Show that the entire function \[ \frac1{\Gamma(z)} \] has order \(1\), but that there is no constant \(C>0\) such that \[ \frac1{\Gamma(z)}=O(e^{C|z|}). \]
Solution
The function \(1/\Gamma\) has order exactly \(1\), but it is not of finite exponential type.
1. The Weierstrass product formula writes \(1/\Gamma\) as a genus-onecanonical product times an exponential factor.
The formula is \[ \frac1{\Gamma(z)} =ze^{\gamma z} \prod_{n=1}^{\infty} \left(1+\frac zn\right)e^{-z/n}. \] The infinite product is the canonical product of genus one associated to the zero sequence \[ -1,-2,-3,\ldots. \] Its critical exponent is \(1\): indeed \[ \sum_{n=1}^{\infty}\frac1{n^{1+\varepsilon}}<\infty \qquad(\varepsilon>0), \] whereas the harmonic series diverges.
2. The order of \(1/\Gamma\) is at most \(1\).
By Borel’s theorem, the canonical product in 1 has order equal to the critical exponent of its zero sequence, namely \(1\).
The factor \(e^{\gamma z}\) also has order \(1\), while multiplication by the polynomial \(z\) does not increase order. Since the order of a product of finite-order entire functions is at most the maximum of their orders, \[ \rho(1/\Gamma)\le1. \]
3. The order of \(1/\Gamma\) is at least \(1\).
The entire function \(1/\Gamma\) has zeros at \[ 0,-1,-2,-3,\ldots, \] all simple. The nonzero zeros therefore have critical exponent \(1\). By Jensen’s formula, the critical exponent of the zero set of an entire function of finite order is at most its order. Hence \[ 1\le\rho(1/\Gamma). \] Together with 2 this gives \[ \boxed{\rho(1/\Gamma)=1}. \]
4. Along the negative half-integers, \(1/\Gamma\) grows like a Gammavalue on the positive real axis.
Euler’s reflection formula is \[ \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}. \] For \[ z_n=\frac12-n, \qquad n=1,2,3,\ldots, \] we have \(|\sin(\pi z_n)|=1\), so \[ \left|\frac1{\Gamma(z_n)}\right| =\frac{\Gamma(n+1/2)}{\pi}. \]
5. No estimate \(1/\Gamma(z)=O(e^{C|z|})\) can hold for any fixed\(C>0\).
Stirling’s formula gives, for real \(s\to+\infty\), \[ \Gamma(s) \sim \sqrt{\frac{2\pi}{s}} \left(\frac{s}{e}\right)^s. \] Taking \(s=n+1/2\) and logarithms, \[ \log\Gamma(n+1/2) =(n+1/2)\log(n+1/2)-(n+1/2)+O(\log n). \] Therefore \[ \frac{\log\Gamma(n+1/2)}{n}\longrightarrow\infty. \] Since \(|z_n|=n-1/2\), for every fixed \(C>0\) we have, for all sufficiently large \(n\), \[ \log\left|\frac1{\Gamma(z_n)}\right| >C|z_n|. \] Equivalently, \[ \frac{|1/\Gamma(z_n)|}{e^{C|z_n|}}\longrightarrow\infty. \] Thus no global bound of the form \[ \frac1{\Gamma(z)}=O(e^{C|z|}) \] is possible.