I'm doing a test from previous years of my school and one of the problems goes as follows;
Let $(X,Y)$ be a couple of random variables with density $f_{X,Y}$ defined as :
$$\forall (x,y)\in\mathbb{R}^2,\ f_{X,Y}(x,y)=\frac{1}{2\pi} \exp^{-\frac{x^2}{2}+xy+y^2}$$
First question is to check that $f_{X,Y}$ is indeed a density function. So we need $$\iint_{\mathbb{R}^2} f_{X,Y}(x,y)\mathrm{d}x \mathrm{d}y = 1$$
My problem is that I find that this double integrale diverges. Indeed, noting that $-\frac{x^2}{2}+xy+y^2 = -\frac{1}{2}(x-y)^2 + \frac{3y^2}{2}$: \begin{align*} &\iint_{\mathbb{R}^2} f_{X,Y}(x,y)\mathrm{d}x \mathrm{d}y &\\ &= \frac{1}{2\pi} \iint_{\mathbb{R}^2} \exp^{-\frac{x^2}{2}+xy+y^2}\mathrm{d}x \mathrm{d}y&\\ &= \frac{1}{2\pi} \iint_{\mathbb{R}^2} \exp^{-\frac{1}{2}(x-y)^2 + \frac{3y^2}{2}}\mathrm{d}x \mathrm{d}y&\\ &= \frac{1}{2\pi} \int_{\mathbb{R}} \exp^{\frac{3y^2}{2}}\int_{\mathbb{R}} \exp^{-\frac{1}{2}(x-y)^2}\mathrm{d}x \mathrm{d}y&(1)\\ &= \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \exp^{\frac{3y^2}{2}} \mathrm{d}y. & (2) \end{align*}
At (1) we recognize the density function of a normal distribution $\mathcal{N}(y,1)$ and at (2) we have a diverging integrale.
Am I doing something wrong or is the problem statement incorrect?