Let $$ D=\{z \in \mathbb{C} : 0<\Re(z)<\frac{1}{2} \text{ or } \frac{1}{2}<\Re(z)<1 \} $$ that is the critical strip without critical line.

I have to find if the following equation, with Gamma Euler function, has any root in $D$ $$ \Gamma(z) = \dfrac {\pi^z} {\cos \left( \dfrac{\pi}{2} \cdot z \right) \cdot 2^{1-z} } \cdot \dfrac {1-2^{1-z}} {1-2^z} $$



Imprecise answer: if you denote by $f(z)$ the log of the ratio of the two sides, we have $f(1-z)=-f(z)$ (I assume you constructed your function in that way). One now uses an old theorem of Hermite, unfortunately I don't remember the exact statement and reference (he states it for polynomials, but it is easily generalized) which shows that all the zeros have real part $\ge1/2$, hence by the functional equation also $\le1/2$, so are on the critical strip. Sorry to be so imprecise (same proof applies to functions such as $\Lambda(s+a)+\Lambda(s-a)$ with $a\ge1/2$ for instance, with $\Lambda(s)=\pi^{-s/2}\Gamma(s/2)\zeta(s)$).

  • $\begingroup$ I'm interested to proof if my equation has got any root in $D$. By your answer I don't understand if there are surely roots in $D$ or it's just a chance. $\endgroup$ – Matey Math Mar 4 at 22:11
  • $\begingroup$ If my observation of the use of Hermite's theorem is correct, this implies that there are NO roots in $D$ since you exclude the critical line. $\endgroup$ – Henri Cohen Mar 5 at 9:20
  • $\begingroup$ thanks Henri Cohen for your answer and your further explanations $\endgroup$ – Matey Math Mar 5 at 12:25

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