(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 $i!$?

  • 3
    $\begingroup$ Nice one! Never heard of that before. $\endgroup$ – Patrick Da Silva Jul 6 '13 at 5:30
  • $\begingroup$ @PatrickDaSilva Says someone with almost 15k reputation as of this writing. Now I'm worried I won't get an answer. $\endgroup$ – Lee Sleek Jul 6 '13 at 5:31
  • $\begingroup$ @LeeSleek Reputation does not always correspond to mathematical ability. I may be a prime example of this... $\endgroup$ – Potato Jul 6 '13 at 5:44
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    $\begingroup$ @Potato: It's not even a matter of ability. I remember I once flabberghasted a professor of mine with the limit $\lim_{n\to\infty}n/\sqrt[n]{n!}$. He guessed 1 and introduced it in Mathematica, upon which the program broke its teeth. It's a very slowly converging limit, but if you know about Stirling's formula, it's peanuts. But he was specialized in algebra and algebraic geometry. So, it's no surprise that even if he came across the gamma function, he would have forgotten it by that time. Anyway, I also think that Mathematica can handle that limit nowadays. $\endgroup$ – Raskolnikov Jul 6 '13 at 6:11
  • $\begingroup$ @Lee Sleek : No one knows everything, and this is not a matter of reputation. You need to assume that before you start doing mathematics, otherwise you're gonna be crying all the time. $\endgroup$ – Patrick Da Silva Jul 6 '13 at 6:52

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 $$|\Gamma(1+i)| = \sqrt{\pi \, \mathrm{csch}(\pi)}.$$

  • $\begingroup$ My second question was: can we find the exact values of the real and imaginary parts of $i!$ knowing this value? $\endgroup$ – Lee Sleek Jul 6 '13 at 15:12
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    $\begingroup$ @LeeSleek I don't think so. Maybe they are known somehow but I don't see how knowing the absolute value will help $\endgroup$ – Cocopuffs Jul 6 '13 at 17:15

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 $$|\Gamma(i)|=\left|\frac{-ie^{-\gamma i}}{\prod_{n=1}^\infty \left(1 + \frac{i}{n}\right) e^{-\frac{i}{n}}}\right|=\frac{1}{\prod_{n=1}^\infty \left|\left(1 + \frac{i}{n}\right)\right|}=\frac{1}{\sqrt{\prod_{n=1}^\infty\left(1 + \frac{1}{n^2}\right)}}=\sqrt{\frac{1}{\frac{\sinh (\pi )}{\pi }}}=\sqrt{\pi \operatorname{csch} \pi}$$

  • 2
    $\begingroup$ Хорошо господин Карандаш! $\endgroup$ – Raskolnikov Jul 6 '13 at 6:16

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