How can we calculate or estimate the series $\sum_{n=1}^{\infty}{\frac{\sin(nx)}{n-a}}$ from above? I thought that $\sum_{n=1}^{\infty}{\frac{\sin(nx)}{n-a}}$ could be represented as something involving $\sum_{n=1}^{\infty}{\frac{\sin(nx)}{n}}$, for which the value is known, but I didn't manage to represent it in a form that I would be able to calculate. Is there a way to estimate it from above or calculate it precisely without using difficult methods?
It's important to mention that $a \in \mathbb R$ and $0 < a < 1$.
EDIT.
Difficult methods include computing special functions such as, for example, the Lerch transcendent function or the Clausen function, mentioned in answers.
 A: $$\sum_{n=1}^{\infty}{\frac{\sin(nx)}{n-a}}=\Im\left(\sum_{n=1}^{\infty}{\frac{e^{inx}}{n-a}}\right)=\Im\left(\sum_{n=1}^{\infty}{\frac{[e^{ix}]^n}{n-a}}\right)=\Im\left(\sum_{n=1}^{\infty}{\frac{t^n}{n-a}}\right)$$ with $t=e^{ix}$
There is a special function
$$\sum_{n=1}^{\infty}{\frac{t^n}{n-a}}=t\, \Phi (t,1,1-a)$$ where appears  the Hurwitz-Lerch transcendent function and if $a=0$ it gives $-\log(1-t)$.
So,
$$\color{blue}{\sum_{n=1}^{\infty}{\frac{\sin(nx)}{n-a}}=\Im\Big[e^{i x} \Phi \left(e^{i x},1,1-a\right)\Big]}$$
Let us try it for $x=\frac \pi {12}$ to see the impact of $a$
$$\left(
\begin{array}{cc}
a & \sum_{n=1}^{\infty}{\frac{\sin \left(n\frac{\pi  }{12}\right)}{n-a}}\\
 0.0 & 1.43990 \\
 0.1 & 1.50531 \\
 0.2 & 1.58060 \\
 0.3 & 1.66924 \\
 0.4 & 1.77679 \\
 0.5 & 1.91305 \\
 0.6 & 2.09718 \\
 0.7 & 2.37321 \\
 0.8 & 2.87165 \\
 0.9 & 4.24105
\end{array}
\right)$$
Edit
If you cannot use the Hurwitz-Lerch transcendent function, you can use the expansion
$$\sum_{n=1}^{\infty}{\frac{t^n}{n-a}}=-\log(1-t)+\sum_{k=1}^\infty \text{Li}_{k+1}(t)\,a^k$$
