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Question Number 161919 by talminator2856791 last updated on 24/Dec/21

$$$$$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\sin (x^2+x+1)dx$$$$$$

Answered by Lordose last updated on 24/Dec/21

$$$$$$\mathrm{\Omega }={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\sin ({\mathrm{x}}^2+\mathrm{x}+1)\mathrm{dx}$$$$\mathrm{\Omega }={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\sin ({(\mathrm{x}+\frac{1}{2})}^2+\frac{3}{4})\mathrm{dx}$$$$\mathrm{\Omega }=\cos \left(\frac{3}{4}\right){\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\sin \left({(\mathrm{x}+\frac{1}{2})}^2\right)\mathrm{dx}+\sin \left(\frac{3}{4}\right){\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\cos \left({(\mathrm{x}+\frac{1}{2})}^2\right)\mathrm{dx}$$$$\mathrm{\Omega }\stackrel{\mathrm{x}=\mathrm{x}+\frac{1}{2}}{=}\cos \left(\frac{3}{4}\right){\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\sin \left({\mathrm{x}}^2\right)\mathrm{dx}+\sin \left(\frac{3}{4}\right){\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\cos \left({\mathrm{x}}^2\right)\mathrm{dx}$$$$\mathrm{\Omega }=2\cos \left(\frac{3}{4}\right){\int }_0^{\mathrm{\infty }}\sin \left({\mathrm{x}}^2\right)\mathrm{dx}+2\sin \left(\frac{3}{4}\right){\int }_0^{\mathrm{\infty }}\cos \left({\mathrm{x}}^2\right)\mathrm{dx}$$$${\int }_0^{\mathrm{\infty }}\mathbf{sin}\left({\mathbf{x}}^2\right)\mathbf{dx}={\int }_0^{\mathrm{\infty }}\mathbf{cos}\left({\mathbf{x}}^2\right)\mathbf{dx}=\sqrt{\frac{\mathit{\pi }}{8}}$$$$\mathrm{\Omega }=2\cos \left(\frac{3}{4}\right)\cdot \sqrt{\frac{\mathit{\pi }}{8}}+2\sin \left(\frac{3}{4}\right)\sqrt{\frac{\mathit{\pi }}{8}}$$$$\mathrm{\Omega }=\sqrt{\frac{\mathit{\pi }}{2}}(\cos \left(\frac{3}{4}\right)+\sin \left(\frac{3}{4}\right))$$$$$$

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