2
$\begingroup$

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

Materials that will help you

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

Riemann–Siegel theta function

Details about what I did

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.

$\endgroup$
2
$\begingroup$

Claim: Consider the principal square root function, for any $x\in\Bbb C$, $$\left(e^{i\,\text{gd}\ x}\cos\text{gd}\ x\right)^{\frac12}=\frac{1+i}{e^{x}+i}e^{\frac x2}.$$

Recall the Euler's formula

$$e^{ix}=\cos x+i\sin x,$$

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*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.