$(1)$ Compute the value of $$\frac{1}{2\pi} \int_{-\infty}^{\infty}\int_{-\infty}^{y} e^{-\frac{1}{2}(x^2+y^2)} dx dy $$
What I tried :$$I=\frac{1}{2\pi} \int_{-\infty}^{\infty}\int_{-\infty}^{y} e^{-\frac{1}{2}(x^2+y^2)} dx dy $$ $$=\frac{1}{2\pi} \int_{-\infty}^{\infty}(e^{\frac{-y^2}{2}}\int_{-\infty}^{y} e^{\frac{-x^2}{2}}dx) dy $$
Now, let $$t=\frac{x^2}{2}$$ $$\Rightarrow dt= x dx$$ $$\Rightarrow dx= x^{-1} dt$$ $$\Rightarrow dx= (2t)^{\frac{-1}{2}} dt$$ Again, $$x\in (-\infty,y)$$ $$\Rightarrow x^2\in (0,\infty)$$ So, $$I=\frac{1}{2\pi} \int_{-\infty}^{\infty}(e^{\frac{-y^2}{2}}\int_{-\infty}^{y} e^{\frac{-x^2}{2}}dx) dy $$ $$=\frac{1}{2\pi} \int_{-\infty}^{\infty}(e^{\frac{-y^2}{2}}\int_{0}^{\infty} e^{-t}(2t)^{\frac{-1}{2}}dt) dy $$ $$=\frac{1}{2^{\frac{1}{2}}}\frac{1}{2\pi} \int_{-\infty}^{\infty}(e^{\frac{-y^2}{2}}\int_{0}^{\infty} e^{-t}(t)^{\frac{-1}{2}}dt) dy $$ $$=\frac{1}{2^{\frac{1}{2}}}\frac{1}{2\pi} \int_{-\infty}^{\infty}e^{\frac{-y^2}{2}}(\Gamma{\frac{1}{2}}) dy $$ $$=\frac{\sqrt{\pi}}{\sqrt{2}}\frac{1}{2\pi} \int_{-\infty}^{\infty}e^{\frac{-y^2}{2}}dy $$ $$=\frac{\sqrt{\pi}}{\sqrt{2}}\frac{1}{\sqrt{2\pi}} (\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{\frac{-y^2}{2}}dy) $$ $$=\frac{\sqrt{\pi}}{\sqrt{2}}\frac{1}{\sqrt{2\pi}} \times 1 $$ (The integrand being a standard normal density)
$$I=\frac{1}{2}$$
Am I correct ?
Here is a quick way.
Observe that $ \int_{-\infty}^{\infty}\int_{-\infty}^{y}$ defines exactly half of the $(x,y)$-plane. Since the integrand depends on $x^2 + y^2 = r^2$ only, the result is $1/2$ of the integral over the full $(x,y)$-plane, and this can be favorably computed in spherical coordinates:
$$ \frac{1}{2\pi} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-\frac{1}{2}(x^2+y^2)} dx dy = \\ \frac{1}{2\pi} \int_{0}^{\infty} e^{-\frac{1}{2} r^2} r dr \int_{0}^{2 \pi}d\phi = \\ \int_{0}^{\infty} e^{-\frac{1}{2} r^2} d(\frac{r^2}{2}) = -e^{-\frac{1}{2} r^2} |_0^{\infty} = 1 $$ So, yes, your integral equals $1/2$.