Calculate the convolution of two functions I've been given a problem, to calculate the convolution of the functions $h$ and $x$ where:
$$h(t) = \begin{cases}t  &\text{if $0<t<2T$}\\ 0& \text{otherwise}\end{cases}$$
     $$x(t) = \begin{cases} 1 & \text{if $0<t<T$}\\ 0 &\text{otherwise.}\end{cases}$$
($T$ is a parameter).
How can I do that? A detailed explanation/solution is preferred. Thanks!
 A: The computation of this convolution $g:=x*h$ is quite tedious since both $x$ and $h$ are defined piecewise. Formally $g$ is defined by
$$g(t):=\int_{-\infty}^\infty x(t-\tau)h(\tau)\>d\tau\qquad(-\infty<t<\infty)\ .$$
Now $x(t-\tau)\ne0$ (in fact $=1$) only if $0<t-\tau<T$, or $t-T<\tau<t$. It follows that we may write
$$g(t)=\int_{t-T}^t h(\tau)\>d\tau\qquad(-\infty<t<\infty)\ .$$
We now have to distinguish cases according to how the interval of integration $[t-T, t]$ intersects the interval $[0,2T]$ where $h$ has an "interesting definition". From a figure we deduce that we have to distinguish the following five cases:
$$\eqalign{g(t)&=\int_{t-T}^t 0\>d\tau=0\qquad(t<0)\ ,\cr
g(t)&=\int_{0}^t \tau\>d\tau=\ldots\qquad(0<t<T)\ ,\cr
g(t)&=\int_{t-T}^t \tau\>d\tau=\ldots\qquad(T<t<2T)\ ,\cr
g(t)&=\int_{t-T}^{2T} \tau\>d\tau=\ldots\qquad(2T<t<3T)\ ,\cr
g(t)&=\int_{t-T}^t 0\>d\tau=0\qquad(t>3T)\ .\cr}$$
I leave it to you to compute the three nonzero integrals. In the end you should obtain a  function $t\mapsto g(t)$ which is continuous on all of ${\mathbb R}$.
