Prove the Identity $\frac{2\pi}{N}\int_{-\pi}^{\pi}\int_{-N}^{N} \left(1-\frac{\xi}{N}\right)e^{i \xi t}d\xi dt=1$

Question

Does anyone know what the best approach for this integral identity is? $$ \frac{2\pi}{N}\int_{-\pi}^{\pi}\int_{-N}^{N} \left(1-\frac{\xi}{N}\right)e^{i \xi t}d\xi dt=1,\quad \forall N \in \mathbb{R}, \quad N >0$$

Answer 1

This is not true $$ \int_{-\pi}^{\pi} \exp(i \xi t) \; dt = \frac{2\sin(\pi \xi)}{\xi} , $$ $$ I =\frac{4\pi}{N}\int_{-N}^N\left(1-\frac{\xi}{N}\right)\frac{\sin(\pi \xi)}{\xi} \; d\xi, $$ by symmetry $\sin(\pi \xi)=-\sin(-\pi \xi)$ so one of the terms cancels to give $$ I =\frac{4\pi}{N}\int_{-N}^N\frac{1}{\xi}\sin(\pi \xi) \; d\xi = \frac{8\pi}{N}\int_{0}^N\frac{\sin(\pi \xi)}{\xi} \; d\xi = \frac{8\pi}{N}\mathrm{Si}(\pi N), $$ where $\mathrm{Si}(x)$ is the Sine-integral function. The point where the original identity is true is at $N\approx39.487806651487979$.