Let $x(t) \stackrel{F}{\longleftrightarrow}X(j\omega)$ be a Fourier transform pair. Is it true that the real part of $X(j\omega)$ equals the even part of $x(t)$?
If it is true, then does the imaginary part of $X(j\omega)$ equal the odd part of $x(t)$?
I'm in a EE class that used the first of these identities in a problem I found the solution to online. The problem was 4.25 f from Signals and Systems by Oppenheim and Willsky, ed. 2. I can't find this anywhere. Thanks!!!!!!
Defining the Fourier transform as $$X(j \omega) = \mathcal{F}\left[x(t)\right] = \int_{\mathbb{R}} e^{-j \omega t} x(t) \mathrm{d} t$$ and the composition of $x$ into even and odd parts as $$ x(t) = x_{even}(t) + x_{odd}(t) = \frac{x(-t) + x(t)}{2} + \frac{x(-t) - x(t)}{2}$$ one can write the transform as $$ X(j \omega) = \mathcal{F}\left[x_{even}(t) + x_{odd}(t)\right] = \mathcal{F}\left[x_{even}(t)\right] + \mathcal{F}\left[x_{odd}(t)\right] \mathrm{.}$$
Generally for the complex conjugate of a Fourier transform holds $$\overline{X(j \omega)} = \int_{\mathbb{R}} \overline{e^{-j \omega t}} x(t) \mathrm{d} t = \int_{\mathbb{R}} e^{j \omega t} x(t) \mathrm{d} t = \int_{\mathbb{R}} e^{-j \omega \tau} x(-\tau) \mathrm{d} \tau $$ and if $x(t)$ is even ($x(\tau) = x(-\tau)$) $$\overline{X(j \omega)} = \int_{\mathbb{R}} e^{-j \omega \tau} x(-\tau) \mathrm{d} \tau = \int_{\mathbb{R}} e^{-j \omega \tau} x(\tau) \mathrm{d} \tau = X(j \omega) $$ so $X(j \omega)$ is real. Equivalently for odd $x$ $$\overline{X(j \omega)} = \int_{\mathbb{R}} e^{-j \omega \tau} x(-\tau) \mathrm{d} \tau = - \int_{\mathbb{R}} e^{-j \omega \tau} x(\tau) \mathrm{d} \tau = - X(j\omega) $$ so $X(j \omega)$ is purely imaginary.
Therefore $$ X(j \omega) = \mathcal{F}\left[x_{even}(t)\right] + \mathcal{F}\left[x_{odd}(t)\right] $$ where $$\mathcal{F}\left[x_{even}(t)\right] = \mathrm{Re}\left[ X(j\omega) \right]$$ and $$\mathcal{F}\left[x_{odd}(t)\right] = \mathrm{Im}\left[ X(j\omega) \right] \mathrm{.}$$