Consider a function $f(x)$ with support on $I = [-L/2, L/2]$. We can expand $f$ into a Fourier series over the interval $I$:
$$ f(x) = \sum_{n=-\infty}^{\infty} a_n e^{2 \pi i n x / L}, $$
where the Fourier coefficients $a_n$ are given by
$$ a_n = \frac{1}{L} \int_{-L/2}^{L/2} f(x) e^{-2 \pi i n x / L} \, dx. $$
Now, consider what happens as $L$ tends to infinity. Introduce the variable
$$ \xi = \frac{n}{L} \quad \text{so that} \quad n = \xi L. $$
We can rewrite the Fourier coefficients in terms of $\xi$:
$$ a_n = \frac{1}{L} \int_{-L/2}^{L/2} f(x) e^{-2 \pi i n x / L} \, dx = \frac{1}{L} \int_{-L/2}^{L/2} f(x) e^{-2 \pi i \xi x} \, dx. $$
Given that $f$ is supported in $[-L/2, L/2]$, the inverse Fourier integral reduces to:
$$ \hat{f}(\xi) = \int_{-L/2}^{L/2} f(x) e^{-2 \pi i \xi x} \, dx. $$
Now observe that the Fourier series for $f$ can be written as:
$$ f(x) = \sum_{n=-\infty}^{\infty} \left( \frac{1}{L} \int_{-L/2}^{L/2} f(x') e^{-2 \pi i \xi x'} \, dx' \right) e^{2 \pi i \xi x}. $$
How do we go from here to obtaining the integral instead of the summation? I am unclear how this transition happens by letting $L$ go to infinity.
$$a_n=\frac{1}{L}, C_L\left(\frac{n}{L}\right)\tag{1}$$
where
$$C_L(\omega)=\int\limits_{-\frac{L}{2}}^{\frac{L}{2}} f(x) , e^{-i 2 \pi \omega x} , dx\tag{2}$$
is a Fourier transform with a truncated integration range and note that
$$F(\omega)=\mathcal{F}{x}f(x)=\int\limits{-\infty}^{\infty} f(x), e^{-i 2 \pi \omega x} , dx=\lim\limits_{L->\infty} C_L(\omega)\tag{3}.$$
– Steven Clark Jun 03 '24 at 16:55