Properties of Periodic Functions
Sources:
Periodic Functions
A periodic function \(x(t)\)1 with fundamental period2 \(T_0\) has the property \[ f(x)=f\left(x+T_0\right) \quad \text { for all } \quad x . \]
A periodic function takes the same values at intervals of \(T_0\).
The area under a periodic signal \(x(t)\) over any interval of duration \(T_0\) is the same; that is, for any real numbers \(a\) and \(b\) \[ \color{grey} {\int_a^{a+T_0} f(x) d x=\int_b^{b+T_0} f(x) d x } . \]
This result follows from the fact that a periodic function takes the same values at intervals of \(T_0\). Hence, the values over any segment of duration \(T_0\) are repeated in any other interval of the same duration. For convenience, the area under \(x(t)\) over any interval of duration \(T_0\) will be denoted by \[ \color {orange} {\int_{T_0} f(x) d x} \]
The proof is simple.
First step: We need to prove that, for any contant \(a\), \[ \int_a^{a+{t_0}} f(x) d x=\int_0^{T_0} f(x) d x . \]
Then, we can substitude \(b\) into a. The result is: \[ \int_b^{b+T_0} f(x) d x = \int_0^{T_0} f(x) d x = \int_a^{a+T_0} f(x) d x . \] Q.E.D.
Now we prove step 1 (-->Source):
For simplicity, here we use period \(T\) instead of minimal period \(T_0\). Let \(a \in [0, T]\) and \(N=\left\lfloor\frac{a}{T}\right\rfloor\), such that \(0 \leq a-N T<T\), we then do the variable change \(u=x-N T, f\) being periodic then \(f(u)=f(x)\) and \[ \int_a^{a+T} f(x) d x=\int_{a-N T}^{a-N T+T} f(u) d u \]
Now, we know that \(0 \leq a-N T \leq T \leq a-N T+T\). So, \(a-N T+T\) is greater than \(T\), thus we need to cut the "excedent", shift it again, and restick it to your main integral : \[ \int_{a-N T}^{a-N T+T} f(x) d x=\int_{a-N T}^T f(x) d x+\int_T^{a-N T+T} f(x) d x \]
Do a variable change of \(v=x-T\) again in the second integral, it's still invariant by \(f\) being just a translation by a multiple of the period, and restick the two parts : \[ \int_{a-N T}^T f(x) d x+\int_T^{a-N T+T} f(x) d x=\int_{a-N T}^T f(x) d x+\int_0^{a-N T} f(v) d v=\int_0^T f(x) d x \]