Compute an integral related with gaussian

Question

Can anyone help me to solve this integral, I think we can try polar coordinate and some property of Gaussian density, but I stuck for long time. Also the WolframAlpha cannot compute this integral. $$ \int_{0}^{\infty} \int_{-2^{n / 2}x}^{\infty} \frac{1}{2 \pi\sqrt{(j-1)}} \exp \left(-\frac{x^{2}}{2(j-1)}-\frac{y^{2}}{2}\right) d y d x $$ Thanks a lot!

Answer 1

The inner integral is simple and we are left with $$2 \sqrt{2 \pi(j-1) }\,I=\int_0^\infty e^{-\frac{x^2}{2(j-1)}}\,dx+\int_0^\infty e^{-\frac{x^2}{2(j-1)}}\text{erf}\,\left(2^{\frac{n-1}{2}} x\right)\,dx$$ The first integral is simple $$\int_0^\infty e^{-\frac{x^2}{2(j-1)}}\,dx=\sqrt{\frac{\pi (j-1)}{2} }$$ For the second one let $2^{\frac{n-1}{2}} x=t$ and $k=\frac{2^{-n}}{j-1} $ to make it $$2^{-\frac {n-1}2}\int_0^\infty e^{-k t^2}\text{erf}(t)\,dt=\frac{2^{-\frac {n-1}2}}{\sqrt{\pi k} }\tan ^{-1}\left(\frac{1}{\sqrt{k}}\right)$$

Look here (formula $4.3.2$)

As a result

$$I=\frac 14+\frac 1 {2\pi} \tan ^{-1}\left(\sqrt{2^n(j-1)}\right)$$

Answer 2

If $Z$ and $Y$ are independent standard Gaussian random variables and $X=\sqrt{j-1}\cdot Z$, then the integral is $$P(X \ge 0, Y \ge -2^{n/2} X) = P(Z \ge 0, Y \ge -2^{n/2} \sqrt{j-1} \cdot Z)$$

If you plot the region of the $(z,y)$ plane defined by the constraints on $(Z,Y)$, you will see it is a sector whose boundaries are given by the lines $z=0$ and $y=-2^{n/2} \sqrt{j-1} \cdot z$. (I strongly encourage you to draw a picture of this region.) Because the distribution of $(Z,Y)$ is radially symmetric, the probability that $(Z,Y)$ lies in this sector is the angle of this sector divided by $2\pi$.