-4
$\begingroup$

I ask about an idea to prove this formula:

$Γ(1/2-iβ)=((\sqrt{π})/(\sqrt{\coshπβ}))\exp(-i(2ϑ(β)+βln2π+\arctan(\tanh(1/2)πβ)))$

where $ϑ(β)$ is the Riemann Siegel function.

$\endgroup$
3
  • $\begingroup$ I'm curious about this unusual phenomenon (yet expected, by the "Reversal" badge) of +15 points difference between question and answer! Some hints? $\endgroup$ Jan 28, 2015 at 17:41
  • $\begingroup$ @PietroMajer: Myabe the question is very trivial and you can see that the answer is not trivial!!. $\endgroup$ Jan 28, 2015 at 17:47
  • 2
    $\begingroup$ Yes, but if a question originates a non trivial and interesting answer, it shouldn't be that bad. Anyway, I respect your sportsman-like attitude. $\endgroup$ Jan 28, 2015 at 18:04

2 Answers 2

7
$\begingroup$

I know two proof, the first uses $$\cos\frac{\pi s}{2}=\frac{1}{\sqrt{2}}\sqrt{\cosh(\pi t)}\,e^{-i\arctan(\tanh\frac{\pi t}{2})}.\qquad (1)$$ and $$\Gamma(\frac12+i\frac t 2)=|\Gamma(\frac14+i\frac t2)|\,e^{i(\vartheta(t)+\frac t 2\log\pi)},\qquad (2)$$

Since $$\Gamma(z)\Gamma(z+1/2)=2^{1-2z}\sqrt{\pi}\Gamma(2z);\quad \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin\pi z}$$ we get $$\Gamma(2z)=2^{2z-1}\pi^{-1/2} \Gamma(z)\frac{\pi}{\cos\pi z}\frac{1}{\Gamma(1/2-z)}.$$ We put now $z=\frac14+i\frac t 2$, $t$ real $$ \Gamma(\frac12+it)=2^{-\frac12+it}\frac{\pi^{\frac12}}{\cos\pi(\frac14+i\frac{t}{2})} \frac{\Gamma(\frac14+i\frac{t}{2})}{\Gamma(\frac14-i\frac{t}{2})}. $$ From (1) and (2) we get $$\Gamma(1/2+it)=2^{it}\frac{\pi^{1/2}}{\sqrt{\cosh \pi t}\; e^{-i \arctan\tanh(\pi t/2)}}\pi^{it} e^{2i\vartheta(t)}. $$ so that $$ \Gamma(1/2+it)=\sqrt{\frac{\pi}{\cosh\pi t}}\exp\bigl\{i(2\vartheta(t)+t\log(2\pi)+\arctan\tanh(\pi t/2))\bigr\} $$

The other proof I know uses the functional equation of the zeta function.

$\endgroup$
2
  • $\begingroup$ @Juan: Note that all the identities concerning arguments holds only modulo factors of 2π if the argument is being restricted to (−π,π] . How you can deal with arguments figured in this proof. This is a problem since these arguments depend on the variable t . $\endgroup$ Dec 16, 2012 at 9:38
  • $\begingroup$ @RH The argument of $\Gamma(s)$ is well defined and harmonic on the plane with a cut along the negative real axis. $\endgroup$
    – juan
    Dec 16, 2012 at 15:25
1
$\begingroup$

I extended juan's proof to a formula for all complex numbers.

Consequently, we get the Euler chi function $\chi(z):=\zeta(1-z)/\zeta(z)$.

(substitution)

$\tan^{-1}(\tanh(\pi t/2))\longrightarrow\frac{\mathrm{gd}(\pi t)}2$, where $\mathrm{gd}(z)$ is the Gudermannian function.

$\sqrt{\frac{\pi}{\cosh(\pi t)}}\longrightarrow\frac{\sqrt{\pi}}{\cosh(\pi t)}\left(\sinh(\frac{\pi t}2)\sin(\frac{\mathrm{gd}(\pi t)}{2})+\cosh(\frac{\pi t}{2})\cos(\frac{\mathrm{gd}(\pi t)}2)\right)$

$t\longrightarrow z$

(simplification) If we covert the above result(trigonometric functions) to exponentials, then we get the following simplified formula.

$\Gamma(\frac12+iz)=\frac{\sqrt{\pi}(1+i)(2\pi)^{iz}}{e^{\pi z}+i}e^{\frac{\pi z}2+2i\vartheta(z)}/;z\in\mathbb{C}$

Therefore,

$\Gamma(z)=\frac{(2\pi)^z}{1+e^{i\pi z}}e^{i(2\vartheta(\frac{i}2-iz)+\frac{\pi z}2)}=\frac{(2\pi)^z}{2\cos(\frac{\pi z}2)}e^{2i\vartheta(\frac{i}2-iz)}=\frac{(2\pi)^z}{2\cos(\frac{\pi z}2)}\chi(z)$

$\endgroup$
6
  • 1
    $\begingroup$ @ Sangkyu Kim: But the gamma function has no closed form. $\endgroup$ Jul 14, 2014 at 8:04
  • $\begingroup$ @ Sangkyu Kim: It has no an explicit formula. $\endgroup$ Jul 16, 2014 at 8:24
  • $\begingroup$ @China-HongKong So do you mean his answer is somewhere wrong? $\endgroup$ Sep 22, 2018 at 12:03
  • $\begingroup$ What is Euler chi function? $\endgroup$ Sep 27, 2018 at 13:14
  • $\begingroup$ How do you prove that $\sqrt{\frac{\pi}{\cosh(\pi t)}}\longrightarrow\frac{\sqrt{\pi}}{\cosh(\pi t)}\left(\sinh(\frac{\pi t}2)\sin(\frac{\mathrm{gd}(\pi t)}{2})+\cosh(\frac{\pi t}{2})\cos(\frac{\mathrm{gd}(\pi t)}2)\right)$ ? $\endgroup$ Oct 1, 2018 at 5:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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