How to prove this relation between definite integral and the sum of definite integrals

Question

When there is a periodic function $f:[0,\infty]\to\mathbb{R}$ with a period of $M>0$. How can I prove the following equality (assuming that the integral exists of course):

$$\int_0^\infty f(t)e^{-st}dt=\sum_{p=0}^\infty\int_{pM}^{(p+1)M}f(t)e^{-st}dt$$

Answer 1

Is this very naive calculation of any use?

If $ |\int_0^M f(t)e^{-st}dt| \le V $ is bounded, then

$\begin{array}\\ |\int_{pM}^{p(M+1)} f(t)e^{-st}dt| &=|\int_{0}^{M} f(t+pM)e^{-s(t+pM)}dt|\\ &=|e^{-spM}\int_{0}^{M} f(t)e^{-st}dt|\\ &\le|e^{-spM}V|\\ &\to 0 \qquad\text{as } p \to \infty\\ \end{array} $

so that

$\begin{array}\\ |\sum_{p=P}^{\infty}\int_{pM}^{p(M+1)} f(t)e^{-st}dt| &\le\sum_{p=P}^{\infty}|e^{-spM}V|\\ &=\dfrac{Ve^{-sPM}}{1-e^{-sM}}\\ \end{array} $

Therefore

$\begin{array}\\ \int_0^\infty f(t)e^{-st}dt &=\lim_{n \to \infty}\int_0^{nM} f(t)e^{-st}dt\\ &=\lim_{n \to \infty}\sum_{p=0}^{n-1}\int_{pM}^{(p+1)M}f(t)e^{-st}dt\\ &=\lim_{n \to \infty}\sum_{p=0}^{n-1}\int_{0}^{M}f(t+pM)e^{-s(t+pM)}dt\\ &=\lim_{n \to \infty}\sum_{p=0}^{n-1} e^{-spM}\int_{0}^{M}f(t)e^{-st}dt\\ &=\lim_{n \to \infty}\dfrac{1-e^{-sMn}}{1-e^{-sM}}\int_{0}^{M}f(t)e^{-st}dt\\ &=\dfrac{1}{1-e^{-sM}}\int_{0}^{M}f(t)e^{-st}dt\\ \end{array} $