Laplace s Résultat Remarquable[4] and its Ramifications

Transcription

1 Laplace s Résultat Remarquable[4] and its Ramifications Mark Bun 3 June 29 Contents Introduction 2 2 Laplace s Integral 2 3 Laplace s Identity 5 3. A Few Estimates The Reflection Formula A Simple Proof of the Identity Some Summation Formulas 9 4. Confluent Hypergeometric Functions Maass Formula Lipschitz s Formula Conclusion 2

2 Introduction One of the most important and widely studied special functions in mathematics is the gamma function Γz) = t z e t dt, Re z >, Figure : A domain-colored plot of Γz), rendered with Sage along with its meromorphic extension to the entire complex plane. However, in studying the gamma function, it is sometimes useful to consider its reciprocal. Like the gamma function itself, there are numerous definitions for this mathematical object. One common form is: Γz) = zeγz k= + z ) e z/k, z C, k where γ.5772 denotes the Euler-Mascheroni constant. For the derivation, see [3] p But occasionally more useful is Pierre-Simon Laplace s 82 integral formula 2π a + it) z dt = Γz). In [7], Wladimir de Azevedo Pribitkin explores Laplace s integral and applies it to the proofs of several properties of the gamma function. This paper will highlight some of the major results of Pribitkin s exposition, culminating in the derivations of the Maass and Lipschitz summation formulas. 2 Laplace s Integral We will begin by defining Laplace s Integral and recording a few of its essential properties. 2

3 Definition 2.. We define Laplace s Integral as where a, Re z >. L a z) = 2π a + it) z dt, We will use the convention z α = e α log z where we take the principal branch of the logarithm, with π arg z < π. Almost immediately clear from the definition is that L a z) converges absolutely and uniformly for Re z + ɛ and Im z C for any C. Let z = x + iy. We estimate the integrand: a + it) z = expa + it z loga + it)) = expa x log a + it + y arga + it)) e a+πc a + it x. For t [ a, a], the integrand is bounded from above by 2e a+πc a x, and is hence proper. Elsewhere, it is bounded by 2e a+πc t x. Since a t x and a t x converge, L a z) converges absolutely and uniformly by the Weierstrass M-Test. Since C was arbitary, we have absolute and uniform convergence for all compact subsets of x >. We leave it to the reader to show that L a z) also converges uniformly on compact subsets of x >, and hence defines an analytic function on this half-plane. Next we will demonstrate the functional equation L a z) = zl a z + ). ) As in the proof of the analogous property for the gamma function, we will integrate by parts: L a z) = 2π = 2π = zl a z + ). a + it) z dt a + it) z+ ) z a + it) z+ We can use this recurrence relation to perform an analytic continuation See [3], pp ) of L a z) to arbitrary z. As a result, we see that L a z) can be extended to be an entire function. Thirdly, we will show that L a z) is in fact independent of a. We accomplish this by formally differentiating with respect to a: 3

4 2π a a + it) z dt = 2π = 2π a + it) z za + it) z a + it) 2z dt a + it) z + zea+it a + it) = L a z) zl a z + ). z+ dt Since these integrals converge uniformly, and by ) their difference is zero, we find that L a z) = Lz) is independent of our choice of positive a. For convenience in calculations, we will generally choose a = in the remainder of this paper. Figure 2: Lz), rendered with Sage Finally, we demonstrate the connection between Lz) and the factorial function. To do so, we evaluate L 2) via contour integration. Let C R be the semicircular contour in the upper half-plane with radius R >. Then by the Residue Theorem, we have: R R e +it + it) CR 2 dt + e +iζ dζ = 2πi Res + iζ) 2 = 2πi lim ζ i = 2π. [ ] e +iζ + iζ) 2, i ζ i) 2 e +iζ d dζ By the ML-estimate, we have: e CR +iζ + iζ) 2 dζ e R 2 πr, + iζ) 2 ) 4

