Why Fourier transform is derived from Fourier series coefficient multiplied by period?

Question

In derivation of fourier transform from fourier series coefficient presented here the first step is to multiply the fourier series coefficient by $T$. Could anyone explain why? What's the purpose of doing it?

Answer 1

A few lines below, I read: $$x(t)=\sum_{n=-\infty}^\infty c_ne^{jn\omega t}=\sum_{n=-\infty}^\infty Tc_ne^{jn\omega t}\frac{1}{T}.$$ He wants to transform a non-periodic function, meaning $T\to\infty$. He sees he can easily calculate the limits of $Tc_n$ and $\frac{1}{T}$ as $T\to\infty$, so he goes ahead. The idea behind this is that $c_n$ is approaching 0, so calculating the limit of $c_n$ directly doesn't lead us anywhere. Instead, $Tc_n$ has a finite nonzero value, and $\frac{1}{T}$ becomes our differential. The sum becomes an integral in $\mathrm{d}\omega$ because sums often have integrals as limits: the integral itself can be seen as a limit of Riemann sums.

In summary, he wants to isolate something finite depending on an integration variable, and the differential of that variable. $\frac{1}{T}$ becomes that differential. The original expression of the sum doesn't have $\frac{1}{T}$, so you have to produce it. You multiply and divide by $T$, so on one side you get the differential, and the other side is precisely $Tc_n$. So there you go: you must calculate $Tc_n$ and its limit.