Given an "L" periodic Fourier series
$$\phi(x) = \sum_{1}^{\infty} \left( \frac{2}{l} \int_{0}^{l}\phi(x) sin\left(\frac{n \pi x}{l}\right)\, \mathrm dx \right) sin\left(\frac{n \pi x}{l}\right)$$
Is there a simple way to derive the Fourier Sine Transform? I've found derivations of $2 \pi$ periodic Sine transforms, but was hoping to better understand the "L" periodic case.
Let $f,f'$ be $L^1$ and $$\hat{f}(\xi) = \int_{-\infty}^\infty f(x)e^{-2i \pi \xi x}dx$$ For any $L$ let $$F_L(x)=\sum_n f(x+nL)$$ Then $F_L$ is $L$-periodic and $F_L,F_L'$ are integrable on one period and we have the uniformly convergent Fourier series $$F_L(x)= \sum_k \frac{1}{L}\hat{F}_L(k) e^{2i \pi xk/L}=\sum_k \frac{1}{L}\hat{f}(k/L) e^{2i \pi xk/L}$$ Locally uniformly $$f(x)=\lim_{L \to \infty}F_L(x)$$
And $$f(x)=\lim_{L \to \infty}F_L(x) = \lim_{L \to \infty}\sum_k \frac{1}{L}\hat{f}(k/L) e^{2i \pi xk/L} = \int_{-\infty}^\infty \hat{f}(\xi) e^{2i \pi \xi x}d\xi$$ where the series and the integral converge absolutely.
For the sine transform it is the same starting with $f$ odd, so that grouping the positive and negative frequencies together give series in $\sin$.