I'm currently trying to derive Fourier transform from Fourier series. Please have a look and see whether there are any problem in my deriving process.
If there are, please let me know what is the problem. Thank you very much.
$$g(t)=\lim_{T\to\infty}\sum_{n=-\infty}^\infty\left[\int_{-\frac{T}{2}}^\frac{T}{2}g(t)e^{-i2\pi nft}dt\right]e^{i2\pi nft}\cdot f$$
$$=\sum_{n=-\infty}^\infty\left[\int_{-\infty}^\infty g(t)e^{-i2\pi n\cdot df\cdot t}dt\right]e^{i2\pi n\cdot df\cdot t}\cdot df$$
$$=\int_{-\infty}^\infty\left[\int_{-\infty}^\infty g(t)e^{-i2\pi ft}dt\right]e^{i2\pi ft}df$$
where $T$ is period, $f$ is fundamental frequency, $t$ is time.
By the definition, $f = \frac{1}{T}$.
Explanation for $t$ in $e^{i2\pi ft}$ : By definition,
$$G(f)=\int_{-\infty}^\infty g(t)e^{-i2\pi ft}dt$$
and this is Fourier transform. I believe Fourier transform can be calculated if $f$ is given. By definition,
$$g(t)=\int_{-\infty}^\infty G(f)e^{i2\pi ft}df$$
and this is inverse Fourier transform. I believe inverse Fourier transform can be calculated if $t$ is given.
I wrote $f$ in the first expression and I changed it to $df$ in the second expression. The reason I did that is I thought $f$ can become $df$ because of $T\to\infty$, therefore $f\to 0$ by definition.
As far as I understand, $nf$ is nth harmonics of fundamental frequency. Since $f\to 0$, the interval between $nf$ and $(n+1)f$ approaches to $0$. It means the spectrum is becoming dense and eventually a continuum.
Meanwhile, let $a$ be $1$ and $b$ can be $0$, $1$, $2$, $3$. (It's a different story and not directly related to the question. I'm trying to make a metaphor). Then, possible combinations of $ab$ are as following: $0$, $1$, $2$, $3$.
This time, let $a$ be $0.1$ and $b$ can be $0$ ~ $30$. Then, possible combinations of $ab$ are $0$, $0.1$, ... $1$, ..., $2$, ..., $3$.
Therefore, as far as I understand, $n\cdot df$ can be any real number, because $df$ is very small number ($f\to 0$) and $n$ can be $(-\infty, \infty)$. So, I think it is possible to write $n\cdot df$ as just $f$, since $f$ can be any frequency. This is why I changed $n\cdot df$ in the second expression to $f$ in the third expression.
And in the third expression sigma and $df$ disappear and integral newly appears. I believe it is reasonable because every sum of width times height between an interval is integral... (in this case the interval is $(-\infty, \infty)$)