TITLE: Proving $\prod_{n=1}^{\infty}\left(1+\frac{1}{n^{3}}\right)=\frac{1}{\pi}\cosh\frac{\pi\sqrt{3}}{2}$
QUESTION [19 upvotes]: First I rewrite $\prod_{n=1}^{\infty}\left(1+\frac{1}{n^{3}}\right)$ as $\prod_{n=1}^{\infty}\left(\frac{1+n^{3}}{n^{3}}\right)$, then by factor out polynomial I get $\prod_{n=1}^{\infty}\left(\frac{(1+n)(n^{2}-n+1)}{n^{3}}\right)$ which is a problem because I can't factor any further which makes me stuck on this step I would hope for any help.

REPLY [9 votes]: Use $$\frac{e^{-\gamma z}}{\Gamma(1+z)}=\prod_{n=1}^{\infty} (1+\frac{z}{n})e^{-z/n}~~~~(1)$$
Next, $$\left(1+\frac{1}{n^3}\right)=\left(1+\frac{1}{n}\right) \left(1+\frac{w}{n}\right) \left(1+\frac{w^2}{n}\right),$$
where $w$ is cube root of unity $w^3=1, w^2+w+1=0, w^2=w^*$
$$P=\prod_{1}^{\infty} \left(1+\frac{1}{n^3}\right)$$
From (1), we get
$$ e^{-(1+w+w^2)\gamma}[\Gamma(2)\Gamma(1+w)\Gamma(1+w^2)]^{-1}= [\Gamma(-w) \Gamma(-w^2)]^{-1}=\prod_{n=1}^{\infty} \left(1+\frac{1}{n^3}\right) e^{(1+w+w^2)/n}$$
So $$P=\frac{1}{|\Gamma(-w)|^2}=$$
$$P=\frac{1}{\Gamma[(1-i\sqrt{3})/2]~ \Gamma[(1+i\sqrt{3})/2]}$$
Use  $\Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin \pi z}$ to evaluate $\Gamma(1/2-it) \Gamma(1/2+it)=\frac{\pi}{\cosh \pi t}$. Then the required $$P=\Gamma(1/2-it) \Gamma(1/2+it), t=\pi\sqrt{3}/2$$
Hence $$P=\frac{1}{\pi} \cosh(\sqrt{3}\pi/2).$$.