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Prove an integral formula for periodic functions

Let f be an integralable function on [0,p], and let f be periodic with period p > 0. Prove

    \[ \int_0^p f(x) \, dx = \int_a^{a+p} f(x) \, dx \]

for all a \in \mathbb{R}.


Proof. For any a, p \in \mathbb{R} with p > 0 we know from a previous exercise that there exists a unique n \in \mathbb{Z} such that

    \[ n \leq \frac{a}{p} < n + 1 \quad \implies \quad np \leq a < (n+1)p \leq a + p. \]

Then, we start by splitting the integral into two pieces. (The goal here is to rearrange the integral so that it starts at np and ends at (n+1)p, and then use the fact that f is periodic to conclude that this integral is the same as the one from 0 to p.)

    \begin{align*}  \int_a^{a+p} f(x) \, dx &= \int_a^{(n+1)p} f(x) \, dx + \int_{(n+1)p}^{a+p} f(x) \, dx \\  &= \int_{a-np}^{(n+1)p - np} f(x+np) \, dx + \int_{(n+1)p - (n+1)p}^{a+p-(n+1)p} f(x+(n+1)p) \, dx \\  &= \int_{a-np}^p f(x) \, dx + \int_0^{a - np} f(x) \, dx & (\text{using periodicity})\\  &= \int_0^p f(x) \, dx & (\text{since } 0 \leq a-np < p ). \end{align*}

This completes the proof. \qquad \blacksquare

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