A complex Fourier series converging on $|x|\le\frac L2$ is:
$$f(x)=\frac1L\sum_{n\in\Bbb Z} e^\frac{2i\pi nx}L\int_{-\frac L2}^\frac L2 e^{-\frac{2\pi i nt}L}f(t)dt$$
Switching the sum and integral gives:
$$f(x)=\frac{\lim\limits_{m\to\infty}}L\int_{-\frac L2}^\frac L2 f(t)\sum_{n=-m}^m \frac{e^\frac{2\pi i nx}L}{e^\frac{2\pi i n t}L}dt$$
We simplify using the complex argument:
$$f(x)= \frac{\lim\limits_{m\to\infty}}L\int_{-\frac L2}^\frac L2 \frac{e^\frac{2\pi i t}L\left(e^\frac{2\pi i(t-x)}L\right)^{2m}-e^\frac{2\pi ix}L}{\left(e^\frac{2\pi i(t-x)}L\right)^m\left(e^\frac{2\pi i t}L-e^\frac{2\pi i x}L\right)}f(t)dt\mathop=^{L,m,t,x\in\Bbb R}= \frac{\lim\limits_{m\to\infty}}L\int_{-\frac L2}^\frac L2 f(t)\csc\left(\frac{\pi(t-x)}L\right)\sin\left(m\arg\left(e^\frac{2\pi i (t-x)}L\right)+\frac{\pi(t-x)}L\right)dt\tag1$$
A test shows a truncated Fourier series equals its integral representation. However, another simplification is:
$$f(x)\mathop=^?_{L,m,t,x>0} \frac{\lim\limits_{m\to\infty}}L\int_{-\frac L2}^\frac L2 f(t)\csc\left(\frac{\pi(t-x)}L\right)\sin\left(\frac{\pi(2m+1)(t-x)}L\right)dt\tag 2$$
Surprisingly, the same numerical test with this integral gives a more accurate result. The main goal of $(1),(2)$ is an explicit integral representation of inverse functions like this one.
Is there any way to simplify and correct restrictions on $(1),(2)$?
From Dirac $\delta(w)$ in DLMF 1.17.iii:
$$f(x)=\frac1L\sum_{n\in\Bbb Z} e^\frac{2\pi i nx}L\int_{-\frac L2}^\frac L2 e^{-\frac{2\pi i nt}L}f(t)dt=\frac1L\int_{-\frac L2}^\frac L2f(t)\sum_{n\in\Bbb Z}e^{i n\frac{2\pi }L(x-t)}dt$$
Invoke the Dirac delta function:
$$f(x)=\frac{2\pi}L\int_{-\frac L2}^\frac L2 \delta\left(\frac{2\pi}L(x-t)\right)f(t)dt$$
Which is just a modified sifting property