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A trigonometric limit

Evaluate the limit:

    \[\ell = \lim_{n \rightarrow +\infty} \int_0^1 \frac{|\sin nx|}{x^2+1}\, \mathrm{d}x\]


Let us begin with the Fourier series of |\sin x | which is of the form:

    \[|\sin x | = \frac{2}{\pi} + \sum_{n=1}^{\infty} a_n \cos nx\]


(1)   \begin{equation*} |\sin n x | = \frac{2}{\pi} + \sum_{m=1}^{\infty} a_m \cos n m x  \end{equation*}

Integrating ( 1 ) we get that:

    \begin{align*} \ell &=\lim_{n \rightarrow +\infty} \int_{0}^{1} \frac{|\sin n x|}{1+x^2} \, \mathrm{d}x \\ &= \lim_{n \rightarrow +\infty} \int_{0}^{1} \frac{1}{x^2+1} \left ( \frac{2}{\pi} + \sum_{m=1}^{\infty} a_m \cos nm x \right ) \, \mathrm{d}x\\ &=\lim_{n \rightarrow +\infty} \left ( \frac{2}{\pi} \int_{0}^{1} \frac{\mathrm{d}x}{x^2+1} + \int_{0}^{1} \frac{1}{x^2+1} \sum_{m=1}^{\infty} a_m \cos nm x \, \mathrm{d}x \right ) \\ &= \frac{1}{2} + \lim_{n \rightarrow +\infty } \sum_{m=1}^{\infty} a_m  \int_{0}^{1}  \frac{\cos nmx}{x^2+1} \, \mathrm{d}x \end{align*}

The Riemann – Lebesgue Lemma implies that

    \[\lim_{n \rightarrow +\infty} \int_{0}^{1}  \frac{\cos nmx}{x^2+1} \, \mathrm{d}x  =0\]

and thus \ell=\frac{1}{2}.

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