I'm practicing Laplace transforms and I stumbled upon one question which I am not exactly sure how to tackle. The question is:
Using Laplace transforms (or otherwise) calculate the convolution of two functions $f(t)$ and $g(t)$ defined as: \begin{align*}f(t)&=\cos t && \text{for } t≥0 \\ f(t)&=0 && \text{for } t<0\end{align*} and \begin{align*} g(t)&=1 && \text{for } t≥0 \\ g(t)&=0 && \text{for } t < 0 .\end{align*}
Can someone please help me with this question?
The Convolution Theorem gives
$$\mathscr{L}\left((f*g)(t)\right)(s)=\mathscr{L}\left(f(t)\right)(s)\mathscr{L}\left(g(t)\right)(s) \tag 1$$
where $\mathscr{L}(f(t))(s)$ is the Laplace Transform of $f$ and
$$(f*g)(t)=\int_{-\infty}^{\infty}f(t')g(t-t')dt'$$
is the convolution (integral) of $f$ and $g$.
Now, we let $f(t)=\cos (t)\,u(t)$ and $g(t)=u(t)$, where $u(t)$ is the unit step function. Note that the Laplace Transform of $\cos (t)$ is
$$\mathscr{L}\left(\cos (t)u(t)\right)(s)=\frac{s}{s^2+1} \tag 2$$
and the Laplace Transform of $u(t)$ is
$$\mathscr{L}\left(u(t)\right)(s)=\frac{1}{s} \tag 3$$
Using $(2)$ and $(3)$ in $(1)$, we see that the Laplace Transform of the convolution of $\cos(t)(t)$ and $u(t)$ is
$$\mathscr{L}\left(\cos (t)*u(t)\right)(s)=\frac{1}{s^2+1} \tag 4$$
Recognizing the right-hand side of $(4)$ is the Laplace Transform of $\sin (t)u(t)$, we conclude that
$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^{\infty}\cos (t')u(t')u(t-t')dt'=\sin(t)u(t) }\tag 5$$
To check to see if $(5)$ is correct, we evaluate the convolution directly. To that end, we have
$$\int_{-\infty}^{\infty}\cos (t')u(t')u(t-t')dt'=\int_0^\infty cos(t')u(t-t')dt' \tag 6$$
For $t<0$, the unit step in the integrand of $(6)$ is always zero and therefore the convolution is zero. For $t>0$, we have
$$\int_{-\infty}^{\infty}\cos (t')u(t')u(t-t')dt'=\int_0^t cos(t')dt'=\sin (t)$$
and therefore find
$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^{\infty}\cos (t')u(t')u(t-t')dt'=\sin(t)u(t)}$$
as expected!