Can someone please explain $|i!|^2 = \frac{\pi}{\sinh(\pi)}$

Question

It links to the gamma function, but how does it link over to pi?

Answer 1

First, $z!=\Gamma(z+1)$. So $i!=\Gamma(1+i)$.

For all $z$, $\Gamma(z)\Gamma(1-z)=\dfrac{\pi}{\sin \pi z}$, and $\Gamma(z+1)=z\,\Gamma(z)$, and $\overline{\Gamma(z)}=\Gamma(\overline z)$.

(see https://en.wikipedia.org/wiki/Gamma_function)

Hence

$$\Gamma(1+i)\Gamma(-i)=\frac{\pi}{\sin (-\pi i)}=\frac{\pi}{-i\sinh \pi}=\frac{i\pi}{\sinh \pi}$$

And

$$\Gamma(-i)=\overline{\Gamma(i)}=\frac{\overline{\Gamma(i+1)}}{\overline i}=i\,\overline{\Gamma(i+1)}$$

Therefore

$$\Gamma(1+i)\Gamma(-i)=i|\Gamma(1+i)|^2=\frac{i\pi}{\sinh \pi}$$

And finally

$$|i!|^2=\frac{\pi}{\sinh \pi}$$