Modulus of Gamma Function of One Half plus Imaginary Number
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Theorem
Let $t \in \R$ be a real number.
Then:
- $\cmod {\map \Gamma {\dfrac 1 2 + i t} } = \sqrt {\pi \map \sech {\pi t} }$
where:
- $\Gamma$ is the Gamma function
- $\sech$ is the hyperbolic secant function.
Proof
| \(\ds \cmod {\map \Gamma {\frac 1 2 + i t} }^2\) | \(=\) | \(\ds \map \Gamma {\frac 1 2 + i t} \overline {\map \Gamma {\frac 1 2 + i t} }\) | Modulus in Terms of Conjugate | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \Gamma {\frac 1 2 + i t} \map \Gamma {\frac 1 2 - i t}\) | Complex Conjugate of Gamma Function | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \Gamma {\frac 1 2 + i t} \map \Gamma {1 - \paren {\frac 1 2 + i t} }\) | applying some algebra | |||||||||||
| \(\ds \) | \(=\) | \(\ds \pi \map \csc {\pi \paren {\frac 1 2 + i t} }\) | Euler's Reflection Formula | |||||||||||
| \(\ds \) | \(=\) | \(\ds \pi \map \sec {\pi i t}\) | Sine of Complement equals Cosine | |||||||||||
| \(\ds \) | \(=\) | \(\ds \pi \map \sech {\pi t}\) | Hyperbolic Cosine in terms of Cosine |
As $\cmod z \ge 0$ for all complex numbers $z$, we can take the non-negative square root of both sides and write:
- $\cmod {\map \Gamma {\dfrac 1 2 + i t} } = \sqrt {\pi \map \sech {\pi t} }$
$\blacksquare$