Simplification of trigonometric integral

Question

I found the following identity but I'm not sure how to prove it using only a change of coordinates. You can prove it explicitly using the Fourier kernel but I'm interested to know if it can be done quickly. $$ \int_{-R}^R \bigl(R - |w|\bigr)\,e^{2 \pi i w \eta} \,dw = \int_0^R \int_{-w}^w e^{2 \pi i \theta \eta} \,d\theta\,dw $$

Answer 1

$$\text{Assuming} R>0: \int_{-R}^R e^{2 \pi i \eta w} (R-w \ \text{sgn}(w)) \, dw =\frac{\sin ^2(\pi \eta R)}{\pi ^2 \eta ^2}$$

$$ = \int_0^R (R-w) e^{2 \pi i \eta w} \, dw+\int_{-R}^0 (R+w) e^{2 \pi i \eta w} \, dw $$

$$= \int_0^R (R-w) e^{-2 \pi i \eta w} \, dw+\int_0^R (R-w) e^{2 \pi i \eta w} \, dw $$

$$= \int_0^R (R-w) \left(e^{-2 \pi i \eta w}+e^{2 \pi i \eta w}\right) \, dw $$

$$= \int_0^R \left(e^{-2 \pi i \eta w}+e^{2 \pi i \eta w}\right) \left(\int_w^R 1 \, d\theta \right) \, dw $$

$$= \int_0^R \left(e^{-2 \pi i \eta w}+e^{2 \pi i \eta w}\right) \left(\int_0^R \Theta (\theta -w) \, d\theta \right) \, dw $$

So finally with Fubini the inner integrand reduces to

$$\int_0^R \left(e^{-2 \pi i \eta w}+e^{2 \pi i \eta w}\right) \theta (\theta -w) \, dw=\frac{\sin (2 \pi \eta \theta )}{\pi \eta }$$