¿Cuál es el valor de ⌊100N⌋⌊100N⌋\lfloor{100N}\rfloor

Nota

$$\begin{eqnarray*} \sum_{k=1}^n\cos(kx)=-\frac{1}{2}+\frac{\sin(\frac{n}{2}+1)x}{2\sin\frac{x}{2}} \end{eqnarray*}$$ y por lo tanto $$\begin{eqnarray*} \sum_{k=1}^n\frac{\sin k}{k}&=&\int_0^1\sum_{k=1}^n\cos(kx)dx\\ &=&-\frac{1}{2}+\int_0^1\frac{\sin(\frac{n}{2}+1)x}{2\sin\frac{x}{2}}dx\\ \end{eqnarray*}$$ Usando el siguiente resultado:

Dejar$f(x)$ser monótono en$[0,A]$. Entonces

$$\lim_{n\to\infty}\int_0^Af(x)\frac{\sin(nx)}{x}dx=f(0)\frac{\pi}{2}.$$

tenemos

$$\begin{eqnarray*} N&=&\lim\limits_{a\to\infty}\sum\limits_{k=-a}^{a}\frac{\sin k}{k}=1+2\lim_{a\to\infty}\sum\limits_{k=1}^{a}\frac{\sin k}{k}\\ &=&1+2\left(-\frac{1}{2}+\lim_{a\to\infty}\int_0^1\frac{\sin(\frac{a}{2}+1)x}{\sin\frac{x}{2}}dx\right)\\ &=&\lim\limits_{a\to\infty}\int_0^1\frac{x}{\sin\frac{x}{2}}\frac{\sin(\frac{a}{2}+1)x}{x}dx\\ &=&\pi \end{eqnarray*}$$ y por lo tanto $$\lfloor{100N}\rfloor=314.$$ Aquí usamos el hecho: la función $\frac{x}{\sin\frac{x}{2}}$es creciente en (0,1] y $$\lim_{x\to 0^+}\frac{x}{\sin\frac{x}{2}}=2.$$

Esta es una respuesta parcial, no puedo justificar el paso final. :(

$$\begin{aligned} \sum_{x=1}^{\infty} \frac{\sin x}{x}=\Im\left(\sum_{x=1}^{\infty} \frac{e^{ix}}{x}\right) &=\Im\left(\int_0^{\infty} \sum_{x=1}^{\infty} e^{-(t-i)x}\,dt\right)\\ &=\Im\left(\int_0^{\infty} \frac{e^i}{e^t-e^i}\,dt\right) \\ &=\Im\left(-\ln(1-e^i)\right)\\ &=\Im\left(-\ln\left(2\sin\left(\frac{1}{2}\right)e^{-i\left(\frac{\pi-1}{2}\right)+i2k\pi}\right)\right) \end{aligned}$$ Si $k=0$, da la respuesta correcta pero no sé cómo justificar esta elección de $k$.

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$$\begin{align} {\large\tt\mbox{Note that}}\qquad\sum_{x = -\infty}^{\infty}{\sin\pars{x} \over x} &=2\sum_{x = 0}^{\infty}{\sin\pars{x} \over x} - 1 \end{align}$$

Con fórmula de Abel-Plana :

$$\begin{align} &\color{#c00000}{\large\sum_{x = -\infty}^{\infty}{\sin\pars{x} \over x}} \\[3mm]&=2\bracks{\!\!\int_{0}^{\infty} {\sin\pars{x} \over x}\,\dd x + \half\lim_{x \to 0}\!\!{\sin\pars{x} \over x} +\ic\int_{0}^{\infty}\!\! {{\sin\pars{\ic x}/\pars{\ic x}} - \sin\pars{-\ic x}/\pars{-\ic x} \over \expo{2\pi x} - 1}\,\dd x\!} - 1 \\[3mm]&=2\pars{{\pi \over 2} + \half} - 1 = \color{#c00000}{\Large\pi} \end{align}$$

$$\color{#66f}{\large\floor{100\sum_{x = -\infty}^{\infty}{\sin\pars{x} \over x}}} =\floor{100\,\pi} = \color{#66f}{\Large 314}$$