As suggested by Reuns in in this question. It is interesting that the problem there boils down to the following limit:

$$\lim_{R\rightarrow +\infty}\int_{-1}^{1}\frac{\sin(2\pi R t)}{t}dt=\frac{1}{\pi}.$$

Anyone has an idea.

Thanks so much.

  • 1
    $\begingroup$ The answer should be $\pi$. $\endgroup$ – Z Ahmed Aug 5 '19 at 2:14

Let $2\pi Rt=x$, then $$I =\lim_{R\rightarrow\infty} \int_{-2\pi R}^{2\pi R} \frac{\sin x}{x} dx= \int_{-\infty}^{\infty} \frac{\sin x}{x} dx= 2 \int_{0}^{\infty} \frac{\sin x}{x} dx=2\frac{\pi}{2}=\pi$$


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