5 which tends to as R. Since the integral defining L2) converges absolutely, we thus have L2) =. By induction on our functional equation ), we have: Ln) = n )! for all n Z +. Our recurrence relation shows that we have simple zeroes at each nonpositive integer. 2) 3 Laplace s Identity The objective of this section is to prove the identity Γz)Lz) =, 3) for all z C, taking limits as appropriate for the nonpositive integers. 3. A Few Estimates An easy result from complex power series is Liouville s Theorem, which states that a bounded entire function is constant. With a little more work, we can prove the following extension. Theorem 3. Extended Liouville Theorem). Let P z) be a polynomial of degree n. Suppose fz) is an entire function and fz) P z ) for z > ρ. Then f is a polynomial of degree at most n. Proof. Suppose P z ) = a n z n + a n z n + + a z + a. Choose R > max{, ρ}. Then: n n P z ) z n a j R n a j. Hence, by the Cauchy estimates, we have f m) ) m! R m Rn n a j. Suppose m > n. Then f m) ) as R, so f m) ) = for all m > n. Hence, the power series expansion of fz) centered at the origin consists only of terms with degree less than or equal to n, so fz) is a polynomial of at most degree n. Example. Suppose fz) is an entire function that satisfies fz) z 4 / log z for z >. For z >, we have log z z, so fz) z 3. Then by the extended Liouville theorem, fz) is a polynomial of at most degree 3. 5

6 A popular and somewhat surprising result characterizing the gamma function is the Bohr-Mollerup Theorem. The theorem states that a continuous function F : R R satisfying F ) =, the functional relation F x + ) = xf x) and the condition of log convexity is the gamma function. In 939, Helmut Wielandt discovered a similar result for the gamma function in the complex plane. The following theorem is taken from [5]. Theorem 3.2 Wielandt s Theorem). Let F z) be an analytic function in the right half-plane A = {z C : Re z > } having the following properties: a) F z + ) = zf z) for all z A b) F z) is bounded in the strip S = {z C : Re z < 2} c) F ) =. Then F z) = Γz). Proof. Let f = F Γ, which is analytic in A. Note that since f) =, we have fz) = z )gz) for gz) analytic at. From a) and the recurrence relation for the gamma function, we can extend f to be analytic at : z + ) )fz + )gz + ) f) lim = g). z z Therefore, by our recurrence formula, we can extend f to the entire function ˆf. Since Γz) ΓRe z), we have that Γ is bounded on S and hence f is bounded by b). Let S = {z C : Re z < }. For all z S with Im z, it is clear that f is bounded. For points with Im z >, the boundedness of f on S follows from fz) = fz + )/z and the boundedness of f on S. Consider the entire function sz) = ˆfz) ˆf z). Since fz) and f z) take on the same values in S, the function sz) is bounded in S. Since ˆfz + ) = z ˆfz) and ˆf z) = ˆf z)/z, it follows that sz + ) = sz). Hence sz) is bounded on C and therfore constant by Liouville s theorem. Since sz) s) = f) ˆf) =, we have ˆfz) so F z) = Γz) on A. Example 2. Let z F z) = π /2 2 z Γ Γ 2) z + Since the gamma function is analytic in the right half-plane, F is also analytic here. We also have ) ) z + z + 2 F z + ) = 2π /2 2 z Γ Γ 2 2 ) z + z = π /2 2 z Γ zγ 2 2) = zf z). 2 ). 6

7 Since t z Re z = t we have that Γz) ΓRe z) in the right half-plane. Hence, Γz) is bounded on vertical strips {z C : a Re z b} for < a < b. In addition, we have that 2 z is bounded on vertical strips, so F z) is bounded on {z C : Re z < 2}. Finally, it is easy to show directly that F ) =. Therefore, by Wielandt s Theorem, F z) = Γz). By a simple change of variables, we have the Legendre Duplication Formula Γ2z) = π /2 2 2z Γz)Γ z + ) The Reflection Formula In this section, we will prove an analog of the Euler reflection formula for Laplace s Integral. We will use this formula to prove Laplace s Identiy, and as a corollary, give an easy proof of the reflection formula for the gamma function. Theorem 3.3. Lz)L z) = sin πz π. Proof. Let Qz) = πlz)l z)/ sin πz. Since the simple zeroes of sin πz coincide with those of Lz), we see that Qz) is an entire function. Further, from our functional equation ), we have Qz) = Qz + ). Let z = x + iy. Then for x 2 and y we have Lz) e 2π = π/2 π/2 e y arg+it) + it x dt e y arctan t + t 2 dt e yt dt = y eπy/2 e πy/2 ). Furthermore, for all x and for all y sin πz = 2i e iz e iz 2 e π y e. π y 7

