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.
$\blacksquare$