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GAMMA FUNCTION 1. 1.1. History and Motivation. in the Early 16Th

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GAMMA FUNCTION 1. 1.1. History and Motivation. in the Early 16Th GAMMA FUNCTION Abstract. In this paper we explore the history and properties of the Gamma function in an analytic number theoretical context. We analyze the behavior of the Gamma function at its critical points and points of discontinuity, and discuss the convergence of the integral. 1. Introduction 1.1. History and Motivation. In the early 16th century, Leonhard Euler and others attempted to expand the domain of the factorial to the real numbers. This cannot be done with elementary functions, however, with the notions of limits and integrals from the calculus, there were a few expressions developed. In 1738, Euler published the following forms of the generalized factorial: Q1 (1+1=k)n ∙ n! = k=1 1+n=k R 1 n ∙ n! = 0 (− log s) ds for n > 0 For the time being, we will refer to these as the product and integral definitions. In the 19th century, Carl Friedrich Gauss rewrote Euler's product definition to extend the domain to the complex plane, rather than simply the real numbers. This expression uses a limiting process on a series of intermediate functions to represent the factorial. r!rx Γ (x) = r x(1 + x)(2 + x) ::: (r + x) Γ(x) = lim Γr(x) r!1 This was also the time period that the notation Γ(x) was applied to the concept. It was named in 1811 by Adrien-Marie Legendre, who also transformed the integral definition in a very simple way to extend its domain: Z 1 n! = (− log s)nds 0 Let t = − log s: s = −et 1 1 dt = − ds = etds s Z 1 Γ(n) = tn−1e−tdt; 0 which is the more common integral definition that we see today. A few years later, Karl Weierstrass reworked Gauss' expression in the following manner: ex log r Γ (x) = r x(1 + x=1)(1 + x=2) ::: (1 + x=r) ex log r = e xex+x=2+x=3+:::+x=r 1 ex ex=2 ex=r Γ (x) = e x ::: r x (1 + x=1) (1 + x=2) (1 + x=r) 1 1 1 Y x = lim = xe− x (1 + )e−x=r; Γ(x) r!1 Γ (x) r r r=1 1 1 1 where = limr!1(log r−1− 2 − 3 −:::− r ) is Euler's gamma constant. This formulation is defined in terms of zeroes rather than poles. This actually inspired Weierstrass to prove the Weierstrass Factoriza- tion Theorem, which says that any entire function can be written as a product of its zeroes over C. In 1922, Bohr and Mollerup confronted the issue of whether these expressions of the gamma function were equivalent, and, relatedly, whether the Gamma function as defined is unique. Their results are summarized in the following theorem: Theorem 1.1 (Bohr-Mollerup Theorem). If Γ1 = Γ2 on the integers and Γ1(x) = Γ2(x) is logarithmically convex for Re(x) > 0, then Γ1 = Γ2 8 x 2 R. 1.2. Log convexity. The previous theorem guarantees uniqueness for the Gamma function if it is logarithmically convex. The following theorem and its proof can be found in Emil Artin's paper, The Gamma Function. Definition 1.2. f(x) is said to be logarithmically convex if the func- tion log f(x) is convex. Theorem 1.3. Let f(x) be a twice differentiable function. Then f(x) is convex if and only if f 00(x) ≥ 0 for all x in the domain. 2 Now we will show that the Gamma function satisfies the above cri- terion for log convexity. Proposition 1.4. Γ(x) is logarithmically convex for all x > 0. Proof: d Γ0(x) (log Γ(x)) = dx Γ(x) 1 1 X 1 1 = − + = ( − ) x r + x r r=1 1 d2 1 X 1 (log Γ(x)) = + dx2 x2 (x + i)2 i=1 It is clear that this expression is strictly positive, so the Gamma func- tion is log convex. □ This establishes the uniqueness of the Gamma function as the ex- tension of the factorial. 