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

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
$$ \lim_{x\to 0^+}\frac{x}{\sin\frac{x}{2}}=2. $$
