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?

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 $$x(t)*h(t) = {1 \over 2}\left( {{{\left( {t - {1 \over 2}} \right)}^2}\mathscr{U}(t - {1 \over 2}) - {{\left( {t - {3 \over 2}} \right)}^2}\mathscr{U}(t - {3 \over 2})} \right).$$
