THE GAMMA FUNCTION AND THE ZETA FUNCTION

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 mathematics. In this paper we will discuss characterizations and properties of each, and how the two are connected. Contents. The Gamma Function.. Convex Functions.. The Gamma Function 4. Values of the Riemann Zeta Function 8 3. Characterization of the Zeta Function 9 Acknowledgments References. The Gamma Function In this paper we will start by characterizing the Gamma function. Its development is motivated by the desire for a smooth extension of the factorial function to R. Recall the recursive definition of the factorial function, namely that (n + )! = (n + )n! and! =. Since it is only defined on the integers, there are many possible continuous extensions to R. However, since we know that e n < n! < n n, 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... Convex Functions. Recall from calculus that a twice differentiable function is called convex if f (x) for all x. We would like an extension of this definition to functions that are not differentiable. Definition.. For a function f(x) define φ f (x, x ) = f(x ) f(x ) x x 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 x 3 on the interval, φ(x, x 3 ) is a monotonically increasing function of x. Date: August 9, 3.

PAUL DUNCAN Alternatively, f(x) is convex if, for all x, x, x 3, Ψ f, where Ψ f is defined as Ψ f (x, x, x 3 ) = φ f (x, x 3 ) φ f (x, x 3 ) (.) x x = (x 3 x )f(x ) + (x x 3 )f(x ) + (x x )f(x 3 ) (.3). (x x )(x x 3 )(x 3 x ) Remark.4. The values of φ f (x, x ) and Ψ f (x, x, x 3 ) are invariant under permutation of their arguments. First we will need some tools to determine whether a given function is convex. Theorem.5. The sum of convex functions, limit function of a convergent sequence 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 Ψ f (x, x, x 3 ) and Ψ g (x, x, x 3 ) then Ψ f+g (x, x, x 3 ) = Ψ f (x, x, x 3 ) + Ψ g (x, x, x 3 ). If we look at the same inequality for a sequence of convex functions f (x), f (x)... that converges to f(x) we see that Ψ f (x, x, x 3 ) = lim Ψ f n (x, x, x 3 ), 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.6. We say that a function is weakly convex on an interval if for every x, x,..., x n on that interval f( x+x+...+xn n ) n (f(x ) + f(x ) +... + f(x n )). Lemma.7. If a function is weakly convex and continuous, then it is convex. Proof. Suppose f(x) is weakly convex. Choose x < x numbers in the interval and p n two integers. Apply the definition of weak convexity in the case where p numbers have the value x and the other n p have the value x. We get f ( p n x + ( p n ) ) x p n f(x ) + ( p n ) f(x ). Assume f(x) is continuous and let t be a real number such that t [, ]. Choose a sequence of rational numbers converging to t. We can write each number as p 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(x ) + ( t)x ) tf(x ) + ( t)f(x ). We want to show that for any x, x, x 3, we have Ψ f (x, x, x 3 ). We can assume x < x 3 < x since we can permute the arguments of Ψ f, so the denominator for expression (.3) is positive. Set t = x3 x x x. We know that t (, ) and t = x x3 x x. So tx + ( t)x = (x 3 x )x + (x x 3 )x x x = x 3. Plugging this into the previous equation we get f(x 3 ) x 3 x x x f(x ) + x x 3 x x f(x ).

