If $a<b<c$ we have $$\int_a^cf(x)\, dx=\int_a^bf(x)\, dx+\int_b^c f(x)\, dx\tag 1$$
But I have also seen $$\int_{\alpha}^{\alpha+T}f(x)\, dx=\int_\alpha^T f(x)\, dx+\int_T^{\alpha+T}f(x)\, dx \tag 2$$
Q1: In $(2)$, does it mean $f$ has to be periodic with $T$, i.e. $f(x+T)=f(x)$?
Q1: Is $(2)$ a special case of $(1)$, I can't see why?
In $(1)$ we have the inequality $a<b<c$, does it mean we have $\alpha<T<\alpha+ T$ in $(2)$?
Following the doubts at the end of your question: can you prove (2) if, say, $\alpha> T>0$ (which requires you to give meaning to $\int_\alpha^T f$).