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Question Number 54573 by Abdo msup. last updated on 07/Feb/19

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

Commented by maxmathsup by imad last updated on 08/Feb/19

$$letI={\int }_0^{\mathrm{\infty }}\frac{cos\left(e^{-x^2}\right)}{1+x^2}dx\Rightarrow I=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(e^{-x^2}\right)}{1+x^2}dx\Rightarrow$$$$2I=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{{ie}^{-x^2}}}{1+x^2}dx\right)letconsiderthecomplexfunction$$$$\phi \left(z\right)=\frac{e^{{ie}^{-x^2}}}{1+z^2}thepolesof\phi areiand-iresidustheoremgive$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,i)but\phi \left(z\right)=\frac{e^{{ie}^{-z^2}}}{(z-i)(z+i)}\Rightarrow$$$$Res(\phi ,i)=\frac{e^{ie}}{2i}\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \frac{e^{ie}}{2i}=\pi (cos\left(e\right)+isin\left(e\right))\Rightarrow$$$$2I=\pi cos\left(e\right)\Rightarrow I=\frac{\pi }{2}cos\left(e\right).$$

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