THE GAMMA FUNCTION AND THE ZETA FUNCTION 3 It is now clear that the numerator of (.3) is also positive, so f(x) must be convex. Remark.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.9. We say f(x) is log convex on an interval if log(f(x)) is defined and convex on that interval. Remark.. If a function is log convex, then it is also convex. From Theorem.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.. 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 x, x or Similarly, We need to show (f( x + x log f( x + x ) (log f(x ) + log f(x )), (f( x + x )) f(x )f(x ). (g( x + x )) g(x )g(x ). ) + g( x + x )) (f(x ) + g(x ))(f(x ) + g(x )). Let S = {(a, b, c) R 3 ac b }. We have (f(x ), f( x + x ), f(x)), (g(x ), g( x + x ), g(x )) S, 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 b ax + bx + c x R, so S = {(a, b, c) R 3 ax + bx + c addition. x R}, which is clearly closed under Theorem.. If φ(x) is a positive continuous function defined on the interior of the integration interval, then b a φ(t)t x dt 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 [a, b] and arbitrary x, and also log convex as a function of x. For n Z and h = (b a)/n, construct F n (x) = h(f(a, x) + f(a + h, x) +... + f(a + (n )h, x)). Since F n (x) is a sum of log convex functions, it is log convex. If we take the limit as n goes to infinity, we get b a f(t, x)dt. If the integral is improper, then it is the limit of a sequence of log convex functions, so it is also log convex. The theorem is the case in which f(t, x) = φ(t)t x. With log convexity, we have enough to define a unique Γ(x)... The Gamma Function. Definition.3. We define Γ(x) = e t t x dt for x >. Note that the integral converges for positive real x. Theorem.4. Suppose a function f(x) satisfies the following three conditions: () f(x + ) = xf(x) () The domain of f(x) contains all x >, and f(x) is log convex for these x (3) f() =. Then n x n! f(x) = lim n x(x + )...(x + n). Moreover, Γ(x) satisfies (), (), (3) for x >. In particular, (), (), (3) characterize 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. Γ(x + ) = lim δ,ɛ δ ɛ e t t x dt = lim δ,ɛ ( e t t x δ ɛ + x = lim δ,ɛ ( e δ δ x + e ɛ ɛ x + x δ ɛ δ e t t x dt) = xγ(x) ɛ e t t x dt) The second condition comes from Theorem.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 (, ], since the first condition then implies that they agree on the full domain. For x (, ] and integer n, we can use the second condition to obtain log f( + n) log f(n) log f(x + n) log f(n) ( + n) n (x + n) n log f( + n) log f(n). ( + n) n By combining the first and third conditions we know that f(n) = (n )!, so we can write log f(x + n) log(n )! log(n ) log n, x which can be rearranged to get log(n ) x (n )! f(x + n) log n x (n )!.

THE GAMMA FUNCTION AND THE ZETA FUNCTION 5 Now, since log is a monotonically increasing function, this implies (n ) x (n )! f(x + n) n x (n )!. From the first condition it is clear that f(x + n) = x(x + )...(x + n)f(x), so (n ) x (n )! x(x + )...(x + n ) f(x) (n) x (n )! x(x + )...(x + n). Since this is true for all n, we can write n x n! x(x + )...(x + n) f(x) n x n! x + n x(x + )...(x + n) n Now we can take the limit as n goes to infinity and get the formula n x n! f(x) = lim n x(x + )...(x + n). But since Γ(x) also satisfies the three conditions, it is equal to f(x).. We use the limit from Theorem.4 to define Γ(x) for x <. Note that Γ(x) is still not defined for non-positive integers. With some manipulation, we can use the formula to find a useful alternative definition for the gamma function. Corollary.5. The gamma function can be expressed as the product Γ(x) = e Cx e x j x + x j for x R \ Z, where C is Euler s constant. Proof. n x n! Γ(x) = lim n x(x + )...(x + n) = lim n ex log n e x/ x/+x/ x/+...+x/n x/n x x + x +... n x + n = lim n ex(log n / /... /n) e x/ e x/ x x/ + x/ +... e x/n x/n +. j= Recall C = lim n + +... + n log n exists and is called Euler s constant. So Γ(x) = e Cx n x lim e x j n + x = e Cx j x j= j= e x j + x j Although differentiability was not one of our criteria for the uniqueness of the gamma function, we would hope that it would be infinitely differentiable. In fact it is. Theorem.6. The function Γ(x) is infinitely differentiable on its domain of definition.

6 PAUL DUNCAN Proof. It suffice to show that log Γ(x) is infinitely differentiable, since the chain rule then tells us that the same is true for Γ(x). Moreover, we only need to consider the case of x > because of the relation Γ(x + ) = xγ(x). We know that log Γ(x) is defined since Γ(x) > for all x >. From the product equation in Corollary. we can write log Γ(x) = Cx log x + ( x j log( + x j )). Now we can differentiate termwise if the series thus obtained is uniformly convergent. The series in question is C x + ( j x + j ) = C x + x j(x + j). j= For x (, r] for arbitrary r j= j= x j(x + j) r j, which is independent of x. Since the right side converges, we can apply the Weierstrass M-test to show that our series converges uniformly, and log Γ(x) is differentiable on that interval. But r is arbitrary, so the same is true for R >. If we differentiate again, we get x j= j= (x + j). Since the jth summand is less than /j, we can again apply the Weierstrass M- test. Repeating the process shows that log Γ(x), and by extension Γ(x) is infinitely differentiable. j= Lemma.7. The gamma function satisfies the relation ( x ) ( ) x + Γ Γ = c x Γ(x) for some constant c. Proof. Consider the function ( x ) ( ) x + f(x) = x Γ Γ. Since f(x) is a product of log convex functions, it is log convex. Moreover, ( ) x + ( x ) f(x + ) = x+ Γ Γ + = xf(x). Since f(x) satisfies the first two conditions of Theorem., we know that f(x) = cγ(x) for some constant c and with slight rearrangement we have the lemma. The product formula for the gamma function is already useful in deriving an alternate formula for sin(x). Theorem.8. The sine function can be expressed as the infinite product sin(πx) = πx ( x j ). j=

