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On an integral of a periodic function

Let f:\mathbb{R} \rightarrow \mathbb{R} be a function such that f(x + \pi ) = f(x -\pi) = f(x) for all x \in \mathbb{R}. Prove that

    \[\int_{-\infty}^{\infty} f(x) \frac{\sin^2 x}{x^2} \, \mathrm{d}x = \int_{0}^{\pi} f(x) \, \mathrm{d}x\]

Solution

We have successively:

    \begin{align*} \int_{-\infty}^{\infty} f(x) \frac{\sin^2 x}{x^2}\, \mathrm{d}x &= \sum_{k=-\infty}^{\infty} \int_{k \pi}^{(k+1) \pi} f(x) \frac{\sin^2 x}{x^2}\, \mathrm{d}x \\ &= \sum_{k=-\infty}^{\infty} \int_{0}^{\pi} f \left ( t + k \pi \right ) \frac{\sin^2 \left ( t + k \pi \right )}{\left ( t + k \pi \right )^2} \, \mathrm{d}t \\ &= \sum_{k=-\infty}^{\infty} \int_{0}^{\pi} f(t) \frac{(-1)^{2k} \sin^2 t}{\left ( t + k \pi \right )^2} \, \mathrm{d}t \\ &=\int_{0}^{\pi} f(t) \sin^2 t \sum_{k=-\infty}^{\infty} \frac{1}{\left ( t + k \pi \right )^2} \, \mathrm{d}t \\ &= \int_{0}^{\pi} f(t) \, \mathrm{d}t \end{align*}

since

    \[\sum_{k=-\infty}^{\infty} \frac{1}{\left ( t + k \pi \right )^2} = \csc^2 t\]

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