connection between integral of perodic function and the sum of integrals

Question

I've a periodic function $y:[0,\infty]\to\mathbb{R}$ with a period of $T>0$. How can I prove the following equality:

$$\int_0^\infty y(t)e^{-st}dt=\sum_{n=0}^\infty\int_{nT}^{(n+1)T}y(t)e^{-st}dt$$

Answer 1

The intervals $(nT,(n+1)T]$ are disjoint and their union is $(0,\infty)$. Periodicity is not involved in this.

Answer 2

Using that $f$ is periodic with a period $T>0$, we have additionally \begin{align} \int_0^\infty f(t) \mathrm{e}^{-st} \, \mathrm{d} t &= \sum_{n=0}^\infty \int_{nT}^{(n+1)T} f(t)\mathrm{e}^{-st} \, \mathrm{d} t \\ &=\sum_{n=0}^\infty \mathrm{e}^{-nTs} \int_0^T f(h) e^{-sh} \mathrm{d} h \\ &= \frac{1}{1-\exp(-Ts)} \int_0^T f(h) e^{-sh} \mathrm{d} h. \end{align} This formula is more interesting, because we have now only the integration over one period involved.

Edit: As already commented, we need additional asumptations to guarantee the existence of the integral (in Lebesgue-sense). For example, we can assume that $f$ is measurable and bounded.