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Question Number 48170 by Abdo msup. last updated on 20/Nov/18

$$calculate{\int }_0^{\mathrm{\infty }}\frac{cos\left(sin\left(x^2\right)\right)}{1+2x^2}dx$$

Commented by maxmathsup by imad last updated on 24/Nov/18

$$letI={\int }_0^{\mathrm{\infty }}\frac{cos\left(sin\left(x^2\right)\right)}{2x^2+1}dx\Rightarrow 2I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(sin\left(x^2\right)\right)}{2x^2+1}dx$$$$=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{isin\left(x^2\right)}}{2x^2+1}dx\right)let\phi \left(z\right)=\frac{e^{isin\left(z^2\right)}}{2z^2+1}\Rightarrow \phi \left(z\right)=\frac{e^{isin\left(z^2\right)}}{(\sqrt{2}z-i)(\sqrt{2}z+i)}=\frac{e^{{isinz}^2}}{2(z-\frac{i}{\sqrt{2}})(z+\frac{i}{\sqrt{2}})}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,\frac{i}{\sqrt{2}})but$$$$Res(\phi ,\frac{i}{\sqrt{2}})={lim}_{z\to \frac{i}{\sqrt{2}}}(z-\frac{i}{\sqrt{2}})\phi \left(z\right)=\frac{e^{isin(-\frac{1}{2})}}{4\frac{i}{\sqrt{2}}}=\frac{e^{-isin\left(\frac{1}{2}\right)}}{i\sqrt{2}}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \frac{e^{-isin\left(\frac{1}{2}\right)}}{4i}\sqrt{2}=\frac{\pi \sqrt{2}}{2}{e}^{-isin\left(\frac{1}{2}\right)}\Rightarrow 2I=\frac{\pi }{\sqrt{2}}cos\left(sin\left(\frac{1}{2}\right)\right)\Rightarrow$$$$I=\frac{\pi }{2\sqrt{2}}cos\left(sin\left(\frac{1}{2}\right)\right).$$

Answered by Abdulhafeez Abu qatada last updated on 24/Nov/18

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