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Bounds for the Logarithm of the Euler Gamma Function and Its Derivatives

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Bounds for the Logarithm of the Euler Gamma Function and Its Derivatives Bounds for the logarithm of the Euler gamma function and its derivatives Harold G. Diamond∗ and Armin Strauby Department of Mathematics University of Illinois at Urbana-Champaign 1409 W. Green St, Urbana, IL 61801, United States August 12, 2015 Abstract We consider differences between log Γ(x) and truncations of certain clas- sical asymptotic expansions in inverse powers of x − λ whose coefficients are expressed in terms of Bernoulli polynomials Bn(λ), and we obtain conditions under which these differences are strictly completely monotonic. In the sym- metric cases λ = 0 and λ = 1=2, we recover results of Sonin, N¨orlundand Alzer. Also we show how to derive these asymptotic expansions using the functional equation of the logarithmic derivative of the Euler gamma function, the representation of 1=x as a difference F (x + 1) − F (x), and a backward induction. 1 Introduction The Euler gamma function Γ(x) is widely regarded as the most important special function. Good accounts and many formulas for Γ(x) can be found in [4], [5, x6.3], 2010 Mathematics subject classification. Primary 33B15, 26D07; Secondary 11B68 Key words and phrases. Gamma function, psi function, inequalities, asymptotic expansions, Bernoulli polynomials, complete monotonicity. ∗[email protected] [email protected] 1 [6, Chapter 1], and [12, Chapter 12]. Various bounds are known for Γ(x) and (x), the logarithmic derivative of Γ(x), also called the digamma function, e.g. [1], [2], [5, 6.3.21], [7], [9, Appendix C], [10]. The logarithm of the gamma function has the classical asymptotic expansion p 1 log Γ(x) ∼ log 2π + x − 2 log(x − λ) − (x − λ)(1) 1 X Bn(λ) + (−1)n (x − λ)1−n; n(n − 1) n=2 which we consider for real x > λ and λ 2 [0; 1]. As usual, Bn(λ) denotes the nth Bernoulli function. Our focus here is on differences (and their derivatives) between log Γ(x) and the truncations p 1 LN (λ; x) := log 2π + x − 2 log(x − λ) − (x − λ) N X Bn(λ) + (−1)n (x − λ)1−n n(n − 1) n=2 of this asymptotic series. In the most basic case, we are interested in conditions under which these truncations provide strict upper or lower bounds. These inequalities together with the functional equations provide an effective method for calculating log Γ(x) and its derivatives with arbitrary accuracy. Example 1.1. A pleasant property of (1) is that we can differentiate both sides to obtain asymptotic expansions for the derivatives of log Γ(x), that is, the derivatives of the truncations LN (λ; x) provide asymptotic expansions for the derivatives of log Γ(x). For instance, in the particularly symmetric case λ = 1=2, we have the asymptotic expansion 1 X B2n(1=2) (2) (x) ∼ log(x − 1=2) − (x − 1=2)−2n: 2n n=1 Our main theorem implies the following result of N¨orlund[11, x56, p. 106]: if the series in (2) is truncated at n = N with N ≥ 0 and even, the resulting sum is a lower bound for (x) valid for all x > 1=2; if N is odd, it is an upper bound. In other words, successive truncations of (2) provide lower and upper bounds valid for all x > 1=2. The first examples of these bounds are (3) (x) > log(x − 1=2) and (4) (x) < log(x − 1=2) + (x − 1=2)−2=24 : 2 A function f is said to be completely monotonic on an interval or half-line I if, for all x 2 I and all integers n ≥ 0, the derivative f (n)(x) exists and satisfies (−1)nf (n)(x) ≥ 0. If this inequality is strict, that is, if (−1)nf (n)(x) > 0, then f is called strictly completely monotonic on I. For an introduction to completely monotonic functions and their properties we refer to [13, Chapter IV]. Our main result, which is proved in Section4, is the following. As indicated in Example 1.4, this is an extension of a result of Alzer [1, Theorem 8], which concerns the case λ = 0. Theorem 1.2. Let N ≥ 1. Suppose that either N is even and λ 2 [0; λ0] where λ0 is the unique root of BN (λ) in [0; 1=2], or N is odd and λ 2 [λ0; 1=2] where λ0 is the unique root of BN+1(λ) in [0; 1=2]. Then, dN=2e (−1) (log Γ(x) − LN (λ; x)) is strictly completely monotonic on (λ, 1). We note that λ0 < 1=4 and that λ0 ! 