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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.

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  • $\begingroup$ 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 '15 at 17:41
  • $\begingroup$ @PietroMajer: Myabe the question is very trivial and you can see that the answer is not trivial!!. $\endgroup$ Jan 28 '15 at 17:47
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    $\begingroup$ 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 '15 at 18:04
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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.

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  • $\begingroup$ @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 '12 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 '12 at 15:25
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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,

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

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    $\begingroup$ @ Sangkyu Kim: But the gamma function has no closed form. $\endgroup$ Jul 14 '14 at 8:04
  • $\begingroup$ @ Sangkyu Kim: It has no an explicit formula. $\endgroup$ Jul 16 '14 at 8:24
  • $\begingroup$ @China-HongKong So do you mean his answer is somewhere wrong? $\endgroup$ Sep 22 '18 at 12:03
  • $\begingroup$ What is Euler chi function? $\endgroup$ Sep 27 '18 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 '18 at 5:03

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