There are two of the same equation given for the Inverse Fourier Transform:
$$ f(t) = {1 \over {2\pi}} \int_{-\infty}^{\infty} X(\omega) \, e^{j \omega t} \operatorname{d\omega} \;\;\;\;\;\text{or}\;\;\;\;\; f(t) = \int_{-\infty}^{\infty} X( f) \, e^{j 2 \pi f t} \operatorname{df}$$
which can be derived from the Fourier Series (1) $ \sum_{n=-\infty}^{\infty} c_n e^{jn\omega_0t} $ or (2) $ \sum_{n=-\infty}^{\infty} c_n e^{jn2\pi f_0t} $:
$$ \begin{align}
(1) \implies x(t) = \sum_{n=-\infty}^{\infty} c_n e^{jn\omega_0t}
\,&=\, \sum_{n=-\infty}^{\infty} \overbrace{\frac{1}{T_0} \int_{-\frac{T_0}{2}}^{\frac{T_0}{2}} x(t) \, e^{-jn\omega_0 t} \operatorname{dt}}^{c_n} \cdot e^{jn\omega_0 t}
\,=\, \sum_{n=-\infty}^{\infty} \overbrace{\frac{\omega_0}{2\pi} \int_{-\frac{T_0}{2}}^{\frac{T_0}{2}} x(t) \, e^{-jn\omega_0 t} \operatorname{dt}}^{c_n} \cdot e^{jn\omega_0 t}
\\
\,&=\, \sum_{n=-\infty}^{\infty} \overbrace{\lim_{\omega_0 \to 0} \frac{\omega_0}{2\pi} \int_{-\infty}^{\infty} x(t) \, e^{-jn\omega_0 t} \,\operatorname{dt}}^{c_n} \cdot e^{jn\omega_0 t}
\,=\, \int_{-\infty}^{\infty} \operatorname{\frac{d\omega}{2\pi}} \Bigg[ \int_{-\infty}^{\infty} x(t) \, e^{-j\omega t} \,\operatorname{dt} \Bigg] \cdot e^{j\omega t}
\\
\,&=\, \int_{-\infty}^{\infty} \Bigg[ \frac{1}{2\pi} \overbrace{\int_{-\infty}^{\infty} x(t) \, e^{-j\omega t} \,\operatorname{dt} \Bigg]}^{X(\omega)} \cdot e^{j\omega t} \,\operatorname{d\omega}
\,=\, \mathcal{F}^{-1}\big\{X(\omega) \big\}
\end{align} $$
$$ \begin{align}
(2) => x(t) = \sum_{n=-\infty}^{\infty} c_n e^{jn 2 \pi f_0 t}
\,&=\, \sum_{n=-\infty}^{\infty} \overbrace{\frac{1}{T_0} \int_{-\frac{T_0}{2}}^{\frac{T_0}{2}} x(t) \, e^{-jn 2 \pi f_0 t} \operatorname{dt}}^{c_n} \cdot e^{jn 2 \pi f_0 t}
\,=\, \sum_{n=-\infty}^{\infty} \overbrace{f_0 \int_{-\frac{T_0}{2}}^{\frac{T_0}{2}} x(t) \, e^{-jn 2 \pi f_0 t} \operatorname{dt}}^{c_n} \cdot e^{jn 2 \pi f_0 t}
\\
\,&=\, \sum_{n=-\infty}^{\infty} \overbrace{\lim_{f_0 \to 0} f_0 \int_{-\infty}^{\infty} x(t) \, e^{-jn 2 \pi f_0 t} \,\operatorname{dt}}^{c_n} \cdot e^{jn 2 \pi f_0 t}
\,=\, \int_{-\infty}^{\infty} \operatorname{df} \Bigg[ \int_{-\infty}^{\infty} x(t) \, e^{-j 2 \pi f t} \,\operatorname{dt} \Bigg] \cdot e^{j 2 \pi f t}
\\
\,&=\, \int_{-\infty}^{\infty} \overbrace{\Bigg[ \int_{-\infty}^{\infty} x(t) \, e^{-j 2 \pi f t} \,\operatorname{dt} \Bigg]}^{X(f)} \cdot e^{j 2 \pi f t} \,\operatorname{df}
\,=\, \mathcal{F}^{-1}\big\{X(f) \big\}
\end{align} $$
Either equation will result in the same $ x(t) $. Choosing $ \omega $ over $ f $, or vice versa, is just a matter of preference. They are identical :
$$ \mathcal{F}^{-1}\{X(\omega)\} \,=\, {1 \over {2\pi}} \int_{-\infty}^{\infty} X(\omega) \, e^{j \omega t} \operatorname{d\omega} \;=\; \int_{-\infty}^{\infty} X(f) \, e^{j 2 \pi f t} \operatorname{df} \,=\, \mathcal{F}^{-1}\{X(f)\}$$
An alternative way of deriving FT from FS is to use the sifting property of an impulse train (or Dirac Comb). Knowing that
