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Question Number 60881 by aliesam last updated on 26/May/19

$$\underset{-\pi }{\stackrel{\pi }{\int }}sin\left(\frac{1}{1-x^2}\right)dx$$

Commented by MJS last updated on 26/May/19

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{we}\mathrm{can}\mathrm{solve}\mathrm{this},\mathrm{not}\mathrm{even}$$$$\mathrm{approximate}.{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{undefined}\mathrm{at}x=\pm 1\mathrm{and}$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{oszillating}\mathrm{very}\mathrm{fast}\mathrm{around}\mathrm{these}\mathrm{values}$$$$\mathrm{of}x$$$$\underset{-\pi }{\stackrel{\pi }{\int }}\sin \frac{1}{1-x^2}dx=2\underset{0}{\stackrel{\pi }{\int }}\sin \frac{1}{1-x^2}dx$$

Commented by aliesam last updated on 26/May/19

$$yes{that}^{\prime }srightandiposteditbecauseitisimproperintegrals$$

Commented by maxmathsup by imad last updated on 27/May/19

$$letI={\int }_{-\pi }^{\pi }sin\left(\frac{1}{1-x^2}\right)dx\Rightarrow 2I={\int }_0^{\pi }sin\left(\frac{1}{1-x^2}\right)dx$$$$={\int }_0^{1}sin\left(\frac{1}{1-x^2}\right)dx+{\int }_1^{\pi }sin\left(\frac{1}{1-x^2}\right)dx=H+K$$$$H{=}_{x=sin\theta }{\int }_0^{\frac{\pi }{2}}sin\left(\frac{1}{{cos}^2\theta }\right)cos\theta d\theta wehavex-\frac{x^3}{6}⩽sinx⩽x\Rightarrow$$$$\frac{1}{{cos}^2\theta }-\frac{1}{6{cos}^6\theta }⩽sin\left(\frac{1}{{cos}^2\theta }\right)⩽\frac{1}{{cos}^2\theta }\Rightarrow {\int }_0^{\frac{\pi }{2}}cos\theta sin\left(\frac{1}{{cos}^2\theta }\right)d\theta$$$$⩾{\int }_0^{\frac{\pi }{2}}\frac{d\theta }{cos\theta }-\frac{1}{6}{\int }_0^{\frac{\pi }{2}}\frac{d\theta }{{cos}^5\theta }lettake{\int }_0^{\frac{\pi }{2}}\frac{d\theta }{cos\theta }$$$${\int }_0^{\frac{\pi }{2}}\frac{d\theta }{cos\theta }{=}_{tan\left(\frac{\theta }{2}\right)=u}{\int }_0^1\frac{2du}{(1+u^2)\frac{1-u^2}{1+u^2}}={\int }_0^1\frac{2du}{1-u^2}$$$$={\int }_0^1(\frac{1}{1+u}+\frac{1}{1-u})du={[ln\mid \frac{1+u}{1-u}\mid ]}_0^1=\mathrm{\infty }sothisintegraldiverge...dont$$$$wastetimetofindit...!$$

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