On deriving the (inverse) Fourier transform from Fourier series

For the complex Fourier series of ##f## we have \begin{align} f(t)&\sim\sum_{n=-\infty}^\infty c_n e^{in\pi t/P}, \tag1 \\ \text{where} \quad c_n&=\frac1{2P}\int_{-P}^P f(t)e^{-in\pi t/P} \ dt. \tag2 \end{align} [...] We define, provisionally, $$\hat{f}(P,\omega)=\int_{-P}^P f(t)e^{-i\omega t} \ dt, \quad \omega\in\mathbb R,\tag3$$ so that ##c_n=\frac{1}{2P}\hat{f}(P,n\pi/P)##. The formula ##(1)## is translated into $$f(t)\sim\frac1{2P}\sum_{n=-\infty}^\infty \hat{f}(P,\omega_n)e^{i\omega_nt}=\frac1{2\pi}\sum_{n=-\infty}^\infty \hat{f}(P,\omega_n)e^{i\omega_nt}\cdot\frac{\pi}{P},\quad \omega_n=\frac{n\pi}{P}.\tag4$$ Because of ##\Delta\omega_n=\omega_{n+1}-\omega_n=\frac\pi{P}##, this last sum looks rather like a Riemann sum. Now we let ##P\to\infty## in ##(3)## and define $$\hat{f}(\omega)=\lim_{P\to\infty} \hat{f}(P,\omega)=\int_{-\infty}^\infty f(t)e^{i\omega t} \ dt\quad \omega\in\mathbb R.\tag5$$ (at this point we disregard all details concerning convergence). If ##(4)## had contained ##\hat{f}(\omega_n)## instead of ##\hat{f}(P,\omega_n)##, the limiting process ##P\to\infty## would have resulted in $$f(t)\sim \frac1{2\pi}\int_{-\infty}^\infty\hat{f}(\omega)e^{i\omega t} \ d\omega.\tag6$$

psie said:

Some say it is taking the period to infinity, but the above derivation shows that it doesn't work.