TITLE: Help evaluating integral (anything simple that I am missing?)
QUESTION [5 upvotes]: This integral is [hopefully] between 0 and 1 as it is supposed to represent a probability.
$$\int_{-\infty}^{\infty} \int_{-2-x}^{2-x} \frac{1}{2\pi} e^{\frac{-x^2-y^2}{2}} dy dx$$
I just wanted to check if anyone saw any easy-ish method to evaluate it that I could be missing. The only idea I can come up with is a few messy steps long:

Express the integrand as a series
Evaluate the inner integral to get two series.
Evaluate the outer improper integral by integrating and finding the limit at both positive and negative infinity.

I don’t even know if the 3rd step is doable, so I am hesitant to do the work leading up to it.

REPLY [9 votes]: Change the variables (rotation by $\pi/4$)
$$
\begin{cases}
u&=&\frac{1}{\sqrt2}(x-y),\\
v&=&\frac{1}{\sqrt2}(x+y)
\end{cases}
$$
to get
$$
\frac{1}{2\pi}\int_{-\infty}^{+\infty}\int_{-\sqrt2}^{\sqrt2}e^{-\frac{u^2+v^2}{2}}\,dvdu=\frac{1}{2\pi}\int_{-\infty}^{+\infty}e^{-u^2/2}\,du\int_{-\sqrt2}^{\sqrt2}e^{-v^2/2}\,dv=\frac{1}{\sqrt\pi}\int_{-1}^1e^{-s^2}\,ds=\color{red}{\operatorname{erf}(1)}.
$$
{"set_name": "stack_exchange", "score": 5, "question_id": 3319765}