Writing Fourier transform integral in terms of real-valued functions I am interested in the following integral.
$$ f(x,t) = \int \limits_{-\infty}^{\infty} \mathrm{d}k e^{-k^2}\left( e^{i(kx-pt-qk^2t)}+e^{i(kx+pt+qk^2t)} \right)$$
One way (for example, by using Mathematica) to solve it is in the following way.
$$ f(x,t) = \frac{\sqrt{\pi}}{2} \frac{(e^{-ipt}e^{\frac{ix^2}{-4i+4qt}}\sqrt{1-iqt}+e^{ipt}e^{\frac{-ix^2}{4i+4qt}}\sqrt{1+iqt})}{\sqrt{1+q^2t^2}} $$
I believe that $f(x,t)$ should be real valued, but it is not clear to me how to simplify given expression so that it consists only of real valued functions.

The two parts of your answer are obviously a complex conjugate pair.
Let's just take one of the terms and double its real part:
$$\begin{align*}f(x,t) &= \frac{\sqrt{\pi}}{2} \frac{(e^{-ipt}e^{\frac{ix^2}{-4i+4qt}}\sqrt{1-iqt}+e^{ipt}e^{\frac{-ix^2}{4i+4qt}}\sqrt{1+iqt})}{\sqrt{1+q^2t^2}}\\
\\
&= \sqrt{\pi}\;\Re\left[\frac{e^{-ipt}e^{-\frac{x^2}{4}\frac{1}{1+iqt}}}{\sqrt{1+iqt}}\right]\\
\\
&= \dfrac{\sqrt{\pi}}{\sqrt{1+q^2t^2}}\;\Re\left[\sqrt{1-iqt}\cdot e^{-ipt}e^{-\frac{x^2}{4}\frac{1-iqt}{1+q^2t^2}}\right] \\
\\
&= \dfrac{\sqrt{\pi}}{\sqrt{1+q^2t^2}}e^{-\frac{x^2}{4}\frac{1}{1+q^2t^2}}\;\Re\left[\sqrt{1-iqt}\cdot e^{-ipt}e^{-\frac{x^2}{4}\frac{-iqt}{1+q^2t^2}}\right] \\
\\
&= \dfrac{\sqrt{\pi}}{\sqrt{1+q^2t^2}}e^{-\frac{x^2}{4}\frac{1}{1+q^2t^2}}\;\Re\left[\sqrt{re^{i(\theta +2n\pi)}}\cdot e^{i\left(\frac{x^2}{4}\frac{qt}{1+q^2t^2}-pt\right)}\right] \\
\\
&= \dfrac{\sqrt{\pi}}{\sqrt{1+q^2t^2}}e^{-\frac{x^2}{4}\frac{1}{1+q^2t^2}}\;\Re\left[{\sqrt[4]{1+q^2t^2}e^{i(\theta/2 +n\pi)}}\cdot e^{i\left(\frac{x^2}{4}\frac{qt}{1+q^2t^2}-pt\right)}\right] \\
\\
&= \dfrac{\sqrt{\pi}}{\sqrt[4]{1+q^2t^2}}e^{-\frac{x^2}{4}\frac{1}{1+q^2t^2}}\;\Re\left[e^{i\left(\frac{\theta}{2}+n\pi+\frac{x^2}{4}\frac{qt}{1+q^2t^2}-pt\right)}\right] \\
\\
&= \dfrac{\sqrt{\pi}}{\sqrt[4]{1+q^2t^2}}e^{-\frac{x^2}{4}\frac{1}{1+q^2t^2}}\;\cos\left(\frac{\theta}{2}+n\pi+\frac{x^2}{4}\frac{qt}{1+q^2t^2}-pt\right) \\
\\
&= \dfrac{\sqrt{\pi}}{\sqrt[4]{1+q^2t^2}}e^{-\frac{x^2}{4}\frac{1}{1+q^2t^2}}\;\cos\left(\frac{\mathrm{arctan2}(-qt,1)}{2}+n\pi+\frac{x^2}{4}\frac{qt}{1+q^2t^2}-pt\right) \\
\\
\end{align*}$$
You can take $n=0$ for the principal value of the square root.  Or you can take $n=1$ and effectively just change the sign of the answer.
