Let $0 < \varepsilon < 1$, how to solve the integral: $$ \int_{\varepsilon}^1 \sin\left( \frac{1}{x} \right) dx $$
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Have you tried substituting 1/x as y and then evaluating it by parts? – user94300 Jun 26 '15 at 15:36
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@stefan I think this question asked before.. did you do enough research ? – Airbag Jun 26 '15 at 15:36
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@Airbag: I could not find it... – StefanH Jun 26 '15 at 15:38
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1http://math.stackexchange.com/questions/660425/integration-of-sin-frac1x – ParaH2 Jun 26 '15 at 15:39
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Maple says this here $-\sin \left( {{\it epsilon}}^{-1} \right) {\it epsilon}+{\it Ci} \left( {{\it epsilon}}^{-1} \right) +\sin \left( 1 \right) -{\it Ci} \left( 1 \right) $ – Dr. Sonnhard Graubner Jun 26 '15 at 15:44
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Let $t=x^{-1}$, then \begin{align*} \int_{\varepsilon}^1\sin\left(\frac{1}{x}\right)dx&=\int_{1/\varepsilon}^1(\sin t)(-t^{-2})dt\\ &=\sum_{k=1}^{\infty}\int_1^{1/\varepsilon}\frac{(-1)^{k-1}t^{2k-1-2}}{(2k-1)!}dt\\ &=\sum_{k=1}^{\infty}(-1)^{k-1}\left[\frac{(1/\varepsilon)^{2k-2}}{(2k-1)!(2k-2)}-\frac{1}{(2k-1)!(2k-2)}\right]dt \end{align*}
Ángel Mario Gallegos
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