The problem is to evaluate:
$\displaystyle\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty}e^{-(19x^2+19y^2+2xy)} \,dx\,dy$
Can anybody help me to solve this type problems? Thanks in advance.
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TheStudent
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MB17
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1$e^{a + b} \neq e^a + e^b$ – rims Jan 20 '20 at 05:22
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oh yeah.. I am sorry.. – MB17 Jan 20 '20 at 05:35
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1There is a formula in my book hope it will help $\displaystyle \int_{-\infty}^{\infty} \frac{e^{-1/2(\frac{x-\mu}{\sigma})^{2}}}{\sigma(\sqrt 2\pi)}=1$ – TheStudent Jan 20 '20 at 05:55
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1See also https://math.stackexchange.com/questions/3501943/solve-int-limits-infty-infty-int-limits-infty-inftye-5x2-5y2/3502067#3502067 – Quanto Jan 20 '20 at 06:20
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yeah.. got it.. Thank you.. – MB17 Jan 20 '20 at 08:31
2 Answers
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$$19(x^2+y^2)+2xy=\frac{19}{2}[(x+y)^2+(x-y)^2]+\frac{1}{2}[(x+y)^2-(x-y)^2]$$ The the give double integral is $$I=\frac{1}{2} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-10u^2} e^{-9v^2} du ~ dv =\frac{4}{2} \int_{0}^{\infty}e^{-10u^2} du \int_{0}^{\infty}e^{-9v^2} dv= 2 \sqrt{\frac{\pi}{40}} \frac{\sqrt{\pi}}{6} =\frac{ \pi}{6\sqrt{10}}. $$
Z Ahmed
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Thank you so much. But I guess it is 1/2 instead of 2 while changing the variables x,y to u,v. – MB17 Jan 20 '20 at 08:30
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But isnt it true that while changing variables we need to take the absolute value of reciprocal of the Jacobian? If so, then in your case, we have the Jacobian to be 2, therefore we have to take 1/2. – MB17 Jan 20 '20 at 08:37
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Since $Q(x,y)=19x^2+2xy+19y^2$ is a positive definite quadratic form we have $$\iint_{\mathbb{R}^2}\exp(-Q(x,y))\,dx\,dy = \frac{\pi}{\sqrt{19^2-1}}.$$
Jack D'Aurizio
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