What is the value of $\lfloor{100N}\rfloor$

What is the value of $\lfloor{100N}\rfloor$ where $\displaystyle N= \lim\limits_{a\,\rightarrow\,\infty}\sum\limits_{x=-a}^{a}\frac{\sin x}{x}$.

This is a part of a bigger problem that I was solving. I need the exact integer value of $\lfloor{100N}\rfloor$ .
I was only able to simplify $\displaystyle \lim\limits_{a\,\rightarrow\,\infty}\sum\limits_{x=-a}^{a}\frac{\sin x}{x}= 1+2\left(\lim\limits_{a\,\rightarrow\,\infty}\sum\limits_{x=1}^{a}\frac{\sin x}{x}\right)$. I don't know how to proceed further. Please help

• Have you yet learned about Fourier series? That one is kind of famous. – Daniel Fischer Jul 2 '14 at 13:06
• @DanielFischer No. Can it be done using that? Please if you could elaborate a little – idpd15 Jul 2 '14 at 13:12
• – lhf Jul 2 '14 at 13:12

Note \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*} and hence \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*} Using the following result:

Let $f(x)$ be monotonic in $[0,A]$. Then $$\lim_{n\to\infty}\int_0^Af(x)\frac{\sin(nx)}{x}dx=f(0)\frac{\pi}{2}.$$

we have \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*} and hence $$\lfloor{100N}\rfloor=314.$$ Here we use the fact: the function $\frac{x}{\sin\frac{x}{2}}$ is increasing in (0,1] and $$\lim_{x\to 0^+}\frac{x}{\sin\frac{x}{2}}=2.$$

This is a partial answer, I can't justify the final step. :(

\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} If $k=0$, it gives the right answer but I don't know how to justify this choice of $k$.

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \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}

With Abel-Plana Formula:

\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}

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