- #1
jostpuur
- 2,116
- 19
I know how to prove
[tex]
\int\limits_0^{\infty} \frac{\sin(x)}{x}dx = \frac{\pi}{2}
[/tex]
using complex analysis, and I know how to prove
[tex]
\lim_{N\to\infty} 2N\sin\Big(\frac{x}{2N}\Big) = x
[/tex]
using series. I have some reason to believe, that if [itex]0<A<\pi[/itex], then
[tex]
\lim_{N\to\infty} \int\limits_0^{NA} \frac{\sin(x)}{2N\sin(\frac{x}{2N})} dx = \frac{\pi}{2},
[/tex]
but don't know how to prove this. Anyone knowing how to accomplish this?
[tex]
\int\limits_0^{\infty} \frac{\sin(x)}{x}dx = \frac{\pi}{2}
[/tex]
using complex analysis, and I know how to prove
[tex]
\lim_{N\to\infty} 2N\sin\Big(\frac{x}{2N}\Big) = x
[/tex]
using series. I have some reason to believe, that if [itex]0<A<\pi[/itex], then
[tex]
\lim_{N\to\infty} \int\limits_0^{NA} \frac{\sin(x)}{2N\sin(\frac{x}{2N})} dx = \frac{\pi}{2},
[/tex]
but don't know how to prove this. Anyone knowing how to accomplish this?