8 By repeated applications of ), on the strip 2 x 3 we have π Qz) = z)2 z)3 z)4 z)l5 z)lz) sin πz = Oy 4 )O y e π y /2 ) 2 Oe π y ) = Oy 2 ), independent of x in this strip. Hence, there are positive constants A and B such that Qz) A z 2 + B on the strip. By our recurrence relation Qz) = Qz + ), this estimate holds on C. By the extended Liouville theorem, Qz) is a polynomial of at most degree 2. However, since Q) = Q2) = Q3), the polynomial Qz) Q) has at least three zeroes. By the fundamental theorem of algebra, Qz) must be constant. Since Qz) lim z desired reflection formula. 3.3 A Simple Proof of the Identity πz sin πz =, we have the In [7], Pribitkin formally gives two rather different proofs of Laplace s Identity 3). Relegated to his closing remarks is an outline of the following proof. With the reflection formula and Wielandt s Theorem in hand, we can fill in the details. Let F z) = /Lz) = πl z)/ sin πz from the reflection formula, and as usual let z = x + iy. We note that where sin πz has simple zeroes, L z) does as well. Hence, the singularities of F are removable, so F is analytic for x >. Furthermore, we have F z + ) = πl z) zπl z) = = zf z). sinπz + π) sin πz Finally, we show that F z) is bounded in the strip x < 2. Making use of our estimates from Theorem 3.3, we have F z) = π z)2 z)3 z)l4 z) sin πz 2π z)2 z)3 z)e πy/2 e πy/2 ) e π y e π y = Oy 2 e π y /2 ). Since the polynomial factor is dominated by the exponential as y, we have that F z) is bounded in this strip. Finally, since F ) = /L) = /L2) =, we can apply the result of Wielandt s theorem to prove that F z) = Γz). Since Γz) has simple poles where Lz) has simple zeroes, it follows that Γz)Lz) on C, taking limits at the nonpositive integers. 8

9 As promised, the following corollary is immediately clear from 3) and Theorem 3.3. Corollary 3.4 Euler s Reflection Formula). Γz)Γ z) = π sin πz. 4 Some Summation Formulas Pribitkin s primary application of Laplace s integral is to the derivation of the Maass and Lipschitz summation formulas. These formulas in turn have a wide range of consequences, especially in the theory of modular forms. To illustrate the power of this field, we present the following surprising result. Theorem 4. Radamacher s Formula). A partition of a positive integer n is an expression of n as the sum of smaller positive integers. Let pn), called the partition function, be the number of partitions of n where order does not matter). Then pn) = π 2 k= A k n) k d sinh dn π k 2 3 n 24 n 24 ) ) ), where and and finally A k n) = m<k,gcd m,k)= sm, k) = fx) = n mod k e πism,k) 2nm/k), fn/k)fmn/k), { x x /2, if x R \ Z;, if x Z. The summation formulas we will derive thus build a bridge between analysis and number theory. For a further discussion, see [6]. 4. Confluent Hypergeometric Functions The coefficients of Maass Formula are most easily expressed in terms of the confluent hypergeometric function of the second kind. Here, we present a few of the basics. Definition 4.2. We define the hypergeometric equation as for a, b, c C. z z)w + c a + b + )z)w abw = 9

10 If we replace z with z/b, the equation has singularities at, b, and. If we let b, then becomes a confluence of two singularities. This gives rise to the confluent hypergeometric equation. Definition 4.3. The confluent hypergeometric equation is given by for a, c C. zw + c z)w aw = The confluent hypergeometric functions are two linearly independent solutions of the confluent hypergeometric equation around z =. Definition 4.4. Suppose Re c > Re a >. Then define the confluent hypergeometric function of the first kind as Φa, c, z) = Γc) Γa)Γc a) t) c a t a e zt dt. Example 3. Here is a quick frustruating example from elementary calculus. For x R erfx = 2 x e t2 dt = 2x Φ/2, 3/2, x). π π Definition 4.5. Suppose Re a >, c C, Re z >. hypergeometric function of the second kind is Ψa, c, z) = Γa) t + ) c a t a e zt dt. Then the confluent Hypergeometric functions have diverse applications in physics and engineering, particularly in the study of waves. For instance, the confluent hypergeometric equation is related to the Bessel equation, which appears in the study of coincidentally) Laplace s equation. Most importantly for this paper, we note that Ψ has an entire analytic continuation. In particular, Ψ, b, z) =. For more information see [] and [2]. 4.2 Maass Formula A function that makes numerous appearances in the theory of automorphic forms is fz) = l= e 2πiλx+l) z + l) s z + l) w, for λ R and s, w, z = x + iy C. To ensure absolute convergence, we suppose y and Res + w) >. Since fz) is periodic with period, we can express it as a Fourier series

