Modulus of Gamma Function of Imaginary Number

Theorem
Let $t$ be a real number, then:


 * $\left\vert{\Gamma\left({i t}\right)}\right\vert = \sqrt{\dfrac {\pi \operatorname {csch} \pi t} t}$

where:
 * $\Gamma$ is the Gamma function
 * $\operatorname{csch}$ is the hyperbolic cosecant function.

Proof
By Euler's Reflection Formula:


 * $\Gamma \left({i t}\right) \Gamma \left({1 - i t}\right) = \pi \csc \left({\pi i t}\right)$

From Gamma Difference Equation:


 * $-i t \Gamma \left({i t}\right) \Gamma \left({-i t}\right) = \pi \csc \left({\pi i t}\right)$

Then:

and:

So:


 * $\left\vert{\Gamma\left({i t}\right)}\right\vert^2 = \dfrac {\pi \operatorname{csch} \left({\pi \left\vert{t}\right\vert}\right)} {\left\vert{t}\right\vert}$

As both sides are positive, we can write:


 * $\left\vert{\Gamma \left({i t}\right)}\right\vert = \sqrt {\dfrac {\pi \operatorname{csch} \left({\pi \left\vert{t}\right\vert}\right)} {\left\vert{t}\right\vert} }$

However, by Hyperbolic Sine Function is Odd:


 * $\dfrac{\pi \operatorname{csch} \left({-\pi t}\right)} {- t} = \dfrac {-\pi \operatorname{csch} \left({\pi t}\right)} {- t} = \dfrac {\pi \operatorname{csch} \left({\pi t}\right)} t$

Hence we can remove the modulus and simply write:


 * $\left\vert{\Gamma\left({i t}\right)}\right\vert = \sqrt {\dfrac{\pi \operatorname{csch} \pi t} t}$