- This exercise gives two simple observations about \(1 / \Gamma\) (a) Show that \(1 / | \Gamma ( s ) |\) is not \(O ( e ^ { c | s | } )\) for any \(c > 0\) . [Hint: If \(s = - k - 1 / 2\) where k is a positive integer, then \(| 1 / \Gamma ( s ) | \geq k ! / \pi . ]\)
- Show that there is no entire function \(F ( s )\) with \(F ( s ) = O ( e ^ { c \left| { s } \right| } )\) that has simple zeros at \(s = 0 , - 1 , - 2 , . . . , - n , . . . ,\) and that vanishes nowhere else.
Solution
(a).
1. For \(s = -k - 1/2\) (with \(k\) a positive integer), \(|1/\Gamma(s)| \ge k!/\pi\).
the hint (using the reflection formula \(\Gamma(s)\Gamma(1-s) = \pi/\sin(\pi s)\), which gives \(|\Gamma(-k-1/2)| \le \pi/k!\)).
2. If \(1/|\Gamma(s)| = O(e^{c|s|})\), then there is \(C\) with \(1/|\Gamma(s)| \le C e^{c|s|}\) for all \(s\).
definition of big-O.
3. At \(s = -k - 1/2\), this gives \(k!/\pi \le C e^{c(k + 1/2)}\) for all \(k\).
1 and 2.
4. But \(k!\) grows faster than any exponential \(e^{ck}\), so this is impossible for large \(k\).
\(k!/e^{ck} \to \infty\) as \(k \to \infty\).
5. Hence \(1/|\Gamma(s)|\) is not \(O(e^{c|s|})\) for any \(c > 0\).
3 and 4.
(b).
1. Suppose such an \(F\) exists, with simple zeros at \(0, -1, -2, \ldots\) and nowhere else.
assume for contradiction.
2. Then \(F(s)/\Gamma(s)\) is entire with no zeros (the zeros of \(F\) match the poles of \(1/\Gamma\), and \(1/\Gamma\) has simple zeros exactly at \(0, -1, -2, \ldots\)).
\(1/\Gamma\) is entire with simple zeros at the non-positive integers, matching the zeros of \(F\).
3. Hence \(G(s) = F(s)/\Gamma(s)\) is an entire function with no zeros, so \(1/G\) is entire.
2.
4. By (a), \(1/|\Gamma(s)|\) is not \(O(e^{c|s|})\), but \(F(s) = O(e^{c|s|})\); this forces \(G\) to grow in a way that contradicts the Hadamard factorization theorem (an entire function of order \(\le 1\) with no zeros must be \(e^{as+b}\), but then \(F = e^{as+b}\Gamma\) would not be \(O(e^{c|s|})\)).
3 and (a); more directly, if \(F = O(e^{c|s|})\) then \(F\) has order \(\le 1\), so by Hadamard factorization \(F(s) = e^{as+b}\prod_n (1 - s/z_n)\); but the product over the zeros \(0, -1, -2, \ldots\) is \(1/\Gamma(s)\) (up to a factor), and \(1/\Gamma\) is not \(O(e^{c|s|})\), contradiction.
5. Hence no such \(F\) exists.
4.
6. Q.E.D.
5 (a) and 5 (b).