Evaluation of series $\sum \limits_{n=1}^{\infty} \frac{\arctan n}{n^2}$ Can we find a closed form for the series:
$$\mathcal{S}=\sum_{n=1}^{\infty} \frac{\arctan n}{n^2}$$
Here is some basic manipulation I did:
\begin{align*}
\sum_{n=1}^{\infty} \frac{\arctan n}{n^2} &=\sum_{n=1}^{\infty} \frac{1}{n^2} \sum_{k=0}^{\infty} (-1)^{k-1} \frac{n^{2k+1}}{2k+1} \\ 
 &= \sum_{k=0}^{\infty}\frac{(-1)^{k-1}}{2k+1} \sum_{n=1}^{\infty} \frac{n^{2k+1}}{n^2}\\ 
 &= \sum_{k=0}^{\infty} \frac{(-1)^{k-1} \zeta(1-2k)}{2k+1}
\end{align*}
Now unfortunately I don't know how to proceed further. Maybe the sum does not admit a closed form and all that I have done so far be in vain. Another way would be :
$$\sum_{n=1}^{\infty} \frac{\arctan n}{n^2} =\mathfrak{Im} \left ( \sum_{n=1}^{\infty} \frac{\ln (1+in)}{n^2} \right )$$
Now I don't know how to sum the second series! Any help?
 A: The given  series admits a closed form in terms of the poly-Stieltjes constants defined here.

Proposition. One has
  $$
\sum_{n=1}^{\infty} \frac{\arctan n}{n^2}=\frac{\pi^3}{12}+\frac1{2i}\gamma'_1(i,0)-\frac1{2i}\gamma'_1(-i,0)
$$ 

where $\gamma'_1(a,0)$ denotes $\displaystyle \left.\partial_b\gamma_1(a,b)\right|_{b=0}$ and where 
$$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right). 
$$
Proof. We are allowed to perform a term by term differentiation$^1$ and to use Theorem 2 obtaining
$$
\begin{align}
\sum_{n=1}^{\infty} \frac{\arctan n}{n^2}&=\sum_{n=1}^{\infty} \frac1{n^2}\left(\frac{\pi}2-\arctan \frac1n \right)
\\&=\frac{\pi}2\sum_{n=1}^{\infty}\frac1{n^2}-\frac1{2i}\sum_{n=1}^{\infty}\frac1{n^2}\left(\log(n+i)-\log(n-i)\right)
\\&=\frac{\pi^3}{12}+\frac1{2i}\left.\partial_b\sum_{n=1}^{\infty}\frac1{n+b}\left(\log(n+i)-\log(n-i)\right)\right|_{b=0}
\\&=\frac{\pi^3}{12}+\frac1{2i}\left.\partial_b\frac{}{}\left(\gamma_1(i,b)-\gamma_1(-i,b)\right)\right|_{b=0}
\\&=\frac{\pi^3}{12}+\frac1{2i}\gamma'_1(i,0)-\frac1{2i}\gamma'_1(-i,0).
\end{align}
$$

$^1$Assume $b\ge0$ and $n\ge1$. Each function $b \mapsto f_n(b):=\frac1{n+b}\left(\log(n+i)-\log(n-i)\right)$ is differentiable over $(-1,\infty)$, the series $\sum f_n(b)$ is uniformly (normally) convergent over $[0,\infty)$, there exists a constant $c_1$ such that: $$\left|f_n(b)\right|=\left|\frac1{n+b}\left(\log(n+i)-\log(n-i)\right)\right|\le\frac{c_1}{n^2}, \quad n \to \infty,$$ and the series $\sum f'_n(b)$ is uniformly (normally) convergent over $[0,\infty)$, there exists a constant $c_2$ such that: $$\left|f'_n(b)\right|=\left|\frac1{(n+b)^2}\left(\log(n+i)-\log(n-i)\right)\right|\le\frac{c_2}{n^3}, \quad n \to \infty.$$

Remark. The proposition above may easily be generalized, giving new closed forms for a vast family of integrals. For example, one deduces a closed form for @nospoon's integral
$$
\int_0^{\infty}\frac{\sin x}{x}\text{Li}_2(e^{-x})dx=\text{Im}\:\gamma'_1(-i,0)
$$ 
using the standard result $\displaystyle \int_0^{\infty} \frac{\sin x}{x}e^{-nx}dx=\arctan \frac1n, \, n>0.$
