I am using this definitions of Fourier transform. Should there be 1/2\pi factor in the inverse one?
\begin{equation} G(\omega) = \mathcal{F}[g(t)] = \int\limits_{-\infty}^{+\infty}g(t) e^{-i \omega t} \mathrm{d} t \end{equation}
\begin{equation} g(t) = \mathcal{F}^{-1}[G(\omega)] = \frac{1}{2\pi} \int\limits_{-\infty}^{+\infty}G(\omega) e^{i \omega t} \mathrm{d} \omega \end{equation}
Why the sum of the FTs of sin and cos is not equal to the FT of the exponential according to Euler's equation? It should be the real and the imaginary part, right?
\begin{equation} \mathcal{F}[e^{i \omega_0 t}] = \int\limits_{-\infty}^{\infty} e^{i \omega_0 t} e^{-i \omega t} \mathrm{d} t = \int\limits_{-\infty}^{\infty} e^{i t (\omega_0 - \omega)} \mathrm{d} t= \mathrm{d}elta(\omega_0 - \omega) = \delta(\omega - \omega_0) \end{equation}
\begin{equation} \begin{aligned} \mathcal{F}[\cos(\omega_0 t)] &= \int\limits_{-\infty}^{\infty} \cos(\omega_0 t) e^{-i \omega t} \mathrm{d} t = \int\limits_{-\infty}^{\infty} \frac{e^{i t\omega_0} + e^{-i t \omega_0}}{2} e^{-i \omega t} \mathrm{d} t =\\ &= \frac{1}{2} \left[ \int\limits_{-\infty}^{\infty} e^{-i t(\omega - \omega_0)} \mathrm{d} t + \int\limits_{-\infty}^{\infty} e^{-i t(\omega + \omega_0)} \mathrm{d} t \right] = \frac{1}{2} \left[ \delta (\omega - \omega_0) + \delta (\omega + \omega_0) \right] \end{aligned} \end{equation}
\begin{equation} \begin{aligned} \mathcal{F}[\sin(\omega_0 t)] &= \int\limits_{-\infty}^{\infty} \sin(\omega_0 t) e^{-i \omega t} \mathrm{d} t = \int\limits_{-\infty}^{\infty} \frac{e^{i t\omega_0} - e^{-i t \omega_0}}{2i} e^{-i \omega t} \mathrm{d} t =\\ &= \frac{1}{2i} \left[ \int\limits_{-\infty}^{\infty} e^{-i t(\omega - \omega_0)} \mathrm{d} t - \int\limits_{-\infty}^{\infty} e^{-i t(\omega + \omega_0)} \mathrm{d} t \right] = \frac{1}{2i} \left[ \delta (\omega - \omega_0) - \delta (\omega + \omega_0) \right] \end{aligned} \end{equation}
Thank you
Edit:
\begin{equation} \mathcal{F} [\delta (t-a)] = \int\limits_{-\infty}^{\infty} \delta (t-a) e^{- i \omega t} \mathrm{d}t = e^{-i \omega a} \end{equation}
\begin{equation} \mathcal{F}^{-1} [e^{-i \omega a}] = \frac{1}{2\pi} \int\limits_{-\infty}^{\infty} e^{-i \omega a} e^{i \omega t} \mathrm{d}\omega = \frac{1}{2\pi} \int\limits_{-\infty}^{\infty} e^{i \omega (t-a)} \mathrm{d}\nu= \frac{1}{2\pi} \delta(t-a) \end{equation}