Let $0\neq\mathbf{w}\in\mathbb{R}^n$ and denote by $H:\mathbb{R}\rightarrow\mathbb{R}$ the Heaviside step function. I need to calculate the Fourier transform of:
$f(\mathbf{x})=H(\langle \mathbf{w},\mathbf{x}\rangle)\cdot e^{-\frac{1}{2}\lVert\mathbf{x}\rVert^2} $
Namely, the Heaviside composed with inner product in the direction of $\mathbf{w}$ w.r.t a Gaussian weight function. I am specifically interested in the effect of the direction $\mathbf{w}$ on the transform.
My initial idea was to calculate the Fourier transform of each function separately, and then calculate the product as the convolution of those transforms. I know how to calculate the Fourier transform of a Gaussian, my problem is that the Heaviside function has a Fourier transform only in the sense of a tempered distribution. On the other hand, $f(\mathbf{x})\in L_2$ means that it should have a Fourier transform in the standard sense.
Because your Gaussian is rotation-invariant (isotropic) and so is the Fourier transform, we may as well assume that $\mathbf{w}=(1, 0,\dots,0)^T$. For the general case, just rotate (and maybe rescale) to align with that axis (see at the end of this answer).
So let's assume that $\mathbf{w}=(1, 0, \dots, 0)^T$. Now, for $\mathbf{x}$ and $\boldsymbol{\omega}$ both in $\mathbb R^n$, write $\mathbf{x}=(x_1, \mathbf{y})$ and $\omega = (\omega_1, \boldsymbol{\xi})$ with $\mathbf{y}$ and $\boldsymbol{\xi}$ both in $\mathbb R^{n-1}$.
$$\begin{split} \int_{\mathbb R^n}H(\langle \mathbf{w}, \mathbf{x}\rangle)e^{-\frac12 \|\mathbf{x}\|^2}e^{-i\langle \boldsymbol{\omega}, \mathbf{x}\rangle}d\mathbf{x} &= \int_{\mathbb R^{n-1}}\int_0^{+\infty}e^{-\frac 12\|\mathbf{x}\|^2}e^{-i\langle \boldsymbol{\omega}, \mathbf{x}\rangle}dx_1\,d\mathbf{y}\\ &=\left(\int_{\mathbb R^{n-1}}e^{-\frac 12\|\mathbf{y}\|^2}e^{-i\langle \boldsymbol{\xi}, \mathbf{y}\rangle}d\mathbf{y}\right)\left(\int_0^{+\infty}e^{-\frac 1 2x_1^2}e^{-i\omega_1 x_1}dx_1\right)\\ &=\left( 2\pi\right)^{\frac {n-1} 2} e^{-\frac 1 2\|\boldsymbol{\xi}\|^2}\int_0^{+\infty}e^{-\frac 1 2x_1^2}e^{-i\omega_1 x_1}dx_1& (1) \end{split}$$ So we're down to evaluating $$f(\omega_1)=\int_0^{+\infty}e^{-\frac 1 2x_1^2}e^{-i\omega_1 x_1}dx_1$$ Now $$f^\prime(\omega_1)=-i\int_0^{+\infty}x_1e^{-\frac 1 2x_1^2}e^{-i\omega_1 x_1}dx_1=i\left[ e^{-\frac 1 2x_1^2}e^{-i\omega_1 x_1}\right]_0^{+\infty}-\omega_1\int_0^{+\infty}e^{-\frac 1 2x_1^2}e^{-i\omega_1 x_1}dx_1$$ Thus $$f^\prime(\omega_1) +\omega_1f(\omega_1)=-i$$ This O.D.E. has solutions of the form $$f(\omega_1)=ce^{-\frac 1 2 \omega_1^2}-i \sqrt{\frac{\pi}2}e^{-\frac 1 2 \omega_1^2}\text{erfi}\left(\frac {\omega_1}{\sqrt 2}\right)$$ See this for the definition of the erfi function. Since $f(0)=\sqrt {\frac{\pi}2}$, we get $$f(\omega_1)=\sqrt{\frac{\pi}2} e^{-\frac 1 2 \omega_1^2}\left [ 1 -i\text{erfi}\left(\frac {\omega_1}{\sqrt 2}\right)\right]\tag{2}$$ Putting all the pieces $(1)$ and $(2)$ together:
$$\boxed{ \int_{\mathbb R^n}H(\langle \mathbf{w}, \mathbf{x}\rangle)e^{-\frac12 \|\mathbf{x}\|^2}e^{-i\langle \boldsymbol{\omega}, \mathbf{x}\rangle}d\mathbf{x}=\frac{\left( 2\pi\right)^{\frac {n} 2}}{2} e^{-\frac 1 2\|\boldsymbol{\omega}\|^2}\left [ 1 -i\text{erfi}\left(\frac {\omega_1}{\sqrt 2}\right)\right] }$$ Now for the general case, where you don't assume that $\mathbf{w}$ is aligned with the unit vector of the first axis, but rather is an arbitrary vector, then you can find a rotation $Q$ such that $$\|\mathbf{w}\|Q(1, 0, \dots, 0)^T = \mathbf{w}$$ You can check that using the result above with $Q\frac{\mathbf{w}}{\|\mathbf{w}\|}$ instead of $\mathbf{w}$ yields $$\boxed{ \int_{\mathbb R^n}H(\langle \mathbf{w}, \mathbf{x}\rangle)e^{-\frac12 \|\mathbf{x}\|^2}e^{-i\langle \boldsymbol{\omega}, \mathbf{x}\rangle}d\mathbf{x}=\frac{\left( 2\pi\right)^{\frac {n} 2}}{2} e^{-\frac 1 2\|\boldsymbol{\omega}\|^2}\left [ 1 -i\text{erfi}\left(\frac {\langle \boldsymbol{\omega}, \mathbf{w}\rangle}{\sqrt{2}\|\mathbf{w}\|}\right)\right] }$$ Note that setting $\boldsymbol{\omega}=0$ returns the right result $\frac 1 2 (2\pi)^{\frac n 2}$, which is half of the Gaussian's integral over all of $\mathbb R^n$.
Finally, note that the same technique can be applied if the Gaussian is not isotropic, i.e. if it has a covariance matrix that is not diagonal. A similar result can be obtained that involves that matrix.