I need to prove:
$$|(ix)!| = \frac{\pi x}{\sinh \pi x}$$
I already have this formula:
$$z!(-z)!= \frac{\pi z}{\sin \pi z}\tag{1}$$
I never used gamma for complex values. It is still valid that $\Gamma(z+1) = z\Gamma(z)$, right? So I should try and see that $(ix)! = \frac{\Gamma(ix+1)}{ix+1}$ but I don't see how that helps. Also, the way that formula $1$ is derived in my book is by using Legendre's formula for $\frac{1}{\Gamma(z)}$, so I don't think my reasoning would work for what I wanted to prove.
Or I should just use formula $1$ and apply it for the value $z = 0+ix$ and take its absolute value...
$$|(0+ix)!(0-ix)!| = \left| \frac{\pi ix}{\sin \pi ix}\right|$$
but I don't quite get $|(ix)!|$ on the left...
Recall that Euler's Reflection Formula is given by
$$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)} \tag 1$$
Letting $z=-ix$, where $x\in \mathbb{R}\setminus \{0\}$, in $(1)$ reveals
$$\begin{align} \Gamma(-ix)\Gamma(1+ix)&=\frac{\pi}{\sin(-i\pi x)}\\\\ &=i\frac{\pi}{\sinh(\pi x)}\tag 2 \end{align}$$
Next, using the functional equation $\Gamma(1+z)=z\Gamma(z)$ in $(2)$, we obtain
$$\Gamma(1-ix)\Gamma(1+ix)=\frac{\pi x}{\sinh(\pi x)}$$
Finally, denoting $\Gamma(z+1)=z!$, we see that
$$|(ix)!|^2=\frac{\pi x}{\sinh(\pi x)}$$
as was to be shown!