Derivation of Fourier Transform? So from the fourier series, we can simplify it further and use trig identities to get the following:
$$ f(t) = \frac{a_0}{2} + \sum^{\infty}_{n=1}
\left(\frac{a_n}{2}+\frac{b_n}{2i}\right)e^{i n \omega t} +
\left(\frac{a_n}{2}-\frac{b_n}{2i}\right)e^{-i n \omega t} $$
So how do you go from the above line to
$$ f(t) = \sum^{\infty}_{n=-\infty} C_n e^{in\omega t} $$
and 
$$ f(t) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty} F(\omega) e^{-i \omega t} d\omega $$
Thanks
 A: There's a mistake. Sin formula is $\sin(\omega nt)=\frac{e^{i \omega nt}-e^{-i\omega nt}}{2i}$
so the serie become:
$ f(t) = \frac{a_0}{2} + \sum^{\infty}_{n=1}
\left(\frac{a_n}{2}+\frac{b_n}{2i}\right)e^{i n \omega t} +
\left(\frac{a_n}{2}-\frac{b_n}{2i}\right)e^{-i n \omega t} $
Well, you call $C_{n}=\left(\frac{a_n}{2}+\frac{b_n}{2i}\right)$ and $C_{-n}=\left(\frac{a_n}{2}-\frac{b_n}{2i}\right)$ and obtain:
$ f(t) = \frac{a_0}{2} + \sum^{\infty}_{n=1}
C_{n}e^{i n \omega t} +
C_{-n}e^{-i n \omega t} $
that can be re-written in
$ f(t) = \sum^{\infty}_{n=-\infty} C_n e^{in\omega t} $
because $e^{i\omega0t}=1$ and you can rename $\frac{a_0}{2}$ in ${c_0}e^{i\omega0t}$.
Notice that if f(t) is real $a_n$ and $b_n$ are reals. so $c_n=\overline c_{-n}$.
For Fourier Transform the question is a little more complicated, because Fourier Transform is an operator that you define in $L^1$, you extend it on $L^2$ and in temperate distributions.
Fourier serie is connected with  a function $L^1$on a interval (that become periodic on $\mathbb{R}$) or an hyperrectangle on $\mathbb{R}^n$.  Imagine you can extend this to all $R^n$...you obtain Fourier transform. The factor  $\frac{1}{2\pi}$ is arbitrary, there are other choices, but all they must be coherent with the choice of the factor of the inverse Fourier Transform. $\frac{1}{\sqrt {2\pi}}$ is the most natural, because it's very similar to normalized complex exponencials that you use in Fourier Serie for the finite interval.
And this choice makes Fourier Transform an Unitary Operator, because $F^{+}=F^{-1}$.
A: The proof of Fourier in based that integration $-\pi$ to $\pi$ $\cos(kt)\cos(nt)dt$ is  zero.
What we basically do is multiply the Fourier series with $\cos(nt)$ on both sides. And integrate both sides from $-\pi$ to $\pi$.
All other terms on the right hand side become zero except $\cos(nt)$(because $\cos(nt)^2 dt$ integrated from $-\pi$ to $\pi$ is not zero but $\pi$). I have derived the Fourier transform (in detail) in the post http://communicationsm.blogspot.in/.
