I wonder if one can express the Fourier transform in the trigonometric approach like, say, in the case of the Fourier series, where we can write it as:
$Sf(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left [ a_{n}cos\frac{n\pi x}{L} + b_{n}sin\frac{n\pi x}{L} \right ]$, where $L=\frac{T}{2}$,
$a_{n}=\frac{1}{L}\int_{-L}^{L}f(x)cos\frac{n\pi x}{L}dx$,
$b_{n}=\frac{1}{L}\int_{-L}^{L}f(x)sin\frac{n\pi x}{L}dx$.
Can we infer from the above formulas the formula for the Fourier transform? If so, how? Thank you in advance.
With
$$F(\omega)=\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt$$
you get
$$F(\omega)=\int_0^{\infty}f(t)e^{-i\omega t}dt+\int_0^{\infty}f(-t)e^{i\omega t}dt$$
With $e^{i\omega t}=cos(\omega t)+i\sin(\omega t)$
you obtain
\begin{align} F(\omega)&=\int_0^{\infty}[f(t)+f(-t)]\cos(\omega t)dt-i\int_0^{\infty}[f(t)-f(-t)]\sin(\omega t)dt\\ & = 2\int_0^{\infty}f_e(t)\cos(\omega t)dt -2i\int_0^{\infty}f_o(t)\sin(\omega t)dt \end{align}
where
$$f_e(t)=\frac{f(t)+f(-t)}{2}\\ f_o(t)=\frac{f(t)-f(-t)}{2}$$
are the even and odd parts of $f(t)$, respectively.