THE GAMMA FUNCTION AND THE ZETA FUNCTION PAUL DUNCAN Abstract. The Gamma Function and the Riemann Zeta Function are two special functions that are critical to the study of many different fields of math- ematics. In this paper we will discuss characterizations and properties of each, and how the two are connected. Contents 1. The Gamma Function 1 1.1. Convex Functions 1 1.2. The Gamma Function 4 2. Values of the Riemann Zeta Function 8 3. Characterization of the Zeta Function 9 Acknowledgments 11 References 11 1. The Gamma Function In this paper we will start by characterizing the Gamma function. Its devel- opment is motivated by the desire for a smooth extension of the factorial func- tion to R. Recall the recursive definition of the factorial function, namely that (n + 1)! = (n + 1)n! and 0! = 1. Since it is only defined on the integers, there are many possible continuous extensions to R. However, since we know that en < n! < nn, we can imagine narrowing our options using some notion of "fast" growth. Fortunately, it turns out that we can find a unique extension using the concept of convexity. 1.1. Convex Functions. Recall from calculus that a twice differentiable function is called convex if f 00(x) ≥ 0 for all x. We would like an extension of this definition to functions that are not differentiable. Definition 1.1. For a function f(x) define f(x1) − f(x2) φf (x1; x2) = x1 − x2 which can be thought of as the slope of the secant line between two points. We say that f(x) is convex on an interval if for every x3 on the interval, φ(x1; x3) is a monotonically increasing function of x1. Date: August 29, 2013. 1 2 PAUL DUNCAN Alternatively, f(x) is convex if, for all x1, x2, x3,Ψf ≥ 0, where Ψf is defined as φf (x1; x3) − φf (x2; x3) (1.2) Ψf (x1; x2; x3) = x1 − x2 (x − x )f(x ) + (x − x )f(x ) + (x − x )f(x ) (1.3) = 3 2 1 1 3 2 2 1 3 : (x1 − x2)(x2 − x3)(x3 − x1) Remark 1.4. The values of φf (x1; x2) and Ψf (x1; x2; x3) are invariant under per- mutation of their arguments. First we will need some tools to determine whether a given function is convex. Theorem 1.5. The sum of convex functions, limit function of a convergent se- quence of convex functions, and sum of a convergent series of convex functions are all convex. Proof. From the second statement of the definition of convexity we can see that for two functions f(x) and g(x) if Ψf (x1; x2; x3) ≥ 0 and Ψg(x1; x2; x3) ≥ 0 then Ψf+g(x1; x2; x3) = Ψf (x1; x2; x3) + Ψg(x1; x2; x3) ≥ 0. If we look at the same inequality for a sequence of convex functions f1(x); f2(x)::: that converges to f(x) we see that Ψf (x1; x2; x3) = lim Ψf (x1; x2; x3) ≥ 0 ; n!1 n so f(x) is also convex. The third part of the theorem then follows from the first two since a series is a sequence of partial sums. Definition 1.6. We say that a function is weakly convex on an interval if for every x1+x2+:::+xn 1 x1; x2; :::; xn on that interval f( n ) ≤ n (f(x1) + f(x2) + ::: + f(xn)). Lemma 1.7. If a function is weakly convex and continuous, then it is convex. Proof. Suppose f(x) is weakly convex. Choose x2 < x1 numbers in the interval and 0 ≤ p ≤ n two integers. Apply the definition of weak convexity in the case where p numbers have the value x1 and the other n − p have the value x2. We get p p p p f x + 1 − x ≤ f(x ) + 1 − f(x ) : n 1 n 2 n 1 n 2 Assume f(x) is continuous and let t be a real number such that t 2 [0; 1]. Choose a p sequence of rational numbers converging to t. We can write each number as n , so we can apply the above equation. But f(x) is continuous, so we can take the limit of the sequence to obtain f(t(x1) + (1 − t)x2) ≤ tf(x1) + (1 − t)f(x2) : We want to show that for any x1; x2; x3, we have Ψf (x1; x2; x3) ≥ 0. We can assume x2 < x3 < x1 since we can permute the arguments of Ψf , so the denominator for expression (1.3) is positive. Set