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Prove that the following integral: $$\int_0^{\infty} \int_0^{\infty} e^{-(x^2+y^2+2xy \cos \theta)} \,dx dy = \frac{\theta}{2\sin\theta}$$

The hints written on the book are beta function and to suppose the slanted $xy$ plane. However, I cannot figure out a thing.

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  • $\begingroup$ Have you tried polar coordinates ? $\endgroup$
    – Lucian
    Dec 28 '14 at 10:27
  • $\begingroup$ Suggestion: Complete the square in the argument of the exponential function: $x^2+y^2+2xy\cos\theta=(x+\cos\theta y)^2+(1-\cos^2\theta)y^2$. Then, it might be easier to do the change of variables... (note that $(1-\cos^2\theta)=\sin^2\theta$) $\endgroup$
    – mickep
    Dec 28 '14 at 10:31
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Assume $\theta\in\left[0,\frac{\pi}{2}\right]$. Then, by setting $z=x+y\cos\theta$ we have: $$ I = \int_{0}^{+\infty}\int_{y\cos\theta}^{+\infty} e^{-(z^2+y^2\sin^2\theta)}\, dz \,dy = \frac{1}{\sin\theta}\int_{0}^{+\infty}\int_{w\cot\theta}^{+\infty}e^{-(z^2+w^2)}\,dz\,dw.$$ Switching now to polar coordinates we get $I=\frac{\theta}{2\sin\theta}$ as stated.

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  • $\begingroup$ Where is beta function here, or was that a red herring? $\endgroup$ Dec 28 '14 at 11:01
  • $\begingroup$ @SuzuHirose: IMHO, it is not needed to use the Euler Beta function here, just a simple geometric argument. The last integral is the integral of $e^{-\rho^2}$ over an angle with vertex in the origin of $\mathbb{R}^2$, hence you just need to compute the amplitude of such angle. $\endgroup$ Dec 28 '14 at 12:18
  • $\begingroup$ @JackD'Aurizio why is it necessary for $\theta\in (0,\pi/2]$? $\endgroup$
    – take008
    Jul 20 '19 at 13:56
  • $\begingroup$ The stated identity does not hold unconditionally (for instance it does not hold if theta is positive but sin theta is negative), so I assumed a reasonable constraint. $\endgroup$ Jul 20 '19 at 15:59

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