1.3. Properties. The Gamma function satisfies the following func- tional equations: ∙ Γ(z + 1) = zΓ(z) ∙ Γ(1) = 1 ∙ Γ(n) = (n − 1)! for n 2 N ∙ Γ(1 − z)Γ(z) = sinz We express the derivatives of the Gamma function in terms of an- other special function, the Polygamma function, Ψ(k). Z 1 m −zt (m) (m+1) t e Ψ (z) = (−1) −t dt 0 1 − e More important in our context is the following form: d Ψ(m)(z) = ( )m+1(log Γ(z)): dz The digamma is the first polygamma function, in the case m = 0. Γ0(z) Ψ(0)(z) = Γ(z) Γ0(z) = Ψ(0)(z)Γ(z) This leads us to use the following expression for the derivatives of gamma: Z 1 Γ(k)(z) = tz−1e−t(log t)ndt: 0 3 The Gamma function is analytic for all values z 2 C except for the non-positive integers. At these points the function has simple poles. (−1)−n The residue for the point at n = 0; −1; −2;::: is (−n)! . 2. Notation Here is some notation that we will use later in this paper: For x, s, and n real numbers, we define: Γ0(x) Ψ(x) := Γ(x) 1 x E+(x) := 1 + x 1 −x E−(x) := 1 − x n X 1 H(s; n) := ks k=1 1 X 1 (s) := ks k=1 3. Properties Here are some properties that we will use throughout this paper: Proposition 3.1. For all real x with jxj > 1, we have: 1 x X 1 log = x − 1 kxk k=1 4 1 Proof. Because jxj > 1, we have jxj < 1. Therefore, we have the following: 1 1 X 1 = 1 − 1 xk x k=0 1 1 X 1 = x2 1 − 1 xk x k=2 1 1 X 1 = x(x − 1) xk k=2 1 1 X 1 − = − x(x − 1) xk k=2 1 1 1 X 1 − = − x x − 1 xk k=2 Now integrating both sides gives us: 1 1 Z 1 1 Z X 1 X Z 1 − dx = − dx = − dx x x − 1 xk xk k=2 k=2 1 X 1 log x − log(x − 1) = kxk k=1 1 x X 1 log = x − 1 kxk k=1 This proves our proposition. □ + 1 x Proposition 3.2. For all real x the function E (x) = 1 + x is increasing on the interval [1; 1). Proof. We have: 1 x E+(x) = 1 + x 1 log E+(x) = x log 1 + x + 0 1 (E (x)) 1 − x2 + = log 1 + + x ⋅ 1 E (x) x 1 + x x + 1 1 = log − x x + 1 5 By proposition 3.1, we have: 0 1 (E+(x)) X 1 1 = − E+(x) k(x + 1)k x + 1 k=1 1 X 1 = > 0 k(x + 1)k k=2 Our proof is done because we have proved that (E+(x))0 > 0 for all x ≥ 1. □ Using this proposition, we can establish the boundaries for E+(x) on [1; 1): 2 ≤ E+(x) < lim E+(x) = e x!1 − 1 −x Proposition 3.3. For all real x the function E (x) = 1 − x is decreasing on the interval (1; 1). Proof. We have: 1 −x E−(x) = 1 − x 1 log E−(x) = −x log 1 − x − 0 1 (E (x)) 1 x2 − = − log 1 − − x ⋅ 1 E (x) x 1 − x x − 1 1 = − log − x x − 1 x 1 = log − x − 1 x − 1 1 X 1 1 = − kxk x − 1 k=1 1 X 1 1 < − xk x − 1 k=1 1 1 = − = 0 x − 1 x − 1 − 0 Since (E (x)) < 0 for all x > 1, our proof is done. □ Note that this proposition can be used to establish the following inequality: E−(x) > lim E−(x) = e x!1 6 Proposition 3.4. For every positive integer n, the function g(n) := Pn 1 log n − k=1 k is increasing. Proof. We have: n + 1 1 g(n + 1) − g(n) = log − n n + 1 1 X 1 1 = − k(n + 1)k n + 1 k=1 1 X 1 = > 0 k(n + 1)k k=2 This proves our proposition. □ 4. The behavior of the gamma function near its points of discontinuity Now let us analyze how the gamma function behaves near its points of discontinuity. First let's prove the following properties: 1 (4.1) lim − Γ(x) = x!0 x 1 (4.2) lim − − Γ(−x) = x!0 x Proof. We have: 1 1 Γ(1 + x) 1 lim − Γ(x) = − lim Γ(x) − = − lim − x!0 x x!0 x x!0 x x Γ(1 + x) − 1 Γ(1 + x) − Γ(1) = − lim = − lim x!0 x x!0 x = −Γ0(1) = 7 Similarly: 1 1 lim − − Γ(−x) = − lim + Γ(−x) x!0 x x!0 x 1 Γ(1 − x) = − lim + x!0 x −x 1 − Γ(1 − x) = − lim x!0 x Γ(1) − Γ(1 − x) = − lim x!0 x = −Γ0(1) = □ These limits tell us that the gamma function behaves almost exactly 1 like y = x when x gets close to 0. Combining these limits gives us: 1 1 lim (Γ(x) + Γ(−x)) = − lim − Γ(x) + − − Γ(−x) = −2 x!0 x!0 x x Now let's see how the gamma function behaves at other points of discontinuity. Using the recursive formula for gamma function gives us: Γ(x + n) Γ(x) = Qn−1 k=0(x + k) (x + n)Γ(x + n) (x + n)Γ(x) = Qn−1 k=0(x + k) Note that from (4.1) we have: 1 lim xΓ(x) = lim x Γ(x) − + 1 = 0 ⋅ (− ) + 1 = 1 x!0 x!0 x In this limit, it does not matter if x goes to zero from the left or right n side.
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