How do we know that $|i!| = \sqrt{\pi \operatorname{csch} \pi}$?

Question

(Source: Wolfram Alpha)

Or, to write it out in full,

$$|i!| = \sqrt{\frac{2\pi e^\pi}{e^{2\pi} - 1}}$$

How is this identity derived? Also, knowing this, could we find the exact values for the real and imaginary parts of $i!$?

Answer 1

Recall the functional equation (Euler's reflection formula) $$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}.$$ Set $z = 1+i$; then, we get $$\Gamma(1+i)\Gamma(-i) = \frac{\pi}{\sin(\pi i)} = \pi \, \mathrm{csch}(\pi).$$ On the other hand, $\Gamma(-i) = -i\Gamma(1-i) = \overline{i \Gamma(1+i)}$, and so $|\Gamma(i+1)| = |\Gamma(-i)|$; therefore $$|\Gamma(1+i)| = \sqrt{\pi \, \mathrm{csch}(\pi)}.$$

Answer 2

First of all one can make use of Gamma function representation: $$|i!|=|\Gamma(i+1)|=|\Gamma(i)\ i|=|\Gamma(i)|$$ Following the definition: $$\frac{1}{\Gamma(z)}=ze^{\gamma z} \prod_{n=1}^\infty \left(1 + \frac{z}{n}\right) e^{-\frac{z}{n}}$$ where $\gamma$ is the Euler–Mascheroni constant.
So $$|\Gamma(i)|=\left|\frac{-ie^{-\gamma i}}{\prod_{n=1}^\infty \left(1 + \frac{i}{n}\right) e^{-\frac{i}{n}}}\right|=\frac{1}{\prod_{n=1}^\infty \left|\left(1 + \frac{i}{n}\right)\right|}=\frac{1}{\sqrt{\prod_{n=1}^\infty\left(1 + \frac{1}{n^2}\right)}}=\sqrt{\frac{1}{\frac{\sinh (\pi )}{\pi }}}=\sqrt{\pi \operatorname{csch} \pi}$$