Gaussian integrals over a half-space

Edit: I shall try to reformulate my question in order to make it -hopefully- more clear.

Let $X$ be a random variable that follows the $n$-dimensional Gaussian distribution. The probability density of $X$ is given by the following function:

$$ f_X(\mathbf{x;\mathbf{\mu}, \Sigma}) = \frac{1}{(2\pi)^{n/2}|\Sigma|^{1/2}} exp\{ -\frac{1}{2} (\mathbf{x}-\mathbf{\mu})^T \Sigma^{-1} (\mathbf{x}-\mathbf{\mu}) \}, $$

where $\mathbf{x},\mathbf{\mu}\in\Re^n$ and $\Sigma \in S_{++}^{n}$.

In addition, let $\mathcal{H}: \mathbf{x}^T\mathbf{w}=0$ be a hyperplane in the $n$-dimensional Euclidean space $\Re^n$, where $\mathbf{w}\in\Re^n$ and $b\in\Re$. The hyperplane $\mathcal{H}$ defines two half-spaces:

$$ \Omega_{+} = \{\mathbf{x} \in \Re^n | \mathbf{x}^T\mathbf{w}+b \geq 0 \}\\ \Omega_{-} = \{\mathbf{x} \in \Re^n | \mathbf{x}^T\mathbf{w}+b < 0 \} $$

What I would like to find out is what's happening when I try to integrate the above gaussian pdf over the region $\Omega_{+}$ (or $\Omega_{-}$), as it's well-known that -by definition- integrating over the whole $\Re^n$ gives 1. This is the spirit! And this is the reason of the existence if the (normalisation) term $\frac{1}{(2\pi)^{n/2}|\Sigma|^{1/2}}$.

Please correct me if I am wrong in something above.

Next, we could observe that if $b=0$, i.e., the hyperplane crosses the origin, then - due to the symmetry of the gaussian pdf - the integral should be equal to $1/2$. But how can I compute the value of the integral when $b\neq0$? Is there some clever trick to compute how the gaussian integral changes with respect to the bias term $b$?

Thanks in advance and apologies for being kinda "newbie" or not strict enough!... Every correction or helpful comment (either on my question or my notation) would be extremely appreciated!

To compute the integral

$$P = \frac{1}{(2\pi)^{n/2}\sqrt{\det \Sigma}} \int_{\Omega^+} \exp \left(-\frac12 (\mathbf{x}-\mu)^T \Sigma^{-1} (\mathbf{x}-\mu)\right)\,d\mathbf{x},$$

we will need several coordinate transforms. We start with a translation to get rid of the mean, $\mathbf{y} = \mathbf{x}-\mu$. Then $\Omega^+ = \{\mathbf{x} : \mathbf{x}^T\mathbf{w}+b \geqslant 0\}$ corresponds to $\Omega_1 = \{\mathbf{y} : \mathbf{y}^T\mathbf{w} + (\mu^T\mathbf{w} + b)\geqslant 0\}$, and we have

$$P = \frac{1}{(2\pi)^{n/2}\sqrt{\det \Sigma}} \int_{\Omega_1} \exp \left(-\frac12 \mathbf{y}^T \Sigma^{-1} \mathbf{y}\right)\,d\mathbf{y}.$$

Next, since $\Sigma$ is positive definite symmetric, there is an orthogonal matrix $U$ and a diagonal matrix $D$ with positive diagonal elements such that $\Sigma = U^TDU$. Then you have $\Sigma^{-1} = (U^TDU)^{-1} = U^{-1}D^{-1}(U^T)^{-1} = U^TD^{-1}U$ since $U^T = U^{-1}$. Then let $\mathbf{z} = U\mathbf{y}$, and $\mathbf{w}_1 = U\mathbf{w}$. With $\Omega_2 = \{\mathbf{z} : \mathbf{z}^T\mathbf{w}_1 + (\mu^T\mathbf{w}+b)\geqslant 0\}$, we obtain

$$P = \frac{1}{(2\pi)^{n/2}\sqrt{\det\Sigma}} \int_{\Omega_2} \exp \left(-\frac12 \mathbf{z}^T D^{-1} \mathbf{z} \right)\,d\mathbf{z},$$

since $\lvert \det U\rvert = 1$.

Now we do a bit of rescaling. Let $\mathbf{a} = \sqrt{D^{-1}}\mathbf{z}$, or $\mathbf{z} = \sqrt{D}\mathbf{a}$, $\mathbf{w}_2 = \sqrt{D}\mathbf{w}_1$, and $\Omega_3 = \{ \mathbf{a} : \mathbf{a}^T\mathbf{w}_2 + (\mu^T\mathbf{w}+b)\geqslant 0\}$. Since $\det \sqrt{D} = \sqrt{\det\Sigma}$, that yields

$$P = \frac{1}{(2\pi)^{n/2}} \int_{\Omega_3} \exp \left(-\frac12 \mathbf{a}^T\mathbf{a}\right)\,d\mathbf{a}.$$

Now find an orthogonal matrix $B$ such that $B\mathbf{w}_2 = \lVert\mathbf{w}_2\rVert\cdot e_n$, and let $\mathbf{b} = B\mathbf{a}$, $\Omega_4 = \{\mathbf{b} : \mathbf{b}^T(\lVert \mathbf{w}_2\rVert\cdot e_n) + (\mu^T\mathbf{w}+b)\geqslant 0\} = \{\mathbf{b} : \lVert\mathbf{w}_2\rVert b_n + (\mu^T\mathbf{w}+b)\geqslant 0\}$. That yields

$$P = \frac{1}{(2\pi)^{n/2}} \int_{\Omega_4} \exp\left(-\frac12 \mathbf{b}^T\mathbf{b}\right)\,d\mathbf{b}.$$

Now $\Omega_4 = \mathbf{R}^{n-1}\times [c,\infty)$, with $c = -\dfrac{(\mu^T\mathbf{w}+b)}{\lVert \mathbf{w}_2\rVert}$, and hence

$$P = \frac{1}{\sqrt{2\pi}} \int_c^\infty \exp\left(-\frac12 x^2\right)\,dx,$$

which can be evaluated in terms of the error function.