Question Number 163349 by smallEinstein last updated on 06/Jan/22
Answered by Ar Brandon last updated on 06/Jan/22
![I=∫_(−∞) ^∞ (∫_(−∞) ^∞ e^(−(3x^2 +2(√2)xy+3y^2 )) dx)dy =∫_0 ^(2π) ∫_0 ^∞ re^(−(3r^2 +2(√2)r^2 sinϑcosϑ)) drdϑ =∫_0 ^(2π) ∫_0 ^∞ re^(−r^2 (3+(√2)sin2ϑ)) drdϑ u=r^2 (3+(√2)sin2ϑ)⇒du=2r(3+(√2)sin2ϑ)dr I=∫_0 ^(2π) ∫_0 ^∞ (e^(−u) /(2(3+(√2)sin2ϑ)))dudϑ =(1/2)∫_0 ^(2π) (dϑ/(3+(√2)sin2ϑ))=∫_(−π) ^π ((sec^2 ϑ)/(3sec^2 ϑ+2(√2)tanϑ))dϑ =∫_(−∞) ^∞ (dt/(3t^2 +2(√2)t+3))=(1/3)∫_(−∞) ^∞ (dt/((t+((√2)/3))^2 +(7/9))) =(1/( (√7)))[arctan(((3t+(√2))/( (√7))))]_(−∞) ^∞ =(π/( (√7)))](Q163365.png)
$${I}=\int_{−\infty} ^{\infty} \left(\int_{−\infty} ^{\infty} {e}^{−\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}{xy}+\mathrm{3}{y}^{\mathrm{2}} \right)} {dx}\right){dy} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \int_{\mathrm{0}} ^{\infty} {re}^{−\left(\mathrm{3}{r}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}{r}^{\mathrm{2}} \mathrm{sin}\vartheta\mathrm{cos}\vartheta\right)} {drd}\vartheta \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \int_{\mathrm{0}} ^{\infty} {re}^{−{r}^{\mathrm{2}} \left(\mathrm{3}+\sqrt{\mathrm{2}}\mathrm{sin2}\vartheta\right)} {drd}\vartheta \\ $$$${u}={r}^{\mathrm{2}} \left(\mathrm{3}+\sqrt{\mathrm{2}}\mathrm{sin2}\vartheta\right)\Rightarrow{du}=\mathrm{2}{r}\left(\mathrm{3}+\sqrt{\mathrm{2}}\mathrm{sin2}\vartheta\right){dr} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{u}} }{\mathrm{2}\left(\mathrm{3}+\sqrt{\mathrm{2}}\mathrm{sin2}\vartheta\right)}{dud}\vartheta \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{d}\vartheta}{\mathrm{3}+\sqrt{\mathrm{2}}\mathrm{sin2}\vartheta}=\int_{−\pi} ^{\pi} \frac{\mathrm{sec}^{\mathrm{2}} \vartheta}{\mathrm{3sec}^{\mathrm{2}} \vartheta+\mathrm{2}\sqrt{\mathrm{2}}\mathrm{tan}\vartheta}{d}\vartheta \\ $$$$\:\:\:=\int_{−\infty} ^{\infty} \frac{{dt}}{\mathrm{3}{t}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}{t}+\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}}\int_{−\infty} ^{\infty} \frac{{dt}}{\left({t}+\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}\right)^{\mathrm{2}} +\frac{\mathrm{7}}{\mathrm{9}}} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}\left[\mathrm{arctan}\left(\frac{\mathrm{3}{t}+\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{7}}}\right)\right]_{−\infty} ^{\infty} =\frac{\pi}{\:\sqrt{\mathrm{7}}} \\ $$