Calculate $\int_{0}^\infty\frac{dx}{\left(1+\frac{x^3}{1^3}\right)\left(1+\frac{x^3}{2^3}\right)\left(1+\frac{x^3}{3^3}\right)\ldots}$ I'm interested in the integral
$$
I=\int_{0}^\infty\frac{dx}{\left(1+\frac{x^3}{1^3}\right)\left(1+\frac{x^3}{2^3}\right)\left(1+\frac{x^3}{3^3}\right)\ldots}.\tag{1}
$$
So far I have been able to reduce this integral to an integral of an elementary function in the hope that it will be more tractable
$$
I=\frac{8\pi}{\sqrt{3}}\int_{-\infty}^\infty\frac{e^{ix\sqrt{3}}\ dx}{\left(e^x+e^{-x}+e^{ix\sqrt{3}}\right)^3},\tag{2}
$$
using the approach from this question. In that question it was also proved that
$$
\int_{-\infty}^\infty\frac{dx}{\left(e^x+e^{-x}+e^{ix\sqrt{3}}\right)^2}=\frac{1}{3},\tag{3}
$$
which gives some indication that the integral in the right hand side of $(2)$ might be calculable.
Also note that the integrand in $(1)$ can be expressed as 
$$
\Gamma(x+1)\left|\Gamma\left(1+e^{\frac{2\pi i}{3}}x\right)\right|^2.
$$
Bending the contour of integration in the integral on the RHS of $(2)$ one obtains an alternative representation
$$
I=8\pi\int_0^\infty\frac{e^{x\sqrt{3}}~dx}{\left(2\cos x+e^{x\sqrt{3}}\right)^3}.\tag{4}
$$
There are some calculable integrals containing the infinite product $\prod\limits_{k=1}^\infty\left(1+\frac{x^3}{k^3}\right)$, e.g.
$$
\int_{0}^\infty\frac{\left(1-e^{\pi\sqrt{3}x}\cos\pi x\right)e^{-\frac{2\pi}{\sqrt{3}}x}\ dx}{x\left(1+\frac{x^3}{1^3}\right)\left(1+\frac{x^3}{2^3}\right)\left(1+\frac{x^3}{3^3}\right)\ldots}=0.
$$

Q: Is it possible to calculate $(1)$ in closed form?

 A: define:
$$ Q^n(x) = \prod_{a=1}^n{1-\left(\frac{x}{a}\right)^3} 
          = \left(1-\left(\frac{x}{a}\right)^3\right)
                 \cdot\prod_{b\ne a}{1-\left(\frac{x}{b}\right)^3} 
          = \left(1-\left(\frac{x}{a}\right)^3\right)\cdot Q_a(x)
$$
using:
$$\begin{align*}
(a) && & 1-\left(\frac{x}{a}\right)^3 = 0 
             \Leftrightarrow x=az_i, \; (z_i)^3=0.\; i=0,1,2 \\
(b) && & \frac{d}{dx}\left(1-\left(\frac{x}{a}\right)^3\right) 
             = -3\frac{x^2}{a^2}; \; (x=az_i)\; -3\frac{\bar{z}_i}{a} \\
(c) && & (b\ne a) \Rightarrow1-\left(\frac{az_i}{b}\right)^3
             = 1-\left(\frac{a}{b}\right)^3 \\
(d) && & \frac{d}{dx}Q^n(x)=\left(1-\left(\frac{x}{a}\right)^3\right)Q'_a(x) 
             - 3\frac{x^2}{a^3}Q_a(x) \\
(a,b,c,d) \Rightarrow (e) && & Q'(az_i)=0-3\frac{\bar{z}_i}{a}Q_a(az_i) 
             = -3\frac{\bar{z}_i}{a}
               \prod_{b\ne a}{\left(1-\left(\frac{a}{b}\right)^3\right)}
             = -3\frac{\bar{z}_i}{a}P(a)
\end{align*}$$

$$\begin{align*}
\frac1{Q^n(x)} && & \stackrel{pfd}{=} 
           \sum_{a=1}^n{\sum_{i=0}^3{\frac1{(x-az_i)\cdot Q'_a(x)}}} \\
&& & \stackrel{(e)}{=} \sum_{a=1}^n{\sum_{i=0}^3{\frac1{-3\bar{z_i}(x-az_i)}
           \cdot \frac{a}{P(a)}}} \\
&& & = \sum_{a=1}^n{\frac{a}{P(a)}\sum_{i=0}^3{\frac{z_i}{-3(x-az_i)}}}\\
&& & = \sum_{a=1}^n{\frac{a}{P(a)}\cdot\frac{a^2}{a^3-x^3}}\\ 
\end{align*}$$

$$\begin{align*} 
\int_0^{\infty}{\frac1{Q_n(-x)}dx} && & = \int_0^{\infty}{
           \sum_{a=1}^n{\frac{a}{P(a)}\cdot\frac{a^2}{a^3+x^3}}dx} \\
&& & = \sum_{a=1}^n{\frac{a}{P(a)}\cdot\int_0^{\infty}{\frac{a^2}{a^3+x^3}dx}}\\
&& & = \sum_{a=1}^n{\frac{a}{P(a)}\cdot\frac{2\pi}{3\sqrt{3}}} \\
\end{align*}$$

$$\begin{align*} 
\lim_{n\to\infty}\int_0^{\infty}{\frac1{Q_n(-x)}dx} && &
  = \lim_{n\to\infty}\sum_{a=1}^n{\frac{a}{P(a)}\cdot\frac{2\pi}{3\sqrt{3}}}\\
&& & = \frac{2\pi}{3\sqrt{3}}\lim_{n\to\infty}\sum_{a=1}^n{\frac{a}{P(a)}}\\
\end{align*}$$

Answer:
$$\frac{2\pi}{3\sqrt{3}}\sum_{a=1}^{\infty}{a\prod_{b\ne a}{\left(1-\left(\frac{a}{b}\right)^3\right)^{-1}}}$$
