$f(t) = f(t+2a) \\[2ex] Here \ f(t) \ is \ periodic \ with \ period \ 2a.\\[2ex] \displaystyle \begin{align*} \therefore{} L\ \left[f(t)\right]&=\dfrac{1}{1-e^{-2as}}\int_0^{2a}e^{-st}\ f(t)\ dt\\[2ex] \displaystyle &= \frac{1}{1-e^{-2as}}\left[\int_0^ae^{-st}\ \left(1\right)dt+\ \int_a^{2a}e^{-st}\ \left(-1\right)dt\ \right]\\[2ex] &=\frac{1}{1-e^{-2as}}\left[{\left\{-\frac{e^{-st}}{s}\right\}}_0^a+{\left\{\frac{e^{-st}}{s}\right \}}_a^{2a}\right]\\[2ex] &= \frac{1}{s}\ \frac{1}{1-e^{-2as}}\left[1-e^{-as}+e^{-2as}-e^{-as}\right]\\[2ex] &= \frac{1}{s}\ \frac{1}{\left(1-e^{-as}\right)\left(1+e^{-as}\right)}{\left[1-e^{-as}\right]}^2\\[2ex] &= \frac{1}{s}\ \frac{\left(1-e^{-as}\right)}{\left(1+e^{-as}\right)}\\[2ex] &= \frac{1}{s}\ \frac{e^{\frac{as}{2}}-e^{\frac{-as}{2}}}{e^{\frac{as}{2}}+e^{\frac{-as}{2}}}\\[2ex] &= \frac{1}{s}\tanh\left(\frac{as}{2}\right)\\[2ex] \therefore{} L\ \left[f\left(t\right)\right]&=\ \frac{1}{s} \tanh \left(\frac{as}{2}\right) \end{align*}$