Fourier transform. Recall the formula
\[
\int_{\mathbb{R}}\mathrm{e}^{\mathrm{i}tx}\tfrac{1}{\sqrt{2\pi\sigma^2}}\mathrm{e}^{-\frac{x^2}{2\sigma^2}}\mathrm{d}x
=\mathrm{e}^{-\frac{\sigma^2 t^2}{2}}.
\] A good way to remember that the dispersion parameter $\sigma^2$ is in the numerator in the right hand side is to have in mind the uncertainty principle : if the signal is localized then its Fourier transform is not, and vice versa. In case of doubt regarding the sign inside the right hand side, remember that the standard Gaussian is an eigenfunction of the Fourier transform.
Moments. The simplest way to compute the even moments of the standard Gaussian is neither integration by parts nor Fourier transform : just take the $n$-th derivative with respect to $\beta$ of
\[
\int_{\mathbb{R}}\mathrm{e}^{-\beta x^2}\mathrm{d}x=\sqrt{\frac{\pi}{\beta}}
\] which gives
\[
(-1)^n\int_{\mathbb{R}}x^{2n}\mathrm{e}^{-\beta x^2}\mathrm{d}x
=(-1)^n\sqrt{\frac{\pi}{\beta^{2n+1}}}\prod_{k=1}^{n}\Bigr(\frac{1}{2}+k-1\Bigr),
\] in other words, with $\beta=\frac{1}{2\sigma^2}$,
\[
\int_{\mathbb{R}} x^{2n}\tfrac{1}{\sqrt{2\pi\sigma^2}}\mathrm{e}^{-\frac{x^2}{2\sigma^2}}\mathrm{d}x
=\sigma^{2n}\prod_{k=1}^n(2k-1)=\sigma^{2n}(2n-1)!!.
\] This trick also works for the Gamma distribution and is an instance of the energy trick in statistical physics : in the same spirit, if $\lambda$ is some reference measure on some space and
\[
Z:=\int\mathrm{e}^{-\beta V(x)}\mathrm{d}\lambda(x),
\]then, denoting $X$ a random variable with density $\tfrac{1}{Z} \mathrm{e}^{-\beta V}$ with respect to $\lambda$,
\[
\partial_\beta\log Z=-\mathbb{E}(V(X))
\quad\text{while}\quad
\partial^2_\beta\log Z=\mathbb{Var}(V(X)).
\]