# Prove that $e^{\frac12i\,\text{gd}(\pi t)}\cos^\frac12(\text{gd}(\pi t))=\frac{1+i}{e^{\pi z}+i}e^{\frac{\pi z}2}$

This question was derived from this post about Gamma function.

Juan said:

$$\Gamma\left(\frac12+it\right)=\sqrt{\frac{\pi}{\cosh\pi t}}\exp\left\{i\left(2\vartheta(t)+t\log(2\pi)+\arctan\tanh\frac{\pi t}2\right)\right\}.$$

Sankyu kim said:

$$\Gamma\left(\frac12+iz\right)=\frac{\sqrt{\pi}(1+i)(2\pi)^{iz}}{e^{\pi z}+i}e^{\frac{\pi z}2+2i\vartheta(z)}\qquad \forall z\in\mathbb{C}.$$

Sankyu Kim derived above by generalizing Juan's answer.

If it is correct, it would help to solving my question, so I tried to follow it myself. However what made me suspicious was that no one voted on his answer. So I added comments to his answer but no gain.

Anyway while I was following Sankyu kim's method I couldn't clarify the missing link between his expressions.

I finally stuck here;

Prove that $$e^{\frac12i\,\text{gd}(\pi t)}\cos^\frac12(\text{gd}(\pi t))=\frac{1+i}{e^{\pi z}+i}e^{\frac{\pi z}2}.$$

Gudermanian function $$\text{gd}(z)$$

Riemann–Siegel theta function

By Gudermanian function's properties, $$\arctan\tanh\dfrac{πt}2=\dfrac {\text{gd}(πt)}2$$. Thus, \begin{align*} \Gamma\left(\frac 12 +i z\right) &=\frac{\sqrt{π}}{\sqrt{\cosh πz}}\exp\left\{i\left(2\vartheta(z)+z \ln{2π} + \frac{\text{gd}(\pi z)}2\right)\right\}\\ &=\frac{\sqrt{π}}{\sqrt{\cosh πz}}e^{2i\vartheta(z)}e^{iz \ln{2π}}e^{\frac12i\,\text{gd}(\pi z)}\\ &=\frac{\sqrt{π}}{\sqrt{\cosh πz}}e^{2i\vartheta(z)}{(2π)}^{iz}e^{\frac12i\,\text{gd}(\pi z)}\\ &=\frac{\sqrt{π}{(2π)}^{iz}}{\sqrt{\cosh πz}}e^{2i\vartheta(z)} e^{\frac12i\,\text{gd}(\pi z)}\\ &=\sqrt{π}{(2π)}^{iz}e^{2i\vartheta(z)}e^{\frac12i\,\text{gd}(\pi z)}{\text{sech}}^{\frac 12}(πz)\\ &=\sqrt{π}{(2π)}^{iz}e^{2i\vartheta(z)}e^{\frac12i\,\text{gd}(\pi z)}{\cos}^{\frac 12}(\text{gd}(πz)). \end{align*}

Then I stuck here.

Anyone who would lead me to the end? Thanks.

Claim: Consider the principal square root function, for any $$x\in\Bbb C$$, $$\left(e^{i\,\text{gd}\ x}\cos\text{gd}\ x\right)^{\frac12}=\frac{1+i}{e^{x}+i}e^{\frac x2}.$$
$$e^{ix}=\cos x+i\sin x,$$
we have \begin{align*} e^{i\,\text{gd}\ x}\cos\text{gd}\ x&=\left(\cos\text{gd}\ x+i\sin\text{gd}\ x\right)\cos\text{gd}\ x\\ &=\left(\ \text{sech}\ x+i\tanh x\right)\,\text{sech}\ x\\ &=\frac{1+i\sinh x}{\cosh^2x}\\ &=\frac{1}{1-i\sinh x}\\ &=\frac{2i}{e^x+2i-e^{-x}}\\ &=\frac{2i}{\left(e^{\frac x2}+ie^{-\frac x2}\right)^2}. \end{align*} Consider the principal square root function, we have $$\sqrt{2i}=1+i$$ and \begin{align*} \left(e^{i\,\text{gd}\ x}\cos\text{gd}\ x\right)^{\frac12}&=\frac{\sqrt{2i}}{\sqrt{\left(e^{\frac x2}+ie^{-\frac x2}\right)^2}}\\ &=\frac{1+i}{e^{\frac x2}+i e^{-\frac x2}}\\ &=\frac{1+i}{e^{x}+i}e^{\frac x2}. \end{align*}