What are the steps to derive the following inverse Fourier transformations I'm reading a text which is an introductory text on Fourier transforms. The author has two expressions:
$$ F(\omega_{o}) =  \frac{1}{\sigma \sqrt {2 \pi} } e^{\Large- \frac{{\omega_{o}}^2}{2\sigma^2}} $$
and
$$G(\omega_{o}) = 2\pi \cos(\omega_{o}t).$$
Then he states that the inverse transformations of the functions are
$$f(t_{o}) = \frac{1}{2 \pi}  e^{\Large-\frac{1}{2}\sigma^2  t_{o}^2}$$
and 
$$g(t_{o}) = 2 \pi \left( \frac{\delta(t_{o} - t)}{2} + \frac{\delta(t_{o} + t)}{2} \right) .$$
Would someone be able to derive how $f(t_{o})$ and $g(t_{o})$ are obtained. I'd appreciate it because I'd probably learn a lot from seeing the steps. Thank you very much.

For the first one, we have
$$
F(\omega)=\frac{1}{\sigma\sqrt{2\pi}} e^{\Large- \frac{\omega^2}{2\sigma^2}}.
$$
It seems that you are using the textbook for physics or engineering, so the inverse Fourier transform of $F(\omega)$ using the 'convention' notation in those fields is
$$
\begin{align}
\mathcal{F}^{-1}[F(\omega)]&=\frac{1}{2\pi}\int_{-\infty}^\infty F(\omega)\ e^{\large it\omega}\ d\omega\\
f(t)&=\frac{1}{2\pi}\cdot\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^\infty e^{\Large- \frac{\omega^2}{2\sigma^2}}\ e^{\large it\omega}\ d\omega\\
&=\frac{1}{2\pi}\cdot\frac{2}{\sigma\sqrt{2\pi}}\int_{0}^\infty e^{\Large- \left(\frac{\omega^2}{2\sigma^2}- it\omega\right)}\ d\omega.
\end{align}
$$
In general
$$
\begin{align}
\int_{x=0}^\infty e^{-(ax^2-bx)}\,dx&=\int_{x=0}^\infty \exp\left(-a\left(\left(x-\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}\right)\right)\,dx\\
&=\exp\left(\frac{b^2}{4a}\right)\int_{x=0}^\infty \exp\left(-a\left(x-\frac{b}{2a}\right)^2\right)\,dx.
\end{align}
$$
Let $u=x-\frac{b}{2a}\;\rightarrow\;du=dx$, then
$$
\begin{align}
\int_{x=0}^\infty e^{-(ax^2-bx)}\,dx&=\exp\left(\frac{b^2}{4a}\right)\int_{x=0}^\infty \exp\left(-a\left(x-\frac{b}{2a}\right)^2\right)\,dx\\
&=\exp\left(\frac{b^2}{4a}\right)\int_{u=0}^\infty e^{-au^2}\,du.\\
\end{align}
$$
The last form integral is Gaussian integral that equals to $\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}$. Hence
$$
\int_{x=0}^\infty e^{-(ax^2-bx)}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}\exp\left(\frac{b^2}{4a}\right).
$$
It yields the same result for $\displaystyle\int_{x=0}^\infty e^{-(ax^2\color{red}+bx)}\,dx$. Thus
$$
\begin{align}
\frac{1}{2\pi}\cdot\frac{2}{\sigma\sqrt{2\pi}}\int_{0}^\infty e^{\Large- \left(\frac{\omega^2}{2\sigma^2}- it\omega\right)}\ d\omega&=\frac{1}{2\pi}\cdot\frac{2}{\sigma\sqrt{2\pi}}\cdot\frac{1}{2}\sqrt{2\pi\sigma^2}\exp\left(\frac{( it)^2\cdot2\sigma^2}{4}\right)\\
f(t)&=\color{blue}{\frac{1}{2\pi}\ e^{\Large- \frac{\sigma^2t^2}{2}}}.
\end{align}
$$
