I have came across the following step, I suspect it is true but I am not sure how it is justified.
$$ I = \int_{-\infty}^{\infty} \frac{dk}{2\pi} \exp{\left(i k (x_0 - x_t) - \frac{k^2}{2} G\right)}$$ where $G$ is another function independent of $k$. The author claims that the integral over $dk$ is a Gaussian integral which when evaluated yields $$I = \frac{1}{\sqrt{2\pi G }}\exp{\left(-\frac{(x_0 - x_t)^2}{2G}\right)}$$
Thank you.
You complete the square in the exponent. That is, write the exponent as $$ \frac{-G}{2}\left( k^2 - \frac{2i(x_0 - x_t)}{G}k \pm \frac{(x_0 - x_t)^2}{G^2}\right) = -\frac{G}{2} \left( k - i\frac{(x_0 - x_t)}{G}\right)^2 - \frac{(x_0 - x_t)^2}{2G}.$$ So in the integral, you can perform the change of variables $k \mapsto k + i \frac{(x_0 - x_t)^2}{G}$ to reduce the integrand to a standard Gaussian integral multiplied by the additional piece $\exp\left( - \frac{(x_0 - x_t)^2}{2G}\right)$ visible as the extra term from completing the square in the exponent.
And this gives your integral. Does that make sense?