Show that $f_R(t)= \frac{1}{\pi^2R}\int_{0}^{+\infty} \frac{\sin^2(\pi Rs)}{s^2}(f(t+s)+f(t-s))ds$ Let $f:\mathbb{R} \to \mathbb{C}$ integrable. for all $R>0$ and $t\in  \mathbb{R}$. Define $ f_R(t) =\int_{-R}^{R}\left(1-\frac{|x|}{R}\right)\hat{f}(x) e^{2\pi itx} dx$.
Then show that $f_R(t)= \frac{1}{\pi^2R}\int_{0}^{+\infty} \frac{\sin^2(\pi Rs)}{s^2}(f(t+s)+f(t-s))ds$.
I'm a beginner in Fourier analysis so I need some hints to prove this statement.
 A: Note that we have
$$\begin{align}
f_R(t)&=\int_{-R}^R \left(1-\frac{|x|}{R}\right)e^{i2\pi tx}\int_{-\infty}^\infty f(s)e^{-i2\pi sx}\,ds\,dx\tag1\\\\
&=\int_{-\infty}^\infty f(s) \int_{-R}^R e^{i2\pi (t-s)x}\left(1-\frac{|x|}{R}\right)\,dx\,ds\tag2\\\\
&=\frac1{\pi^2R}\int_{-\infty}^\infty f(s) \frac{\sin^2(\pi R(t-s))}{(t-s)^2}\,ds\tag3\\\\
&=\frac1{\pi^2R}\int_{-\infty}^\infty f(t+s)\frac{\sin^2(\pi Rs)}{s^2}\,ds\tag4\\\\
&=\frac1{\pi^2R}\int_0^\infty \frac{\sin^2(\pi Rs)}{s^2} (f(t-s)+f(t+s))\,ds\tag5
\end{align}$$
as was to be shown!

NOTES:
In going from $(1)$ to $(2)$, we applied Fubini-Tonelli to justify interchanging the order of integration.
In going from $(2)$ to $(3)$, we evaluated the inner integral and factored out the constant term $\frac1{\pi^2R}$.
In going from $(3)$ to $(4)$, we enforced the substitution $s\mapsto s+t$ and noted that the integrand is an even function.
And in going from $(4)$ to $(5)$, we split the integral into integrals from $-\infty$ to $0$ and from $0$ to $\infty$.  Then, for the integral from $-\infty$ to $0$, we enforced the substitution $s\mapsto -s$.
