Show $\prod\limits_{k=1}^{+\infty}\coth\left({\frac{\pi k}{2}}\right)=\left({\frac{2}{\pi}}\right)^{\frac{1}{4}}\Gamma\left({\frac{3}{4}}\right)$ Prove that $$\prod_{k=1}^{+\infty}\coth\left({\frac{\pi k}{2}}\right)=\left({\frac{2}{\pi}}\right)^{\frac{1}{4}}\Gamma\left({\frac{3}{4}}\right).$$
The best I was able to get: $$\ln(P)=\sum_{k=1}^{\infty}{\frac{1}{k-1/2}}\cdot{\frac{1}{e^{2(k-1/2)\pi}-1}},$$
where $P$ denotes the product. I'm not sure what to do next. Maybe it makes sense to try $${\frac{1}{a}}=\int_{0}^{1}t^{a-1}dt=\int_{0}^{\infty}e^{-a t}dt$$ or something else.
 A: Such infinite products can be evaluated using theta functions and their special values,
Define the theta function for $|{q}| < 1$,
\begin{equation}
\phi(q) = \sum_{n=-\infty}^\infty q^{n^2}
\end{equation}
One of its remarkable identities is that,
\begin{equation}
\phi(-q) = \sum_{n=-\infty}^\infty (-1)^{n}q^{n^2} = \prod_{n=1}^\infty \frac{1-q^n}{1+q^n}
\end{equation}
Now, we can rewrite the infinite product,
\begin{align}
\prod_{n=1}^\infty \coth(n\frac{\pi}{2}) = \prod_{n=1}^\infty \frac{e^{n\pi/2} + e^{-n\pi/2}}{e^{n\pi/2} - e^{-n\pi/2}} = \prod_{n=1}^\infty \frac{1 + e^{-n\pi}}{1 - e^{-n\pi}} = \frac{1}{\phi(-e^{-\pi})}
\end{align}
This is a special value of the theta function and can be written in close form with
$$\phi(-e^{-\pi}) = \Big(\frac{\pi}{2}\Big)^{1/4}\frac{1}{\Gamma\Big(\frac{3}{4}\Big)}$$
It gives us,
\begin{equation}
\prod_{n=1}^\infty \coth\Big(n\frac{\pi}{2}\Big) = \Big(\frac{2}{\pi}\Big)^{1/4}\Gamma\Big(\frac{3}{4}\Big)
\end{equation}
Note :
The special value of $\phi(-e^{-\pi})$ is not explicitly given in the link above, but can be computed from the others using an other identity of the theta function,
\begin{equation}
\phi(q^4) = \frac{1}{2}(\phi(q) + \phi(-q))
\end{equation}
which implies that,
\begin{equation}
\phi(-e^{-\pi}) = 2\phi(e^{-4\pi}) - \phi(e^{-\pi})
\end{equation}
A: The key is using the the obscure formula
$$\Gamma\Big(\frac{1}{4}\Big) = (2\pi)^\frac{3}{4}\prod_{k=1}^\infty\tanh\Big(\frac{\pi k}{2}\Big),$$
together with much more well-known formula
$$\Gamma(1-z)\Gamma(z) = \frac{\pi}{\sin(\pi z)}.$$
For $z=\frac{1}{4}$, the latter can be written as
$$\Gamma\Big(\frac{3}{4}\Big) = \frac{\sqrt{2}\pi}{\Gamma(\frac{1}{4})},$$
then substituting our spooky product for $\Gamma(\frac{1}{4})$ we get
$$\Gamma\Big(\frac{3}{4}\Big) = \frac{\sqrt{2}\pi}{(2\pi)^\frac{3}{4}\prod_{k=1}^\infty\tanh\Big(\frac{\pi k}{2}\Big)} = \Big(\frac{\pi}{2}\Big)^\frac{1}{4}\prod_{k=1}^\infty\coth\Big(\frac{\pi k}{2}\Big). $$