1=4 very rapidly as N increases. This is explained in Example 3.3, which also illustrates that Theorem 1.2 would be false without the conditions on N and λ. On the other hand, the restriction to the case λ 2 [0; 1=2] is made for the sake of exposition, and we invite the interested reader to adjust the details for the case λ 2 [1=2; 1]. Example 1.3. The particular case λ = 1=2 and N = 3 shows that p 1 1 1=24 log Γ(x) − log 2π − x − 2 log x − 2 − 1 + 1 x − 2 is strictly completely monotonic on (1=2; 1). This generalizes the inequality (4). Example 1.4. In the case λ = 0, we find that the functions (x) − L2M (0; x), explicitly given by M p X B2n (x) − log 2π − x − 1 log(x) + x − ; 2 2n(2n − 1)x2n−1 n=1 are strictly completely monotonic on (0; 1) if M is even. Likewise, the functions L2M (0; x) − (x) are strictly completely monotonic on (0; 1) if M is odd. This is a result of Alzer [1, Theorem 8]. This paper is organized as follows. In Section2, we give a first illustration of our approach by proving the claims made in Example 1.1. These bounds for the digamma function are then generalized in Section3, before we prove the main result in Section4. A very brief indication of how to use these bounds for numerical calculations is then given in Section5. 3 2 Proving bounds for the digamma function In this section, we illustrate our approach by proving the claims made in Example 1.1. In particular, we show, by a method we believe new, how to derive asymptotic expansions for (x), such as (2), whose successive truncations provide lower and upper bounds. Assuming there is an approximation of the form (x) ≈ F (x), then the functional equation (x + 1) = (x) + 1=x would give (5) F (x + 1) − F (x) ≈ 1=x: Our method is to find functions F giving upper (resp. lower) bounds in (5) and then show by a \backward induction" that these functions provide bounds for (x). We use one additional fact about the digamma function (x), namely the asymptotic formula (6) (x) = log x + o(1); x ! 1: For completeness, we prove this at the end of the article. Example 2.1. To illustrate our method, we first obtain the lower bound (3). By convexity, Z x+1=2 du 1 log(x + 1=2) − log(x − 1=2) = > : x−1=2 u x This inequality and the functional equation give (x) > (x + 1) − (log(x + 1=2) − log(x − 1=2)) for all x > 1=2, or, equivalently, (7) (x) − log(x − 1=2) > (x + 1) − log(x + 1=2): Now we prove (3) by an induction: repeating (7) with x replaced by x + 1; x + 2;::: and using (6), we see that (x) − log(x − 1=2) > lim inf( (y) − log(y − 1=2)) = 0: y!1 The key to our treatment of the general case is to develop a suitable expression for 1=x with the aid of the Euler{Maclaurin sum formula. The following version of Euler{Maclaurin, given in [9, Appendix B] is convenient for our application. 4 Theorem 2.2 (Euler{Maclaurin). Let K be a positive integer, and suppose that f is K times continuously differentiable on [a; b]. Let Bk(·) denote the k-th Bernoulli function. Then, X Z b f(n) = f(x)dx a<n≤b a K X (−1)k + (B (fbg)f (k−1)(b) − B (fag)f (k−1)(a)) k! k k k=1 K Z b (−1) (K) − BK (fxg)f (x)dx; K! a where ftg := t − btc. We now show the following result of N¨orlund[11, x56, p. 106] stated in Exam- ple 1.1. Theorem 2.3. The asymptotic expansion (2) for (x) holds as x ! 1. Moreover, if the series in (2) is truncated at n = N with N ≥ 0 and even, the resulting sum is a lower bound for (x) valid for all x > 1=2; if N is odd, it is an upper bound. Proof. Take f(t) := 1=(x+t), a = −1=2, b = 1=2 in Theorem 2.2. Because Bk(1=2) = 0 for all odd k, we restrict the series to even indices. Starting with the one-term (!) sum 1 X 1 = ; x x + n −1=2<n≤1=2 we have Z 1=2 N 1 dt X B2n(1=2) 1 1 = − − + R ; x x + t 2n (x + 1=2)2n (x − 1=2)2n N −1=2 n=1 where Z 1=2 dt RN := − B2N (ftg) 2N+1 : −1=2 (x + t) For the claimed inequalities, we establish the sign of RN with the aid of several properties of Bernoulli functions, all of which are given in [9]. For −1=2 < t < 0, B2N (ftg) = B2N (1 + t) = B2N (−t): 5 Thus, we have Z 1=2 1 1 RN = − B2N (t) 2N+1 + 2N+1 dt : 0 (x − t) (x + t) Next, Z y B2N+1(y) = (2N + 1) B2N (t) dt: 0 Integrating by parts and noting that the integrated term vanishes because B2N+1(1=2) = B2N+1(0) = 0, we find Z 1=2 1 1 RN = B2N+1(t) 2N+2 − 2N+2 dt : 0 (x − t) (x + t) N+1 For 0 < t < 1=2; sgn(B2N+1(t)) = (−1) .
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