11 where fz) = c n+λ y, s, w)e 2πinx, 4) n= c n+λ y, s, w) = fz)e 2πinx dx. Since our definition is symmetric about the imaginary axis, we can take y >. Evaluating the coefficients, and making use of the substitution u = x + l: c n+λ y, s, w) = = l= l= l+ = i) s w l e 2πiλx+l) z + l) s z + l) w e 2πinx dx e 2πin+λ)u u + yi) s u yi) w du e 2πin+λ)u y iu) s y + iu) w du. Suppose Re s > and Re w >. Then from the definitions of the gamma function and Laplace s integral, and by Fubini s theorem, we have c n+λ y, s, w) = i) s w = i)s w Γs) Γz) v s e yv = 2π i)s w Γs)Γw) e2πn+λ)y ) e v s e y iu)v 2πin+λ)u dv y + iu) w du e i[v 2πn+λ)]u y + iu) w du dv max{,2πn+λ)} v s v 2πn + λ)) w e 2yv dv. Hence we have and c y, s, w) = 2π i)s w Γs + w ) Γs)Γw)2y) s+w, 5) c n+λ y, s, w) = 2π) s+w i) s+w n + λ s+w e 2π n+λ y { Ψw,s+w,4πn+λ)y) Γs), if n + λ > ; Ψs,s+w, 4πn+λ)y) Γw), if n + λ <. Maass formula is the Fourier series 4) given by coefficients 5) and 6). 6)

12 4.3 Lipschitz s Formula Although Maass formula in its full generality has numerous important uses, we can easily derive from it a relatively simple summation formula with interesting consequences of its own. If we set w = in Maass s formula which is legal, since Ψ, c, z) = ), then for y > we obtain for Re s >. l= e 2πiz+l)λ z + l) s = 2πi)s Γs) n+λ> n + λ) s e 2πinz, 7) Example 4. Take s = 2 and λ = in Lipschitz s formula. We have on the right-hand side 2πi) 2 Γ2) ne 2πinz = 4π 2 = 2πi d dz d dz e 2πiz 2πie 2πiz = 2πi e 2πiz ) 2 = 4π 2 e πiz e 2πiz 2πi e2πinz ) 2 ) = 4π 2 2i sin πz ) 2 = π2 sin 2 πz. As a result, we have the identity 5 Conclusion π 2 sin 2 πz = n= ) z + n) 2. since e 2πiz < for y > Pribitkin s examination emphasizes some of the truly powerful aspects of complex analysis. Just as the gamma function has a way of cropping up in various areas of mathematics, so too have we seen Laplace s integral. While the infinite product form for the reciprocal of the gamma function has its merits, Laplace s integral form has proven to be very convenient for making estimates. Further, its applications to the theory of modular forms reinforces the connection between complex analysis and number theory, much in the spirit of the prime number theorem. Although after two centuries, this result still has yet to catch on as a mainstream mathematical tool, its elegance and versatility still make it an exceptional piece of mathematics. 2

13 References [] G. Andrews, R. Askey and R. Roy, Special Functions, Encyclopedia of Mathematics and its Applications, Cambridge University Press, Cambridge, 999. [2] H. Buchholz, The Confluent Hypergeometric Function, Springer Tracts in Natural Philosophy, Springer-Verlag, Berlin, 969. [3] T. Gamelin, Complex Analysis, Undergraduate Texts in Mathematics, Springer, New York, 2. [4] P.S. Laplace, Théorie Analytique des Probabilités, V. Courcier, Paris, 82. [5] R. Remmert, Wielandt s Theorem about the Γ-function. The American Mathematical Monthly, 33):24-22, Mar., 996. [6] W. Pribitkin, Revisiting Rademacher s formula for the partition function pn). Ramanujan J., 4: , 2. [7] W. Pribitkin, Laplace s Integral, the Gamma Function and Beyond. The American Mathematical Monthly, 93): , Mar., 22. 3