Definite Integral of Periodic Function

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Theorem

Let $f$ be a Darboux integrable periodic function with period $L$.

Let $\alpha \in \R$ and $n \in \Z$.

Then:

$\ds \int_\alpha^{\alpha + n L} \map f x \d x = n \int_0^L \map f x \d x$


Proof

For $n \ge 0$:

\(\ds \int_\alpha^{\alpha + n L} \map f x \d x\) \(=\) \(\ds \int_\alpha^0 \map f x \d x + \sum_{k \mathop = 0}^{n - 1} \int_{k L}^{\paren {k + 1} L} \map f x \d x + \int_{n L}^{\alpha + n L} \map f x \d x\) Sum of Integrals on Adjacent Intervals for Integrable Functions/Corollary
\(\ds \) \(=\) \(\ds \int_\alpha^0 \map f x \d x + \sum_{k \mathop = 0}^{n - 1} \int_{k L}^{\paren {k + 1} L} \map f {x - k L} \d x + \int_{n L}^{\alpha + n L} \map f {x - n L} \d x\) General Periodicity Property
\(\ds \) \(=\) \(\ds \int_\alpha^0 \map f x \d x + \sum_{k \mathop = 0}^{n - 1} \int_0^L \map f x \d x + \int_0^\alpha \map f x \d x\) Integration by Substitution
\(\ds \) \(=\) \(\ds n \int_0^L \map f x \d x\) Reversal of Limits of Definite Integral


For $n < 0$:

\(\ds \int_\alpha^{\alpha + n L} \map f x \d x\) \(=\) \(\ds -\int_{\alpha + n L}^\alpha \map f x \d x\) Reversal of Limits of Definite Integral
\(\ds \) \(=\) \(\ds -\int_{\alpha + n L}^{\alpha + n L + \paren {-n L} } \map f x \d x\)
\(\ds \) \(=\) \(\ds -\paren {-n \int_0^L \map f x \d x}\) by the above; $-n > 0$
\(\ds \) \(=\) \(\ds n \int_0^L \map f x \d x\)

Hence the result.

$\blacksquare$