Closed-form of $\sum_{k=1}^{\infty}\arctan(1/k^3)$ Wolfram said that
$$\sum_{k=1}^{\infty}\arctan\left(\frac{1}{k^2}\right)=\arctan\left(\frac{1-\cot\left(\frac{\pi}{\sqrt 2}\right)\tanh\left(\frac{\pi}{\sqrt 2}\right)}{1+\cot\left(\frac{\pi}{\sqrt 2}\right)\tanh\left(\frac{\pi}{\sqrt 2}\right)}\right).$$
I was wondering about a closed-form of
$$S = \sum_{k=1}^{\infty}\arctan\left(\frac{1}{k^3}\right).$$
A numerical approximation of $S$ is
$$S \approx 0.986791652613071125515794193830247643724471031136456434\dots$$
Is there a closed-form of $S$?

Notice for any positive number $x$, we have
$$\tan^{-1}(x) = \frac{1}{2i}\log\left(\frac{1+ix}{1-ix}\right) = \Im\log(1 + ix)$$
We can rewrite the sum at hand as
$$\sum_{k=1}^\infty \tan^{-1}\frac{1}{k^3} = \Im\left[ \sum_{k=1}^\infty \log\left(1 + \frac{i}{k^3}\right) \right]
$$
For each $k$, we have the factorization
$$1 + \frac{i}{k^3} = 1 - \left(\frac{i}{k}\right)^3 = \prod_{j=0}^2 \left(1 - \frac{i\omega^j}{k}\right)$$
where $\omega = e^{i2\pi/3}$ is a primitive cubic root of unity. 
Recall the infinite product expansion of gamma function
$$\frac{1}{\Gamma(z)} = z e^{\gamma z}\prod_{k=1}^\infty \left(1 + \frac{z}{k}\right) e^{-\frac{z}{k}}
\quad\implies\quad
\prod_{k=1}^\infty \left(1+\frac{z}{k}\right)e^{-\frac{z}{k}} = \frac{e^{-\gamma z}}{\Gamma(1+z)}
$$
If we replace $z$ by $-i \omega^j$ for $j = 0,1,2$ and taking the product, we get
$$\begin{align}
\prod_{k=1}^\infty\left(1 + \frac{i}{k^3}\right)
&= \prod_{k=1}^\infty\prod_{j=0}^2\left[\left(1 - \frac{i\omega^j}{k}\right)e^{i\omega^j/k}
\right]
= \prod_{j=0}^2\prod_{k=1}^\infty\left[\left(1 - \frac{i\omega^j}{k}\right)e^{i\omega^j/k}
\right]\\
& = \frac{e^{-\gamma( -i\sum\limits_{j=0}^2 \omega^j )}}{\prod\limits_{j=0}^2\Gamma(1 - i\omega^j)}
  = \frac{1}{
\Gamma(1-i)
\Gamma\left(1 + \frac{\sqrt{3}}{2} + \frac{i}{2}\right)
\Gamma\left(1 - \frac{\sqrt{3}}{2} + \frac{i}{2}\right)}
\\
\end{align}
$$
Taking logarithm on both sides, we will get something close to what we want. 
The catch is the $\log$ function is not holomorphic over the whole complex plane.
In general, the sum of logarithms is equal to the log of the product only up
to some multiples of $2\pi$. i.e.
$$ \sum_{k=1}^\infty \log\left( 1 + \frac{i}{k^3}\right)
=  \log\left[ \prod_{k=1}^\infty \left( 1 + \frac{i}{k^3}\right) \right] + 2N\pi$$
for some unknown integer $N$. 
Instead of determining what $N$ is, we will solve this problem in a different manner.
We use the fact in the sum of the log, the imaginary part for the $k \ge 2$ terms are 
small enough.
If we remove the $k = 1$ term from the sum, the sum of log will be equal to the log of product.
At the end, we have
$$\begin{align}
\sum_{k=1}^\infty \tan^{-1}\frac{1}{k^3} 
&= \frac{\pi}{4} + \sum_{k=2}^\infty \tan^{-1}\frac{1}{k^3}
= \frac{\pi}{4} + \Im\left[\sum_{k=2}^\infty\log\left(1 + \frac{i}{k^3}\right)\right]\\
&= \frac{\pi}{4} + \Im\log\left[\prod_{k=2}^\infty\left(1 + \frac{i}{k^3}\right)\right]\\
&= \frac{\pi}{4} - \Im\log\left[(1+i)
\Gamma(1-i)
\Gamma\left(1 + \frac{\sqrt{3}}{2} + \frac{i}{2}\right)
\Gamma\left(1 - \frac{\sqrt{3}}{2} + \frac{i}{2}\right)
\right]\\
&= 
\frac{\pi}{4} - \Im\log\left[(-1+i)
\Gamma(-i)
\Gamma\left( \frac{\sqrt{3}}{2} + \frac{i}{2}\right)
\Gamma\left( - \frac{\sqrt{3}}{2} + \frac{i}{2}\right)
\right]\\
\end{align}$$
According to WA, this is approximately 
$$0.986791652613071125515794193830247643724471031136456434028974\ldots$$
as expected.
