures a smooth transition (equal values and equal first derivatives).
$$\int_{t-3/2}^00\ dt+\int_0^{t-1/2}t\ dt=\frac12\left(t-\frac12\right)^2.$$

Aside the smoothed knee, it is essentially a delay.

You can also check directly that $$\int_{t-\frac32}^{t-\frac12}t\ dt=t-1.$$
