Laplace transform $f(t) = f(t + \frac{2\pi }{a}) = \frac{\sin at}{|\sin at|}$ I had done the laplace transform with the period fucntion (T): $F(s) = \frac{1}{1 - e^{ - sT}}\int_o^T {e^{ - st}}f(t) \, dt $
Howerver, in this fuction:
$$f(t) = f \left(t + \frac{2\pi}{a}\right) = \frac{\sin at}{|\sin at|}$$
What is the period? And You can give me detail of solution
 A: Note that $T = 2 \pi/a$, and
$$\frac{\sin{a t}}{|\sin{a t}|} = \begin{cases} \\ +1 & a\, t \in (0,\pi) \\ -1 & a \,t \in (\pi,2 \pi)\end{cases}$$
Then
$$\begin{align}\int_0^{2 \pi/a} dt\, \frac{\sin{a t}}{|\sin{a t}|} \, e^{-s t} &= \int_0^{\pi/a} dt \, e^{-s t} - \int_{\pi/a}^{2 \pi/a} dt \, e^{-s t}\\ &= \frac{1-e^{-\pi s/a}}{s} -\frac{e^{-\pi s/a}-e^{-2 \pi s/a}}{s} \\ &= \frac{(1-e^{-\pi s/a})^2}{s} \end{align}$$
The full Laplace transform is then
$$\int_0^{\infty} dt\, \frac{\sin{a t}}{|\sin{a t}|} \, e^{-s t} = \frac{(1-e^{-\pi s/a})^2}{s(1-e^{-2\pi s/a})} = \frac{1}{s}\tanh{\frac{\pi s}{2 a}}$$
A: I did by let u = at, so the period T = 2pi
$\begin{array}{l}
 f(t) = f(t + \frac{{2\pi }}{a}) = \frac{{\sin at}}{{|\sin at|}} \\ 
 Let:f(u) = (u + 2\pi ) = \frac{{\sin u}}{{|\sin u|}} \\ 
 F(s) = L{\rm{[f(u)]}} = \frac{1}{{1 - {e^{ - 2\pi s}}}}\int_o^{2\pi } {{e^{ - st}}\frac{{\sin {a^2}u}}{{|\sin {a^2}u|}}du}  \\ 
  = \frac{1}{{1 - {e^{ - 2\pi s}}}}{\rm{\{ }}\int_o^\pi  {{e^{ - st}}du}  - \int_\pi ^{2\pi } {{e^{ - st}}du} {\rm{\} }} \\ 
  = \frac{{{e^{\pi s}} - 1}}{{s({e^{\pi s}} + 1)}} \\ 
 L{\rm{[f(u)]}} = \frac{{{e^{\pi s}} - 1}}{{s({e^{\pi s}} + 1)}} \\ 
  =  > L{\rm{[f(at)] = }}\frac{1}{a}F(\frac{s}{a}) = \frac{{{e^{\frac{{\pi s}}{a}}} - 1}}{{a\frac{s}{a}({e^{\frac{{\pi s}}{a}}} + 1)}} = \frac{{{e^{\frac{{\pi s}}{a}}} - 1}}{{s({e^{\frac{{\pi s}}{a}}} + 1)}} =  Yours\\ 
 \end{array}$ 
So did i work corecttly??
