integral calculus problem involving infinity problem 
The value of the integral $\displaystyle \int_{0}^{\infty}\int_{x}^{\infty} \frac{e^{-y}}{y} \, dydx$ is?
The value of the integral $\displaystyle \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(3x^2+2\sqrt{2}xy + 3y^2)} \, dxdy$

I couldn't figure out the trick here. Looks like both of them are similar. Please help if anyone.

Original images:



Here are solutions.
Solution of 1. Apply the Fubini's theorem to interchange the order of integration. The domain of integration is specified by the inequality $0 \leq x \leq y$, which tells you that
$$ \int_{0}^{\infty}\int_{x}^{\infty} \frac{e^{-y}}{y} \, dydx
= \int_{0}^{\infty}\int_{0}^{y} \frac{e^{-y}}{y} \, dxdy
= \int_{0}^{\infty} e^{-y} \, dy
= 1. $$

Solution of 2. There are severay ways of solving this. Let me first show you a solution using the gaussian integral:
$$ \int_{-\infty}^{\infty} e^{-ax^2} \, dx = \sqrt{\frac{\pi}{a}}, \qquad a > 0. \tag{*} $$
Completing the square, you have
$$
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(3x^2+2\sqrt{2}xy + 3y^2)} \, dxdy
= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-3(x+\frac{\sqrt{2}}{3}y)^2 - \frac{7}{3}y^2} \, dxdy. $$
Integrating w.r.t. $x$ first, we may apply the substitution $x+\frac{\sqrt{2}}{3}y = \tilde{x}$. Then
\begin{align*}
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(3x^2+2\sqrt{2}xy + 3y^2)} \, dxdy
&= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-3\tilde{x}^2 - \frac{7}{3}y^2} \, dxdy \\
&= \sqrt{\frac{\pi}{3}} \cdot \sqrt{\frac{3\pi}{7}} \\
&= \frac{\pi}{\sqrt{7}}.
\end{align*}

Solution of 2, ver.2. Let me give an alternative answer which does not require the knowledge on $\text{(*)}$. Using the polar coordinates,
\begin{align*}
&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(3x^2+2\sqrt{2}xy + 3y^2)} \, dxdy \\
&= \int_{-\pi}^{\pi}\int_{0}^{\infty} e^{-r^2(3\cos^2 \theta + 2\sqrt{2}\cos\theta\sin\theta + 3\sin^2 \theta)} \, rdrd\theta \\
&= \frac{1}{2} \int_{-\pi}^{\pi} \frac{1}{3\cos^2 \theta + 2\sqrt{2}\cos\theta\sin\theta + 3\sin^2 \theta} \, d\theta \\
&= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sec^2\theta}{3+ 2\sqrt{2}\tan\theta + 3\tan^2 \theta} \, d\theta \\
&= \int_{-\infty}^{\infty} \frac{dt}{3+ 2\sqrt{2}t + 3t^2}. \qquad (t = \tan\theta)
\end{align*}
This integral can be evaluated multiple ways.
