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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

8 comments

    • Awamoki says:

      Hi jamiehlusko,
      I think your solution is wrong because you don’t know if a0 so you are not allowed to use the translation property of integrals.

      • Anonymous says:

        I guess it is because it exists for [0, p] and since it is periodic with period p it exists for all real values. The function just repeats itself and thus the area following p is congruent to the area following 0.

      • Mihajlo says:

        It seems that the proof is assuming what Vivek Suraiya wrote. For a complete proof it would be nice to see a proof of that too.

        Namely, theorem 1.17 assumes existance of 2 intervals out the 3 (and its proof in the book too). Therefore, it seems that the proof here cannot simply rely on it, because it relaxes that and assumes what Vivek Suraiya wrote.

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