Connection between summation and integration

Question

Why is the following equality true (is there some theorem for this)?

$$\int_0^\infty f(t)e^{-st}dt=\sum_{n=0}^\infty\int_{nT}^{(1+n)T}f(t)e^{-st}dt\tag1$$

For a periodic function $f(t)$, with period $T$.

Answer 1

If $A_0,A_1,\dots$ are disjoint and covering measurable sets and $g$ is integrable on measure space $\langle\Omega,\mathcal A,\mu\rangle$ then:$$\int gd\mu=\int\sum_{n=0}^{\infty}g1_{A_n}=\sum_{n=0}^{\infty}\int g1_{A_n}d\mu=\sum_{n=0}^{\infty}\int_{A_n} gd\mu$$

In your case $A_n=(nT,(1+n)T]$ and $(0,\infty)=\Omega$ and $\mathcal A$ is the collection of Borel subsets of $\Omega$.