THE GAMMA FUNCTION AND THE ZETA FUNCTION 7 Proof. Define a new function φ(x) = Γ(x)Γ( x) sin πx for non-integer arguments. Our first goal is to show that φ(x) is constant. First we want to extend φ(x) to R so that it is periodic with period, continuous, and infinitely differentiable. If we replace x by x + in each of the components, we get Γ(x + ) = xγ(x), Γ( x) = Γ( x), sin π(x + ) = sin πx. x Combining them, we see that φ(x + ) = φ(x). By Lemma.7 we have Γ( x )Γ(x + ) = c x Γ(x) for some constant c. Replace x by x to get Now So φ( x )φ(x + Γ( x )Γ( x ) = cx Γ( x). ) = Γ( x )Γ( x ) sin(πx = c Γ(x)Γ( x) sin πx. 4 )Γ(x + )Γ( x ) cos( πx ) (.9) φ( x )φ(x + ) = c 4 φ(x). We want φ(x) to be continuous and infinitely differentiable on R. It is on nonintegral arguments because Γ(x) and sin(x) are, and to continue those properties to the real line we extend φ(x) by writing φ(x) = Γ( + x)γ( x) sin πx x = Γ( + x)γ( x)(π π3 x 3! + π5 x 4 5!...). Using that expansion, we define φ() = π and since φ(x) is periodic with period, φ(n) = π for n Z. Now φ is continuous and infinitely differentiable. We also know that φ(x) is positive everywhere by combining the original definition, the periodicity, and the fact that φ(n) = π for n Z. Now we want to show that φ(x) is constant. Let g(x) = d dx log φ(x). From (.9) we can derive (.) 4 (g(x ) + g(x + )) = g(x). Since g(x) is continuous on [, ], it is bounded, e.g. g(x) M on that interval for some M. But since φ(x) is periodic, so must g(x) be, so it is bounded everywhere. Using (.) we can write g(x) 4 g(x ) + 4 g(x + ) M 4 + M 4 = M. This process can be repeated an arbitrary number of times, so g(x) =. By the definition of g(x), this means that log φ(x) is linear. However it is also periodic, so it must be constant, which means that φ(x) is also constant. Since we know

8 PAUL DUNCAN φ() = π, we know that φ(x) = π for all x. Now we can rewrite the definition of φ(x) π sin πx = Γ(x)Γ( x) = π xγ(x)γ( x). A straightforward substitution of the formula for the gamma function in Corollary. yields sin(πx) = πx ( x i ). i=. Values of the Riemann Zeta Function The Riemann zeta function is another extraordinarily useful special function. We will only talk about its real part here. Definition.. We define for s >. Definition.. We define for s,..., s k >. ζ(s,..., s k ) = ζ(s) = n= n s n >n >...>n k > n s...ns k k From calculus we know that the formula for ζ converges for s >, but we cannot easily compute its values. At certain integers, however, its values have a simple form. Theorem.3. We can obtain the values of the zeta function and multiple zeta function with arguments of s, namely ζ() = π π4, ζ(, ) = 6. Proof. We want to find some C m (N) such that N ( x N j ) = ( ) j C (N) j x j. j= We expand the product to form N ( x j ) = ( x )( x )...( x N ) j= j= = x( + +... N ) + x ( + 3 +... + N (N ) )... Now we can collect terms and see that C (N) = and for m N C m (N) = j. j...j m j >j >...>j m>

