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Question Number 60783 by aliesam last updated on 25/May/19

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

Commented by maxmathsup by imad last updated on 25/May/19

$$letA={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}sin\left(\frac{1}{1+x^2}\right)dx\Rightarrow A=2{\int }_0^{\mathrm{\infty }}sin\left(\frac{1}{1+x^2}\right)dx{=}_{x=tan\theta }{\int }_0^{\frac{\pi }{2}}sin\left({cos}^2\theta \right)(1+{tan}^2\theta )d\theta$$$$={\int }_0^{\frac{\pi }{2}}\frac{sin\left({cos}^2\theta \right)}{{cos}^2\theta }d\theta wehavex-\frac{x^3}{6}⩽sinx⩽xfor0<x<\frac{\pi }{2}\Rightarrow$$$${cos}^2\theta -\frac{1}{6}{cos}^{6}x⩽sin\left({cos}^{2}x\right)⩽{cos}^{2}x\Rightarrow 1-\frac{1}{6}{cos}^{4}x⩽\frac{sin\left({cos}^{2}x\right)}{{cos}^{2}x}⩽1\Rightarrow$$$${\int }_0^{\frac{\pi }{2}}(1-\frac{1}{6}{cos}^{4}x)dx⩽{\int }_0^{\frac{\pi }{2}}\frac{sin\left({cos}^{2}x\right)}{{cos}^{2}x}dx⩽\frac{\pi }{2}\Rightarrow$$$$\frac{\pi }{2}-\frac{1}{6}{\int }_0^{\frac{\pi }{2}}{cos}^{4}xdx⩽A⩽\frac{\pi }{2}$$$${\int }_0^{\frac{\pi }{2}}{cos}^{4}xdx={\int }_0^{\frac{\pi }{2}}{\left(\frac{1+cos\left(2x\right)}{2}\right)}^{2}dx=\frac{1}{4}{\int }_0^{\frac{\pi }{2}}(1+2cos\left(2x\right)+{cos}^2\left(2x\right))dx$$$$=\frac{\pi }{8}+\frac{1}{2}{\int }_0^{\frac{\pi }{2}}cos\left(2x\right)dx+\frac{1}{8}{\int }_0^{\frac{\pi }{2}}(1+cos\left(4x\right))dx$$$$=\frac{\pi }{8}+0+\frac{\pi }{16}+9=\frac{3\pi }{16}\Rightarrow \frac{\pi }{2}-\frac{\pi }{32}⩽A⩽\frac{\pi }{2}$$$$wecantake{\alpha }_0=\frac{\pi -\frac{\pi }{32}}{2}=\frac{31\pi }{64}asapproximatevalueforthisintegra$$$${\alpha }_0\sim 1,52.$$$$$$$$$$

Commented by maxmathsup by imad last updated on 25/May/19

$$A=2{\int }_0^{\mathrm{\infty }}sin\left(\frac{1}{1+x^2}\right)dx\Rightarrow \pi -\frac{\pi }{16}⩽A⩽\pi \Rightarrow {\alpha }_0=\frac{31\pi }{32}\Rightarrow$$$${\alpha }_0\sim 3,04.$$

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