We can find the Fourier transform as the limit for infinite period of a Fourier series as explained in https://class.ece.uw.edu/235dl/EE235/Project/lesson15old/lesson15.html
I do the derivation they do:
$f_p$ is a $L$ periodic function. I define $$k_n = \frac{2 \pi n}{L}\qquad \text{and}\qquad c_n = \frac{1}{L} \int_{-L/2}^{L/2} f_p(x) e^{-i k_n x}~ dx$$
$$f_p(x)=\sum_{n=-\infty}^{+ \infty} c_n e^{i k_n x}$$
Now, we take the period going to infinity and define $$f(x)=\lim_{L \to + \infty} f_p(x)$$
$$\implies f(x)=\lim_{L \to + \infty} \sum_{n=-\infty}^{+ \infty} (k_{n+1}-k_n) \left( \frac{1}{2 \pi} \int_{-L/2}^{L/2}dx f_p(x) e^{-i k_n x} \right) e^{i k_n x}$$
Up to this point I totally agree and I see that $k_{n+1}-k_n=\delta k$ will play the role of the $dk$ in the integration. But this is more a "feeling" than a proof. Because they say that at this point we recognize a Riemann sum and we thus recognize the Fourier transform:
$$f(x)=\int_{-\infty}^{+ \infty} dk \left( \frac{1}{2 \pi} \int_{-\infty}^{+\infty}dx f(x) e^{-i k x} \right) e^{i k x}$$
This is what I don't understand.
From what I understood from comment on this post, it is not true "in general" but only for some "nice" functions. Indeed, for example, the definition of the integral through Riemann sum is the following:
$$ \int_a^b f(x) dx = \lim_{N \to +\infty} \sum_{n=0}^N \frac{(b-a)}{N} f(a + (b-a)\frac{n}{N})$$
But putting both $a$ and $b$ going to $+ \infty$ and $- \infty$ respectively I don't find an analog expression as the infinite-period Fourier series.
My question:
I would like a proof based on the Riemann summations on which for a set of "nice" functions the equality is rigorously proved.
I tried to prove it for compact support functions but I miss something in my proof as you can see in my very last paragraph.
Thanks to @mathworker21 guidance, I managed to prove it for some class of functions.
I assume I work in an ensemble $\widetilde{\mathcal{D}}(\mathbb{R})$. In this ensemble the functions admit Fourier transform.
Definitions
I call $f$ a function in $\widetilde{\mathcal{D}}(\mathbb{R})$.
I have:
$$f(x)=\int_{-\infty}^{+\infty} \widehat{f}(k) e^{ikx}dk$$ $$\widehat{f}(k)=\frac{1}{2 \pi}\int_{-\infty}^{+\infty} f(x) e^{-ikx}dx$$
I define $f_L$, the function that is build from $f$ periodized on $[-L/2,+L/2]$. I can thus write its Fourier series:
$$f_L(x)=\sum_{n=-\infty}^{+\infty} \frac{1}{L} c_{L}(k_n) e^{i k_n x} \text{ with } k_n=\frac{2 \pi n}{L} \text{ and } c_{L}(k_n) \equiv \int_{-L/2}^{+L/2} f(t)e^{-ik_n t}dt$$
I define a function $\widetilde{f}_L$:
$$\widetilde{f}_L=\sum_{n=-\infty}^{+\infty} \frac{1}{L} c_{\infty}(k_n) e^{i k_n x}$$
Derivation
Now, I want to prove that the infinite period Fourier serie match the Fourier transform, i.e: $\lim_{L \rightarrow + \infty} f_L(x) = f(x)$. It is equivalent to prove that $|f_L(x)-f(x)|$ goes to $0$ for $L$ infinite.
To prove it I do:
$$|f_L(x)-f(x)|=|\left(f_L(x)-\widetilde{f}_L\right) + \left(\widetilde{f}_L-f(x)\right)| \leq |f_L(x)-\widetilde{f}_L|+|\widetilde{f}_L-f(x)|$$
I will prove that for "good" functions $f$, the rhs converge to $0$ which will prove my statement. I start to deal with the second term:
$$\widetilde{f}_L(x) = \sum_{n=-\infty}^{+\infty} \frac{1}{L} c_{\infty}(k_n) e^{i k_n x}=\frac{1}{L} \sum_{n=-\infty}^{+\infty} g_x(\frac{n}{L})$$
Where:
$$g_x(\frac{n}{L})=c_{\infty}(2 \pi \frac{n}{L}) e^{i 2 \pi \frac{n}{L} x}$$
As explained here Riemann sum on infinite interval, if $u \rightarrow g_x(u)$ is such that $\int_{-\infty}^{+\infty} g_x'(u)du \leq + \infty$, then I have a Riemann summation, which means:
$$ \lim_{L \rightarrow +\infty} \frac{1}{L} \sum_{n=-\infty}^{+\infty} g_x(\frac{n}{L}) = \int_{-\infty}^{+\infty} g_x(u)du$$
Now, I have:
$$\lim_{L \rightarrow +\infty} \widetilde{f}_L(x)=\lim_{L \rightarrow +\infty} \frac{1}{L} \sum_{n=-\infty}^{+\infty} g_x(\frac{n}{L})=\int_{-\infty}^{+\infty} g_x(u)du\\=\int_{-\infty}^{+\infty} c_{\infty}(2 \pi u) e^{i 2 \pi u x}du=\frac{1}{2 \pi}\int_{-\infty}^{+\infty} c_{\infty}(k) e^{i k x}dk=f(x)$$
Thus, the second term in the triangular inequality vanishes.
It remains to prove that the first term vanishes.
We have:
$$|f_L(x)-\widetilde{f}_L|\leq \frac{1}{L} \sum_{n=-\infty}^{+\infty} |c_L(k_n)-c_{\infty}(k_n)|$$
We need to take "nice" function that allow this term to vanish for $L$ going to infinity. One simple example is to take functions defined on a compact. For $L$ big enough, we have:
$$c_{L}(k_n)=\int_{-L/2}^{+L/2} f(t)e^{-ik_n t}dt=\int_{-\infty}^{+\infty} f(t)e^{-ik_n t}dt=c_{\infty}(k_n)$$
Thus, for $L$ big enough: $\sum_{n=-\infty}^{+\infty} |c_L(k_n)-c_{\infty}(k_n)|=0$.
And we would have proven that for $L$ going to infinity: $$f_L(x) \rightarrow f(x)$$
In summary, sufficient condition we need
We need two different things to get to the result using this method:
First requirement: $\lim_{L \rightarrow +\infty} \frac{1}{L} \sum_{n=-\infty}^{+\infty} g_x(\frac{n}{L})=\int_{-\infty}^{+\infty} g_x(u)du$, which can be true if the derivative of $g_x$ is integrable.
Second requirement: $|f_L(x)-\widetilde{f}_L|$ , or more restrictively: $\sum_{n=-\infty}^{+\infty} |c_L(k_n)-c_{\infty}(k_n)|$ goes to $0$ for $L$ infinite. It can be done for $f$ having a compact support.
I am however not sure that the first requirement is true if $f$ is has a compact support ?