I am given the impression that the following is true (for at least all positive $\lambda$ - may be even true for any complex $\lambda$)
$$ \left\lvert \frac{\Gamma(i\lambda + 1/2)}{\Gamma(i\lambda)} \right\rvert^2 = \lambda \tanh (\pi \lambda) $$
It would be great if someone can help derive this.
Using the Euler's reflection formula $$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin\pi z},$$ we get (for real $\lambda$) \begin{align} \left|\frac{\Gamma\left(\frac12+i\lambda\right)}{\Gamma(i\lambda)}\right|^2&= \frac{\Gamma\left(\frac12+i\lambda\right)\Gamma\left(\frac12-i\lambda\right)}{\Gamma(i\lambda)\Gamma(-i\lambda)}=\\ &=(-i\lambda) \frac{\Gamma\left(\frac12+i\lambda\right)\Gamma\left(\frac12-i\lambda\right)}{\Gamma(i\lambda)\Gamma(1-i\lambda)}=\\ &=(-i\lambda)\frac{\pi/\sin\pi\left(\frac12-i\lambda\right)}{\pi/\sin\pi i\lambda}=\\ &=-i\lambda\frac{\sin \pi i \lambda}{\cos\pi i \lambda}=\\ &=\lambda \tanh\pi \lambda. \end{align} This will not hold if $\lambda$ is complex.