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A cot integral limit

Evaluate the limit:

    \[\ell = \lim_{x \rightarrow +\infty} \int_{\frac{1}{x+1}}^{\frac{1}{x}} \cot t^2 \, \mathrm{d}t\]

Solution

Recalling the Taylor expansion of \tan around x=0 we have that

    \[\cot x^2 = \frac{1}{x^2} + \mathcal{O}\left ( x^2 \right )\]

Thus,

    \begin{align*} \int_{\frac{1}{x+1}}^{\frac{1}{x}} \cot t^2 \, \mathrm{d}t &= \int_{\frac{1}{x+1}}^{\frac{1}{x}} \left ( \cot t^2 - \frac{1}{t^2} + \frac{1}{t^2} \right ) \, \mathrm{d}t \\ &=\int_{\frac{1}{x+1}}^{\frac{1}{x}} \left ( \cot t^2 - \frac{1}{t^2} \right ) \, \mathrm{d}t + \int_{\frac{1}{x+1}}^{\frac{1}{x}} \frac{\mathrm{d}t}{t^2} \\ &=\mathcal{O} \left ( \int_{\frac{1}{x+1}}^{\frac{1}{x}} t^2 \right ) + 1 \\ &= \mathcal{O} \left ( \frac{1}{x^3} - \frac{1}{(x+1)^3} \right ) +1 \\ &= 1 +\mathcal{O} \left ( \frac{1}{x^4} \right ) \end{align*}

The limit follows to be 1.

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