Math 259: Introduction to Analytic Number Theory More about the Gamma function

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1 Math 59: Introduction to Analytic Number Theory More about the Gamma function We collect some more facts about Γs as a function of a complex variable that will figure in our treatment of ζs and Ls, χ. All of these, and most of the Exercises, are standard textboo fare; one basic reference is Ch. XII pp of [WW 94]. One reason for not just citing Whittaer & Watson is that some of the results concerning Euler s integrals B and Γ have close analogues in the Gauss and Jacobi sums associated to Dirichlet characters, and we shall need these analogues before long. The product formula for Γs. Recall that Γs has simple poles at s =,,,... and no zeros. We readily concoct a product that has the same behavior: let gs := / e s/ + s, s the product converging uniformly in compact subsets of C {,,,...} because e x / + x = + Ox for small x. Then Γ/g is an entire function with neither poles nor zeros, so it can be written as exp αs for some entire function α. We show that αs = γs, where γ = is Euler s constant: That is, we show: γ := lim log N +. Lemma. The Gamma function has the product formulas / Γs = e γs gs = e γs e s/ + s = s s lim N s N. s + Proof : For s,,,..., the quotient gs + /gs is the limit as of s N e / + s s + + s+ = s N + s exp s + + s + N = s N + s + exp log N +. Now the factor N/N + s + approaches, while log N + N γ. Thus gs+ = se γ gs, and if we define Γ? s := e γs gs then Γ? satisfies the same functional equation Γ? s + = sγ? s satisfied by Γ. We are claiming that in fact Γ? = Γ.

2 Consider q := Γ/Γ?, an entire function of period. Thus it is an analytic function of e πis C. We wish to show that q = identically. By the definition of g we have lim s sgs = ; hence lim s sγ? s = lim sgs = = lim sγs, s s and q =. We claim that there exists a constant C such that qσ + it Ce π t / for all real σ, t; since the coefficient π/ in the exponent is less than π, it will follow that q is constant, and thus that Γ? = Γ as claimed. Since q is periodic, we need only prove for s = σ + it with σ [, ]. For such s, we have Γσ + it Γσ by the integral formula and Γ? σ + it Γ? σ = σ + σ + + it = exp = = The summand is a decreasing function of, so the sum is log + t/x dx = t t log + σ +. log + /x dx, which on integration by parts becomes t dx/x + = π t. This proves with C = sup σ qσ, and completes the proof of. Consequences of the product formula. Our most important application of the product formula for Γs is the Stirling approximation to log Γs. Fix ɛ > and let R ɛ be the region {s C : Imlog s < π ɛ}. Then R ɛ is a simply-connected region containing none of the poles of Γ, so there is an analytic function log Γ on R ɛ, real on R ɛ R, and given by the above product formula: We prove: log Γs = lim s log N + log N! Lemma. The approximation holds for all s in R ɛ. = logs +. 3 log Γs = s log s s + logπ + O ɛ s 4 Originally only for n! = Γn +, but we need it for complex s as well.

3 Proof : The estimate holds for small s, say s <, because O ɛ s well exceeds all the other terms. We thus assume s, and estimate the sum in 3 as we did for log x! in obtaining the original form of Stirling s approximation. The sum differs from N+ logs + x dx = N + + s logn + + s s logs N = N + + s log N + N + + s log + N s + s logs N by N+ x + s + x dx ɛ s. We already now that log N! = N + / log N N + A + ON for some constant A. The estimate 4 follows upon taing, except for the value logπ of the constant term. This constant can be obtained by letting s in the duplication formula Γs = π / s ΓsΓs +. One can go on to expand the O ɛ s error in an asymptotic series in inverse powers of s see the Exercises, but 4 is already more than sufficient for our purposes, in that we do not need the identification of the constant term with log π. The logarithmic derivative of our product formula for Γs is Γ s Γs = γ s + [ = lim log N s + = ]. s + Either by differentiating 4 or by applying the same Euler-Maclaurin step to /s + we find that N Remar Γ s Γs = log s s + O ɛ s. 5 The product formula for Γs can also be obtained for real s by elementary means, starting from the characterization of Γ as the unique logarithmically convex function satisfying the recursion Γs + = sγs and normalized by Γ = the Bohr-Mollerup theorem, see for instance [Rudin 976, p.93]. The theorem for complex s can then be obtained by analytic continuation. The method used here, though less elegant, generalizes to a construction of product formulas for a much more general class of functions, as we shall see next. While real asymptotic series cannot in general be differentiated why?, complex ones can, thans to Cauchy s integral formula for the derivative. The logarithmic derivative of Γs is often called ψs in the literature, but alas we cannot use this notation because it conflicts with ψx = n<x Λn... 3

4 Exercises On the product formula:. Verify that the duplication formula for Γs yields the correct constant term in 4. Apply Euler-Maclaurin to the sum in 3 to show that the O ɛ s error can be expanded in an asymptotic series in inverse powers of s.. Use to obtain a product formula for ΓsΓ s, and deduce that ΓsΓ s = π/ sin πs. 6 This can also be obtained from ΓsΓ s = Bs, s by using the change of variable x = y/y in the Beta integral and evaluating the resulting expression by contour integration. Use this together with the duplication formula and Riemann s formula for ζ s to obtain the equivalent asymmetrical form ζ s = π s s Γs cos πs ζs of the functional equation for ζs. Note that the duplication formula, and its generalization n Γns = π n n ns Γ s +, n can also be obtained from. = 3. Show that log Γs has the Taylor expansion log Γs = γs + n ζns n n n= about s =. Recover from this the Laurent expansion of Γs about s =. Behavior of Γs on vertical lines: Γs = s γ + γ + π s 6 + Os 4. Deduce from 4 that for fixed σ R Re log Γσ + it = σ log t π t + C σ + O σ t as t. Chec that for σ =, / this agrees with the exact formulas obtained from 6. Γit = π t sinh πt, Γ/ + π it = cosh πt 4

5 5. For a, b, c >, determine the Fourier transform of fx = expax be cx, and chec your answer by using contour integration to calculate the Fourier transform of ˆf. Now apply Poisson summation, let a and C = e c >, and describe the behavior of n= zcn as z from below. What does n z n = z z + z 4 z 8 + z 6 + n= do as z? Use this to prove that Z m=[ m, m+ is an explicit example of a set of integers that does not have a logarithmic density. An alternative proof of the functional equation for ζs: 6. Prove that ζs = Γs u s du e u for σ >, and that when s is not a positive integer an equivalent formula is ζs = e πis Γ s u s du πi C e u where C is a contour coming from +, going counterclocwise around u =, and returning to + : C u Show that this gives the analytic continuation of ζ to a meromorphic function on C; shift the line of integration to the left to obtain the functional equation relating ζs to ζ s for σ <, and thus for all s by analytic continuation. References [WW 94] Whittaer, E.T., Watson, G.N.: A Course of Modern Analysis... 3 fourth edition. Cambridge University Press, 94 reprinted 963. [HA 9.4 / QA95.W38] [Rudin 976] Rudin, W.: Principles of Mathematical Analysis 3rd edition. New Yor: McGraw-Hill, The full title is 6 words long, which was not out of line when the boo first appeared in 9. You can find the title in Hollis. 5