\begin{equation*}
\mathscr{L}\{f(t)\} = \int\limits_{0}^{\infty}e^{-st} f(t)\,dt
\end{equation*}
\begin{equation*}
= \int\limits_{0}^{T}e^{-st} f(t)\,dt+ \int\limits_{T}^{2T}e^{-st} f(t)\,dt+ \int\limits_{2T}^{3T}e^{-st} f(t)\,dt+\cdots
\end{equation*}
\begin{equation*}
=\sum\limits_{n=0}^{\infty}\int\limits_{nT}^{(n+1)T}e^{-st} f(t)\,dt
\end{equation*}
If we put \(t=u+nT\text{,}\) then \(f(u+nT)=f(u)\) as \(T\) is period of the given function. Thus
\begin{equation*}
\mathscr{L}\{f(t)\} = \sum\limits_{n=0}^{\infty}\int\limits_{0}^{T}e^{-s(u+nT)} f(u)\,du
\end{equation*}
\begin{equation*}
= \sum\limits_{n=0}^{\infty}e^{-snt}\int\limits_{0}^{T}e^{-su}f(u)\,du
\end{equation*}
\begin{equation*}
=1+e^{-st}+e^{-2st}+\cdots+\int\limits_{0}^{T}e^{-su}f(u)\,du
\end{equation*}
\begin{equation*}
= \frac{1}{1-e^{-sT}}\int\limits_{0}^{T}e^{-su}f(u)\,du
\end{equation*}
\([\because \quad 1+a+a^{2}+a^{3}+\cdots=\frac{1}{1-a};\) if \(a \lt 1]\text{.}\)
\begin{equation*}
\therefore \quad \mathscr{L}\{f(t)\} = \frac{\int\limits_{0}^{T}e^{-st} f(t)\,dt}{1-e^{-sT}}
\end{equation*}
by changing \(u\) into \(t\text{.}\)
