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 $$ \begin{align} \frac{1}{2\pi}\cdot\frac{2}{\sigma\sqrt{2\pi}}\int_{0}^\infty e^{\Large- \left(\frac{\omega^2}{2\sigma^2}- it\omega\right)}\ d\omega&=\frac{1}{2\pi}\cdot\frac{2}{\sigma\sqrt{2\pi}}\cdot\frac{1}{2}\sqrt{2\pi\sigma^2}\exp\left(\frac{( it)^2\cdot2\sigma^2}{4}\right)\\ f(t)&=\color{blue}{\frac{1}{2\pi}\ e^{\Large- \frac{\sigma^2t^2}{2}}}. \end{align} $$