# The derivative of the Gamma function

The Gamma function satisfies the relation $$z\Gamma(z)=\Gamma(z+1)$$, whence $$|\Gamma(z+1)|>|\Gamma(z)|$$ whenever $$|z|>1$$ (and $$z$$ is not a non-positive integer). This naturally leads us to the idea that for a fixed $$y>1$$ the function $$|\Gamma(x+iy)|$$ of the variable $$x$$ could be increasing. Is this true? I would much appreciate a reference or a proof. The area $$\{x+iy: \ x>1, \ y<1\}$$ is of interest as well.

• What is $G$? Is it supposed to be $\Gamma$? – Alex Ortiz Oct 29 '18 at 18:34
• Yes. Edited. Thanks. – Durac Oct 29 '18 at 18:36
• $$|\Gamma(x+iy)| \sim \sqrt{2 \pi} \, |y|^{x-1/2} \exp{(-\frac{\pi}{2}|y|)} , \quad y \to \infty$$ can be found in Gradshteyn and Ryzhik, eq. 8.328, and a reference given therein. Eq. 8.326.2 might be of interest to you as well. – skbmoore Oct 29 '18 at 18:43
• @skbmoore: IMO your hint does not apply here, because $y$ should be fix. – gammatester Oct 29 '18 at 18:48
• – AccidentalFourierTransform Oct 29 '18 at 18:57

This is wrong. Here a plot for $$|\Gamma(x+1.01 \cdot i)|$$
$$|\Gamma(0.2+1.01\cdot i)|= 0.516403048428670750862809525289$$ $$|\Gamma(0.6+1.01\cdot i)|= 0.510624247985962860913065928364$$
• Thanks, you are right. But in fact I am interested in this property for sufficiently large $y$. What should I do then? Accept this answer and open a new question? – Durac Oct 29 '18 at 19:08
• It is your decision. (The critical $|y|$ seems to be very close to $1$). – gammatester Oct 29 '18 at 19:14
• I'd say there is a convex curve $x+i\, f(x)$ with $f(-\infty) = +\infty, f(+\infty) = 0$ such that for every $z, \Im(z) \ge f(\Re(z))$ then $|\Gamma(z+x)|$ is strictly increasing in $x > 0$. Thus a question would be the asymptotic of the optimal $f$. @Durac – reuns Oct 29 '18 at 22:07