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Question Number 163349 by smallEinstein last updated on 06/Jan/22

Answered by Ar Brandon last updated on 06/Jan/22

$$I={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\left({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-(3x^2+2\sqrt{2}xy+3y^2)}dx\right)dy$$$$={\int }_0^{2\pi }{\int }_0^{\mathrm{\infty }}{re}^{-(3r^2+2\sqrt{2}{r}^2\sin \vartheta \cos \vartheta )}drd\vartheta$$$$={\int }_0^{2\pi }{\int }_0^{\mathrm{\infty }}{re}^{-r^2(3+\sqrt{2}\sin 2\vartheta )}drd\vartheta$$$$u=r^2(3+\sqrt{2}\sin 2\vartheta )\Rightarrow du=2r(3+\sqrt{2}\sin 2\vartheta )dr$$$$I={\int }_0^{2\pi }{\int }_0^{\mathrm{\infty }}\frac{e^{-u}}{2(3+\sqrt{2}\sin 2\vartheta )}dud\vartheta$$$$=\frac{1}{2}{\int }_0^{2\pi }\frac{d\vartheta }{3+\sqrt{2}\sin 2\vartheta }={\int }_{-\pi }^{\pi }\frac{{\sec }^2\vartheta }{{3\sec }^2\vartheta +2\sqrt{2}\tan \vartheta }d\vartheta$$$$={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dt}{3t^2+2\sqrt{2}t+3}=\frac{1}{3}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dt}{{(t+\frac{\sqrt{2}}{3})}^2+\frac{7}{9}}$$$$=\frac{1}{\sqrt{7}}{\left[\arctan \left(\frac{3t+\sqrt{2}}{\sqrt{7}}\right)\right]}_{-\mathrm{\infty }}^{\mathrm{\infty }}=\frac{\pi }{\sqrt{7}}$$

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