t = x3−x2 . We know that t 2 (0; 1) and x1−x2 1 − t = x1−x3 . So x1−x2 (x3 − x2)x1 + (x1 − x3)x2 tx1 + (1 − t)x2 = = x3 : x1 − x2 Plugging this into the previous equation we get x3 − x2 x1 − x3 f(x3) ≤ f(x1) + f(x2) : x1 − x2 x1 − x2 THE GAMMA FUNCTION AND THE ZETA FUNCTION 3 It is now clear that the numerator of (1.3) is also positive, so f(x) must be convex. Remark 1.8. The continuous condition is necessary, but in order to construct an example of a function that is weakly convex, but not convex, one must use the Axiom of Choice. The converse of the lemma is also true. Definition 1.9. We say f(x) is log convex on an interval if log(f(x)) is defined and convex on that interval. Remark 1.10. If a function is log convex, then it is also convex. From Theorem 1.5 we know that the product of log convex functions is log convex. More surprisingly, the same turns out to be true for their sum. Theorem 1.11. Let f(x) and g(x) be log convex functions defined on a common interval. Then f(x) + g(x) is also log convex. Proof. It is enough to prove the statement for weakly convex functions, since we can add continuity to then obtain the theorem. Assume f(x) and g(x) are weakly log convex. Then both are positive and for every x1; x2 x + x 1 log f( 1 2 ) ≤ (log f(x ) + log f(x )) ; 2 2 1 2 or x + x (f( 1 2 ))2 ≤ f(x )f(x ) : 2 1 2 Similarly, x + x (g( 1 2 ))2 ≤ g(x )g(x ) : 2 1 2 We need to show x + x x + x (f( 1 2 ) + g( 1 2 ))2 ≤ (f(x ) + g(x ))(f(x ) + g(x )) : 2 2 1 1 2 2 Let S = f(a; b; c) 2 R3jac − b2 ≥ 0g. We have x + x x + x (f(x ); f( 1 2 ); f(x2)); (g(x ); g( 1 2 ); g(x )) 2 S; 1 2 1 2 2 and we want to show their sum is in S. We will show this by showing that S is closed under addition. By the quadratic formula we know that ac − b2 ≥ 0 () ax2 + bx + c ≥ 0 8x 2 R ; so S = f(a; b; c) 2 R3jax2 + bx + c ≥ 0 8x 2 Rg, which is clearly closed under addition. Theorem 1.12. If φ(x) is a positive continuous function defined on the interior of the integration interval, then Z b φ(t)tx−1 dt a is a log convex function of x for every interval on which the proper or improper integral exists. 4 PAUL DUNCAN Proof. Suppose f(t; x) is defined and continuous for t 2 [a; b] and arbitrary x, and also log convex as a function of x. For n 2 Z and h = (b − a)=n, construct Fn(x) = h(f(a; x) + f(a + h; x) + ::: + f(a + (n − 1)h; x)) : Since Fn(x) is a sum of log convex functions, it is log convex. If we take the limit as n goes to infinity, we get Z b f(t; x)dt : a If the integral is improper, then it is the limit of a sequence of log convex functions, x−1 so it is also log convex. The theorem is the case in which f(t; x) = φ(t)t . With log convexity, we have enough to define a unique Γ(x). 1.2. The Gamma Function. R 1 −t x−1 Definition 1.13. We define Γ(x) = 0 e t dt for x > 0. Note that the integral converges for positive real x. Theorem 1.14. Suppose a function f(x) satisfies the following three conditions: (1) f(x + 1) = xf(x) (2) The domain of f(x) contains all x > 0, and f(x) is log convex for these x (3) f(1) = 1. Then nxn! f(x) = lim : n!1 x(x + 1):::(x + n) Moreover, Γ(x) satisfies (1), (2), (3) for x > 0. In particular, (1), (2), (3) charac- terize Γ(x) uniquely. Proof. First, we will show that Γ(x) satisfies the three conditions. Integration by parts shows that Γ(x) satisfies the first condition. Z δ Z δ −t x −t x δ −t x−1 Γ(x + 1) = lim e t dt = lim (−e t j + x e t dt) δ!1,!0 δ!1,!0 Z δ = lim (−e−δδx + e−x + x e−ttx−1dt) = xΓ(x) δ!1,!0 The second condition comes from Theorem 1.8 and the third is easily checked. Since the gamma function satisfies the requirements, we know that such a function exists. Let f(x) be a function that satisfies these conditions. Then it is enough to show that f(x) = Γ(x) for x 2 (0; 1], since the first condition then implies that they agree on the full domain.