This question follow up from a discussion on this post Prove that there exists a constant $c>0$ such that for $R>0$ and $t\in \Bbb R$ we have, $$\left|\int_{|x|<R} \frac{\sin x}{x}e^{-ixt}dx\right|<c$$
This aim to rpove that the Fourier trnasform of $g: x\mapsto \frac{\sin x}{x}$
Given by $$\widehat{g}(t) = \lim_{R\to \infty}\int_{|x|<R} \frac{\sin x}{x}e^{-ixt}dx$$ is bounded. Please I would like to see a prove with elementary tools. I mean a prove that can ignore the Fourier analysis here.
$$\int_{-R}^R 2\frac{\sin x}{x} e^{-ixt}dx =\int_{-R}^R \int_{-1}^{1}e^{i x u}du e^{-ixt}dx =\int_{-1}^{1}\int_{-R}^R e^{ix(u-t)}dx du = \int_{-1}^1 2\frac{\sin (R(u-t))}{u-t}du=\int_{-1-t}^{1-t} 2\frac{\sin (Rv)}{v}dv=\int_{-R(1+t)}^{R(1-t)} 2\frac{\sin w}{w}dw$$
Thus for $|t| \ne 1$ writing $$\int_{-R(1+t)}^{R(1-t)} (...)= 1_{|t|\le 1}\int_{-R(1+t)}^{R(1-t)}(...)+ 1_{|t|>1}\int_{-R(1+t)}^{R(1-t)}(...)$$ we have, $$ \lim_{R \to \infty} 1_{|t|>1}\int_{-R(1+t)}^{R(1-t)}(...)=0 $$ Since $|t|>1$ implies $t+1$ and $t-1$ are of the same sign.
On the other hand $|t|<1$ we have, $1+t>0$ and $1-t>0$. whence, $$\lim_{R \to \infty} \int_{-R}^R 2\frac{\sin x}{x} e^{-ixt}dx = \lim_{R \to \infty} \int_{-R(1+t)}^{R(1-t)} 2\frac{\sin w}{w}dw \\= 1_{|t|<1}\lim_{R \to \infty} \int_{-R(1+t)}^{R(1-t)} 2\frac{\sin w}{w}dw = 1_{|t|<1}\int_{-\infty}^{\infty} 2\frac{\sin w}{w}dw\\ =2\pi1_{|t|<1}$$
The same method works for showing the Fourier inversion theorem for any $f \in L^1,f' \in L^1$ or $f'$ piecewise $L^1$ as here.