Show that the Laplace transform of $f(t-a) H(t-a)$, where $a \geq 0$, is $e^{-a s} f(s)$ and that, if $g(t)$ is a periodic function of period $T, \bar{g}(s)$ can be written as
$$
\frac{1}{1-e^{-s T}} \int_{0}^{T} e^{-s t} g(t) d t
$$
(a) Sketch the periodic function defined in $0 \leq t \leq T$ by
$$
g(t)=\left\{\begin{array}{cl}
2 t / T & 0 \leq t<T / 2 \\
2(1-t / T) & T / 2 \leq t \leq T
\end{array}\right.
$$
and, using the previous result, find its Laplace transform.
(b) Show, by sketching it, that
$$
\frac{2}{T}\left[t H(t)+2 \sum_{n=1}^{\infty}(-1)^{n}\left(t-\frac{1}{2} n T\right) H\left(t-\frac{1}{2} n T\right)\right]
$$
is another representation of $g(t)$ and hence derive the relationship
$$
\tanh x=1+2 \sum_{n=1}^{\infty}(-1)^{n} e^{-2 n x}
$$.