Limit of the sum $\sum_{k=1}^\infty\frac{\sin(kx)}{kx}$ The sum
$$S(x)=\sum_{k=1}^\infty\frac{\sin(kx)}{kx}$$ 
can be written in a closed form:
$$S(x)=\frac{1}{x}\left(\frac{1}{2}i\left(\ln(1-\exp(ix)\right)-\ln\left(1-\exp(-ix)\right)\right)$$
I am in trouble in calculating the limit:
$$S_0=\lim_{x\to\infty}S(x)$$
Transforming the $S(x)$ in trigonometric functions I get:
$$S(x)=-\frac{1}{2x}\arctan\left(-\sin(x),1-\cos(x)\right)+\frac{1}{2x}\arctan(\sin(x),1-\cos(x))$$
but this doesn't help me to evaluate $S_0$.
I would like to have some suggestion useful to solve the problem. Thanks.
 A: In general let us assume the series $$S_n(x)=\sum_{k=1}^n\frac{\sin(kx)}{k}$$
for $0<x<2\pi$
$$
\begin{eqnarray}
S_n(x)&=&\int_0^x \sum_{k=1}^n \cos(kt) \;\mathrm dt=-\frac{x}{2}+\int_0^x \frac{\sin\left((2n+1)\frac{t}{2}\right)}{2\sin\left(\frac{t}{2}\right)}\;\mathrm dt=\\
&=&-\frac{x}{2}+\int_0^x \frac{\sin\left((2n+1)\frac{t}{2}\right)}{2\sin\left(\frac{t}{2}\right)}\;\mathrm dt\pm \int_0^x \frac{\sin\left((2n+1)\frac{t}{2}\right)}{t}\;\mathrm dt=\\
&=&-\frac{x}{2}+\underbrace{\int_0^x \left[\frac{1}{2\sin\left(\frac{t}{2}\right)}-\frac{1}{t}\right]\sin\left((2n+1)\frac{t}{2}\right)\;\mathrm dt}_{S_{1n}(x)}+\underbrace{\int_0^x \frac{\sin\left((2n+1)\frac{t}{2}\right)}{t}\;\mathrm dt}_{S_{2n}(x)}
\end{eqnarray}
$$
According to the Riemann-Lebesgue Lemma $\lim_{n\to\infty}S_{1n}(x)=0$
Making substitution $u=(2n+1)\frac{t}{2}$ in $S_{2n}(x)$ will lead to
$$S_{2n}=\int_0^{(2n+1)\frac{x}{2}}\frac{\sin(u)}{u}\mathrm du$$
And $\lim_{n\to\infty}S_{2n}(x)=\frac{\pi}{2}$. Hence $\lim_{n\to\infty}S_{n}(x)=\lim_{n\to\infty}\sum_{k=1}^n\frac{\sin(kx)}{k}=\frac{\pi-x}{2}$.
So you sum will be $$S(x)=\frac{\pi-x}{2x}$$
Will this help you to find $S_0$?
