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The question is almost the same as here.

What is the upper bound for a complex Beta function $$\DeclareMathOperator{\Im}{Im}\DeclareMathOperator{\Re}{Re} \displaystyle B(s,z)=\frac{\Gamma(s) \Gamma(z)}{\Gamma(s+z)} $$ where $0<\Re(s)<1$, $0<\Re(z)<1$, $\Im(s)>10$ and $\Im(z)>10$ ?

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  • $\begingroup$ Why the choice of 10? I assumed that you wanted an upper bound or asymptotics as a function of Im(s) and Im(z) $\endgroup$
    – Yemon Choi
    Commented Oct 18, 2021 at 21:53
  • $\begingroup$ @Yemon Choi values start at ~14. non-trivial zeros of the Riemann zeta function $\endgroup$
    – user363337
    Commented Oct 18, 2021 at 22:04

1 Answer 1

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The command of Mathematica

NMaximize[{ComplexExpand[Abs[Gamma[s + I*t]*Gamma[a + I*b]/Gamma[s + a + I*(t + b)]]],s > 0 && a > 0}, {s, t, a, b}]

outputs "NMaximize::cvdiv: Failed to converge to a solution. The function may be unbounded." and

{4.7857*10^8, {s -> 0.32798, t -> 0.720557, a -> 0., b -> -2.08956*10^-9}}

The command of Mathematica

s = 1/2; t = a; MaxLimit[ ComplexExpand[Abs[Gamma[s + I*t]*Gamma[a + I*b]/Gamma[s + a + I*(t + b)]]], {t,b} -> {0, 0},Direction -> "FromAbove"]

results in $\infty$, confirming it.

Addition. Here is the answer with Mathematica to the modified question (Thanks to Brendan McKay who paid my attention to the edit.)

NMaximize[{ComplexExpand[ Abs[Gamma[s + I*t]*Gamma[a + I*b]/
Gamma[s + a + I*(t + b)]]],  s > 0 && a > 0 && s < 1 && a < 1 && 
b > 10 && t > 10}, {s, t, a, b}]

{1.121, {s -> 3.70155*10^-7, t -> 10., a -> 3.57795*10^-7, b -> 10.}}

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  • $\begingroup$ Thanks, but I'm not asking about the limit, but about the absolute value. $\endgroup$
    – user363337
    Commented Oct 18, 2021 at 7:13
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    $\begingroup$ @user363337: The second result means that $$ \left| \frac{\Gamma (a+b i) \Gamma (s+i t)}{\Gamma (a+(b+t) i+s)}\right|$$ is unbounded on the set $a>0,a<1,s>0,s<1$ . Don't hesitate to ask for further explanation in need. $\endgroup$
    – user64494
    Commented Oct 18, 2021 at 7:30
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    $\begingroup$ While I don't know if Mathematica is right or wrong, I don't think the results of executing commands like this form a proof. $\endgroup$
    – Brendan McKay
    Commented Nov 8, 2023 at 12:59
  • $\begingroup$ @BrendanMcKay: Can you ground your claim? Are you aware of Four colors theorem? $\endgroup$
    – user64494
    Commented Nov 8, 2023 at 15:40
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    $\begingroup$ I use computers to prove things all the time. Symbolic algebra systems make plenty of mistakes, especially when the problem does not have a clean deterministic algorithm such as this 4-dimensional optimisation. Anyway, the OP added lower bounds on the imaginary parts (I think after your answer) that you haven't used. $\endgroup$
    – Brendan McKay
    Commented Nov 8, 2023 at 23:05

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