jostpuur
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I know how to prove
<br /> \int\limits_0^{\infty} \frac{\sin(x)}{x}dx = \frac{\pi}{2}<br />
using complex analysis, and I know how to prove
<br /> \lim_{N\to\infty} 2N\sin\Big(\frac{x}{2N}\Big) = x<br />
using series. I have some reason to believe, that if 0<A<\pi, then
<br /> \lim_{N\to\infty} \int\limits_0^{NA} \frac{\sin(x)}{2N\sin(\frac{x}{2N})} dx = \frac{\pi}{2},<br />
but don't know how to prove this. Anyone knowing how to accomplish this?
<br /> \int\limits_0^{\infty} \frac{\sin(x)}{x}dx = \frac{\pi}{2}<br />
using complex analysis, and I know how to prove
<br /> \lim_{N\to\infty} 2N\sin\Big(\frac{x}{2N}\Big) = x<br />
using series. I have some reason to believe, that if 0<A<\pi, then
<br /> \lim_{N\to\infty} \int\limits_0^{NA} \frac{\sin(x)}{2N\sin(\frac{x}{2N})} dx = \frac{\pi}{2},<br />
but don't know how to prove this. Anyone knowing how to accomplish this?