THE GAMMA FUNCTION AND THE ZETA FUNCTION 9 If for all m, lim N C m (N) = C m, then ( x j ) = ( ) j C (N) j x j. j= j= Since every term in the summation of C (N) is positive, it is clear that C m (N) < (C (N) ) m. We can see that C m (N) increases as N does and we know that C is finite from calculus, so each C m (N) must converge. We can also see that C m = ζ(,,..., ) with m arguments. Now let f(x) = ( x i ) = ζ()x + ζ(, )x.... i= Now, by Theorem.8, we can write sin πx = πxf(x ) = πx ζ()πx 3 + ζ(, )πx 5.... But from calculus we know that sin πx = πx π3 x 3 + π5 x 5.... 3! 5! Equating coefficients, we have a formula for zeta values involving, in particular ζ() = π π4 6 and ζ(, ) =. Theorem.4. We can also compute ζ(4) = π4 9. Proof. Notice that ζ() = ( n ) = n 4 + n + n >n n n = ζ(4) + ζ(, ). n <n n n= Solving for ζ(4), we get n= ζ(4) = ζ() ζ(, ) = π4 36 π4 = π4 9. 3. Characterization of the Zeta Function Soon we will be able to see that the gamma and zeta functions are related, but first we must extend the zeta function. From its definition, it is meaningless to evaluate in the interval (, ), but we can use a similar function to derive a useful continuation. Notation 3.. For r Z >, we define ζ r (s) := n> Theorem 3.. For s >, b n n b n = ζ(s) = Moreover, ζ r (s) is defined for s >. { : r n r : r n. rs r s ζ r(s).

PAUL DUNCAN Proof. Notice that r s ζ(s) = n= (rn) s = n= and that b n = ra n. Now we can write So, { a n : r n n s a n = : r n ζ r (s) = ζ(s) r( r s ζ(s)) = ζ(s)( r r s ) = rs r s ζ(s). ζ(s) = rs r s ζ r(s). Now we want to show that ζ r (s) is defined for s >. Let B N = N n= b n. It is easy to see that B N r for all N. Let s >. We want to show that the series that defines ζ r (s) is Cauchy. Using the fact that B k B k = b k, we can see that for any N > M N k=m b k k s = B N N s B N M M s + B k ( k s (k + ) s ). The first two terms go to zero, so it suffices to show that the sum converges. Using calculus we can see that k s (k + ) s = s k k+ k=m x s+ dx s (k + ) s+ and we can combine that with our bound on B N to yield N k=m B k ( k s (k + ) s ) rs (k + ) s+ = rs k which we know converges because s >. We therefore extend ζ(s) to R > by defining k (k + ) s+, ζ(s) = rs r s ζ r(s). Now we have enough for a functional equation relating the zeta and gamma functions. Theorem 3.3. Let then ξ(s) = ξ( s) for s R > \ {}. Proof. Let θ(u) = n= ξ(s) = π s/ Γ( s )ζ(s) e πnu = + (e πu + e 4πu + e 9πu...). We need the following lemma from Fourier analysis: Lemma 3.4. θ(u) satisfies the functional equation θ(/u) = u / θ(u).

THE GAMMA FUNCTION AND THE ZETA FUNCTION We will not prove this here, but a proof can be found on page 5 of Elkies notes[]. Notice for every integer n we have e πnu u s/ du u = = e t π s/ n t s/ dt πn t = π s/ Γ( s ) n. e t s/ dt t t By summing over n we get (θ(u) )u s/ du u = ξ(s). Now we can write ξ(s) = We can substitute /u for u to get (θ(u) )u s/ du u + (θ(u) )u s/ du u = s + θ(u)u s/ du u + (θ(u) )u s/ du u. θ(u)u s/ du u = θ(u )u s/ du u = θ(u)u ( s)/ du u = s + (θ(u) )u ( s)/ du u using the functional equation. Combining these we find ξ(s) = s s + (θ(u) )(u s/ + u ( s)/ ) du u. Since the right side is symmetric in s and s, we can see that ξ(s) = ξ( s). Acknowledgments. It is a pleasure to thank my mentors, Preston Wake and Yun Cheng, for their unceasing help and considerable patience. References [] Emil Artin. The Gamma Function. Holt, Reinhart, and Winston. 964. [] Noam Elkies. The Riemann zeta function and its functional equation. http://www.math.harvard.edu/ elkies/m59./zeta.pdf. [3] Tobias Oekiter, Hubert Partl, Irene Hyna and Elisabeth Schlegl. The Not So Short Introduction to LATEX ε. http://tobi.oetiker.ch/lshort/lshort.pdf.