# Inequality for a gamma function

Let $s=\sigma+it$ and $\Gamma(s)$ be the Euler gamma function. Does the inequality hold? $$\left|\frac{\Gamma(s)}{\Gamma(2-s)}\right|\leq |s|^{2(\sigma-1)},\, 1<\sigma<2,\, t\in \mathbb{R}.$$ Difficulties to prove inequality appears when $\sigma$ approximates 1.

Such inequality appeared studying a zeta functions of a second order. Namely, comparing the values of the Selberg zeta-function for the modular subgroup $PSL(2,\mathbb{Z})$ across the critical line: |Z(1-s)|>|Z(s)| (|Z(1-s)|<|Z(s)| ?), $1/2<\sigma<1$.

We can show that $$\left|\frac{\Gamma(s)}{\Gamma(2-s)}\right|\leq \left|s-2-\frac{\sqrt{2}}{2}\right|^{2(\sigma-1)},\, 1<\sigma<2,\, t\in \mathbb{R}.$$ See the Lemma 7 in http://link.springer.com/article/10.1007%2Fs00025-015-0486-7#page-1

However, the same technic doesn't work for the first inequality.

• The answer is yes. I have googled "gamma function inequality complex argument" and was pointed to a paper (arxiv.org/pdf/1301.1749.pdf) by Ismail and Muldoon (2011). The inequality (in slightly modified form) is given in Eq.(5.1). The proof (according to the paper) can be found in "Topics on Analytic Number Theory" by Hans Rademacher (Springer, 1973) on pp 68-70. Mar 25, 2014 at 11:20
• The logarithmic derivative of the Selberg Zeta function grows like $CT^2$ as $\Im z = T \rightarrow \infty$, which can be seen from the Weyl law. More important for its growth is the Barnes-G-function. $\Gamma$ contributes at most $T \log(T)$ in the non-compact setting. Apr 8, 2014 at 14:55

1. I am not sure this estimate is true. In the cited preprint signs of s are the same, yours are opposite.
2. Standard inequalities gives not power but exponential growth $$|\frac{\Gamma(s)}{\Gamma(2-s)}|\le \frac{1}{\pi} \sinh(\pi |s|).$$ Really the better estimate is true?
• In Eq. (5.1) of arxiv.org/pdf/1301.1749.pdf the $c$ has to be replaced by $2(\sigma-1)$ (note that this matches the condition $0\leq c\leq 1$), then replace $s\rightarrow s-2(\sigma-1)$ and note that $s=\sigma + i t$ and (most importantly) that $|\Gamma(\sigma - i t)| = |\Gamma(\sigma + i t)|$. Apr 14, 2014 at 10:59
• In fact $$|\frac{\Gamma(s)}{\Gamma(2-s)}|=|\frac{\Gamma(s)\Gamma(s-1)}{\Gamma(2-s)\Gamma(s-1)}|=\frac{1}{\pi}|\sin\pi s||\Gamma(s)||\Gamma(s-1)|.$$ Due to the well-known inequality $$|\Gamma(x+iy)|\le|\Gamma(x)|$$ both gammas are $\le 1$. But why sinus is power-bounded? Apr 14, 2014 at 15:50
• The proof is more complicated, making use of a generalization of the Phragmén–Lindelöf principle (or theorem as it is also called). (links are en.wikipedia.org/wiki/… or encyclopediaofmath.org/index.php/…) Unfortunately, the above cited book of Rademacher with the proof of the inequality is hard to get. If I find some time, I will post a sketch of the proof here. Apr 17, 2014 at 14:18

There is a genuine asymptotic $\Gamma(s+a)/\Gamma(s)\sim s^a$ for $s$ in a half-plane to the right of $0$, as $|s|\to \infty$, for bounded $a$. (This is proven in many places, as a corollary of Watson's lemma, much easier than Stirling-Binet-Laplace, and not using the latter. E.g., in course notes at http://www.math.umn.edu/~garrett/m/mfms/ named "asymptotics of integrals, including the Gamma function".

Then use $|\Gamma(\sigma-it)|=|\Gamma(\sigma+it)|$, so $|\Gamma(2-s)|=|\Gamma(2-\sigma+it)|$, and $|\Gamma(\sigma+it)/\Gamma(2-\sigma+it)|\sim |t|^{2\sigma-2}$. An asymptotic, not an inequality.

Write $$s=\sigma+it$$. Let $$\ell(\sigma)$$ be the line such that $$\ell(1)=1$$ and $$\ell(2)=0$$, namely $$\ell(\sigma)=2-\sigma$$. On the line $$\sigma = 1$$, we have

$$|\frac{\Gamma(s)}{\Gamma(2-s)}| = |\frac{\Gamma(1+it)}{\Gamma(1-it)}|=1=:M_1$$.

On the line $$\sigma = 2$$, we have (by $$s\Gamma(s)=\Gamma(s+1)$$)

$$|\frac{\Gamma(s)}{\Gamma(2-s)}| = |\frac{\Gamma(2+it)}{\Gamma(-it)}| = |t|\cdot|t+i|=:M_2$$.

The Phragmen-Lindelof principle now tells us that for $$1\leq \sigma\leq 2$$, we have

$$|\frac{\Gamma(s)}{\Gamma(2-s)}|\leq M_1^{\ell(\sigma)} M_2^{1-\ell(\sigma)}=(|t|\cdot|t+i|)^{\sigma-1}$$.

This is (almost) your desired inequality.