So, I tried to find the Fourier transform of $e^{-2\pi t^2}\cos(2\pi t)$, and now I am stuck.
So far, I have gotten to $$\frac{1}{2} \left[\int_{-\infty}^\infty e^{-2\pi(t+\frac{i}{2})^2 +\frac{\pi}{2}}e^{-i2\pi st}dt+\int_{-\infty}^\infty e^{-2\pi(t-\frac{i}{2})^2 +\frac{\pi}{2}}e^{-i2\pi st} \, dt\right],$$ but I do not know as to what to do next.
How do I solve for the rest of this?
It is faster to exploit the convolution theorem (the Fourier transform of a product is a convolution of Fourier transforms) since the Fourier transform of a cosine is a sum of two Dirac deltas and $e^{-2\pi t^2}$ is a fixed point of the Fourier transform (or almost a fixed point, depending on the chosen normalization). The final outcome is
$$ \int_{-\infty}^{+\infty}e^{-2\pi t^2}\cos(2\pi t)e^{-2\pi i t x}\,dt = \color{red}{ \frac{1+e^{2\pi x}}{2\sqrt{2}} e^{-\frac{\pi}{2}(x+1)^2}}. $$