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$$\frac{1}{2\pi}\int\limits_{0}^{\infty}\int\limits_{-x}^{\infty}e^{-\frac{(x^2+y^2)}{2}}dydx$$

Is there a particularly nice way of working this to an exact value? The -x on the limits of integration makes this a little different from how I am used to solving these.

I was told to look at differentiating it this way, but I didn't see where this was heading:

$$=\frac{1}{2\pi}\int_{0}^{\infty}e^{-x^2/2}\int_{-x}^{\infty}e^{-y^2/2}dydx$$ Let $g(x) =\int_{-x}^{\infty}e^{-y^2/2}dy$

$$=\frac{1}{2\pi}\int_{0}^{\infty}e^{-x^2/2}g(x)dx$$

Let $2\pi*F(t)=\int_{0}^{t}e^{-x^2/2}g(x)dx$.

Thus $2\pi*F'(t)=e^{-t^2/2}g(t)-g(0)=e^{-t^2/2}g(t)-\frac{\sqrt{2\pi}}{2}$

At this point I don't see which move to make.

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Use polar coordinates & the integral becomes \begin{eqnarray*} \int_0^{\frac{3 \pi}{4}} d \theta \int_0^{\infty} e^{r^2/2} r dr. \end{eqnarray*}

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Hint: The $x$ and $y$ axes, together with the lines $y=x$ and $y=-x$, divide the plane into $8$ regions. Now use the rotational symmetry of the integrand, together with the fact that the integral over the whole plane is $1$.

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