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Periodicity and integral

Let f:\mathbb{R} \rightarrow \mathbb{R} be a continuous and periodic function with period T \neq 0. If \kappa, \mu, \nu \in \mathbb{N} then prove that:

  1. \displaystyle \int_{0}^{\kappa \mathrm{T}} f(x) \, \mathrm{d}x = \kappa \int_0^{\mathrm{T}} f(x) \, \mathrm{d}x
  2. \displaystyle \int_{\alpha + \mu \mathrm{T}}^{\beta + \nu \mathrm{T}} f(x) \, \mathrm{d}x = \int_{\alpha}^{\beta} f(x) \, \mathrm{d}x + \left ( \nu - \mu \right ) \int_{0}^{\mathrm{T}} f(x) \, \mathrm{d}x where \alpha, \beta \in \mathbb{R}.

Solution

  1. We have successively:

        \begin{align*} \int_{0}^{\kappa T} f(x) \, \mathrm{d}x &= \sum_{n=0}^{\kappa-1} \int_{nT}^{(n+1) T} f(x) \, \mathrm{d}x \\ &= \sum_{n=0}^{\kappa-1} \int_{0}^{T} f \left ( u + nT \right )\, \mathrm{d}u \\ &= \sum_{n=0}^{\kappa-1} \int_{0}^{T} f(x) \, \mathrm{d}x \\ &=\kappa \int_{0}^{T} f(x) \, \mathrm{d}x \end{align*}

  2. We have successively:

        \begin{align*} \int_{\alpha + \mu \mathrm{T}}^{\beta + \nu \mathrm{T}} f(x) \, \mathrm{d}x - \int_{\alpha}^{\beta} f(x) \, \mathrm{d}x &= \int_{\alpha + \mu \mathrm{T}}^{\beta + \nu \mathrm{T}} f(x) \, \mathrm{d}x - \\ & - \left ( \int_{\alpha}^{\alpha + \mu T} f(x) \, \mathrm{d}x + \int_{\alpha +\mu T}^{\beta } f(x) \, \mathrm{d}x \right )\\ &=\int_{\beta }^{\beta + \nu \mathrm{T}} f(x) \, \mathrm{d}x - \int_{\alpha}^{\alpha + \mu T} f(x) \, \mathrm{d}x \\ &=\nu \int_{0}^{\mathrm{T}} f(x) \, \mathrm{d}x - \mu \int_{0}^{\mathrm{T}} f(x) \, \mathrm{d}t \\ &= \left ( \nu - \mu \right ) \int_{0}^{\mathrm{T}} f(x) \, \mathrm{d}x \end{align*}

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