$$ \delta(at)=\frac{1}{|a|}\delta(t) \,\,\implies\,\, \delta(f) = \delta \Big(\frac{\omega}{2\pi} \Big) = 2\pi \, \delta(\omega) \;\;\;\;\;\text{and}\;\;\;\;\; \operatorname{df} = \operatorname{d \frac{\omega}{2\pi}} = \frac{1}{2\pi} \operatorname{d\omega} $$
I can rewrite the FS $ \sum_{n=-\infty}^{\infty} c_n e^{jn2\pi f_0t} $ in terms of an impulse train :
$$ \begin{align}
(2) \implies x(t) \,&=\, \sum_{n=-\infty}^{\infty} c_n \, e^{jn \omega_0 t}
\,=\, \sum_{n=-\infty}^{\infty} c_n \, e^{jn 2 \pi f_0 t}
\,=\, \sum_{n=-\infty}^{\infty} c_n \, \bigg[ \, \int_{-\infty}^{\infty} e^{j2\pi f_0t} \cdot \delta(f-nf_0) \,\operatorname{df} \,\bigg] \\
&= \int_{-\infty}^{\infty} \overbrace{\sum_{n=-\infty}^{\infty} c_n \, \delta(f-nf_0)}^{X(f)} \cdot e^{j2\pi ft} \operatorname{df} \,=\, \int_{-\infty}^{\infty} X(f) \cdot e^{j2\pi ft} \operatorname{df} \,=\, \mathcal{F}^{-1} \big\{X(f)\big\} \\
\end{align}
$$
All good and make sense…. until I did a change of variable from $ f $ to $ \omega $ :
$$
\begin{align}
(1) &\implies
\int_{-\infty}^{\infty} \sum_{n=-\infty}^{\infty} c_n \, \delta\bigg(\frac{\omega}{2\pi}-n\frac{\omega_0}{2\pi}\bigg) \cdot e^{j\omega t} \operatorname{d}\bigg(\frac{\omega}{2\pi} \bigg) \\ &=\,
\int_{-\infty}^{\infty} \sum_{n=-\infty}^{\infty} c_n \, 2\pi \,\delta\bigg(\omega-n \omega_0 \bigg) \cdot e^{j\omega t} \bigg(\frac{1}{2\pi} \bigg) \operatorname{d \omega}
\\
&=
\int_{-\infty}^{\infty} \overbrace{\sum_{n=-\infty}^{\infty} c_n \, \,\delta\bigg(\omega-n \omega_0 \bigg)}^{X(\omega)} \cdot e^{j\omega t} \operatorname{d \omega}
\,=\, \underbrace{\int_{-\infty}^{\infty} X(\omega) \cdot e^{j\omega t} \operatorname{d\omega}}_{\text{where is my}\; {1 / (2\pi)} \;\text{???}}
\,=\, \mathcal{F}^{-1} \big\{X(\omega)\big\}
\end{align} $$
We know there has to be a $ \frac{1}{2\pi} $ scaling factor whenever we write a FT in terms of $ \omega $, but that factor disappears if I derive FT from FS using $ \delta $ impulse train instead. Could someone explain what went wrong? I always feel dirty whenever I see a $ \delta $ in my equation.
Best Answer
In the first part of the derivation, it is explicitly shown that
$$ x(t) = \int_{-\infty}^{\infty} \overbrace{\Bigg[ \frac{1}{2\pi}\int_{-\infty}^{\infty} x(t) \, e^{-j\omega t} \,\operatorname{dt} \Bigg]}^{X(\omega)} \cdot e^{j\omega t} \,\operatorname{d\omega} $$
Which means
$$ x(t) = \int_{-\infty}^{\infty} X(\omega) \cdot e^{j\omega t} \,\operatorname{d\omega} $$
And this is exactly the same expression in the second part.
So, I don't understand why one should expect an extra $2\pi$ when using Dirac Combs.
In fact if one defines
$$X(\omega) = \frac{1}{2\pi}\int_{-\infty}^{\infty} x(t) \, e^{-j\omega t} \,\operatorname{dt} $$ then we have $$x(t) = \int_{-\infty}^{\infty} X(\omega) \, e^{j\omega t} \,\operatorname{d\omega}.$$
Notice the absence of $2\pi$ in the inverse expression.