I try to find the following
$$\int_{\mathbb{R}}^{}\int_{\mathbb{R}}^{} e^{-y(x+z)-(x^2+z^2)} dxdz$$ and I change variables $x=r\cos(\theta)$ and $z=r\sin(\theta)$ and the integral becomes:
$$\int_{0}^{2\pi}\int_{0}^{+\infty} e^{r(-y\cos(\theta)-ysin(\theta)-1)}r drd\theta$$
Then I find the following
$$\int_{0}^{+\infty}{e^{-ax}x} dx = \left[-\frac{1}{a}e^{-ax}x-\frac{1}{a^2}e^{-ax}\right]_0^{+\infty} = \left[0-0-\left(0-\frac{1}{a^2}\right)\right]=\frac{1}{a^2} $$
now we can find the inner integral of the last double integral if we set $a=y\cos(\theta)+y\sin(\theta)+1$, so
$$\int_{0}^{+\infty} e^{r(-y\cos(\theta)-ysin(\theta)-1)}r dr = \frac{1}{(y(\cos(\theta)+sin(\theta))+1)^2}$$
and replace it to
$$\int_{0}^{2\pi}\int_{0}^{+\infty} e^{r(-y\cos(\theta)-ysin(\theta)-1)}r drd\theta=\int_{0}^{2\pi} \frac{1}{(y(\cos(\theta)+\sin(\theta))+1)^2} d\theta$$ after that I don't know what can I do. Any idea ?
Why do you need polar coordinates? It seems that the FT is separable:
$$\int_{-\infty}^{\infty} dx \, e^{-x^2-xy} \, \int_{-\infty}^{\infty} dz \, e^{-z^2-zy} = \left (\int_{-\infty}^{\infty} dx \, e^{-x^2-xy} \right )^2$$
Now,
$$\int_{-\infty}^{\infty} dx \, e^{-x^2-xy} = e^{y^2/4}\int_{-\infty}^{\infty} dx \, e^{-(x+y/2)^2} = \sqrt{\pi} \, e^{y^2/4}$$
Thus, the double integral is equal to $\pi \, e^{y^2/2}$.