THE ZERO-DISTRIBUTION AND THE ASYMPTOTIC BEHAVIOR OF A FOURIER INTEGRAL. Haseo Ki and Young One Kim

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1 THE ZERO-DISTRIBUTION AND THE ASYMPTOTIC BEHAVIOR OF A FOURIER INTEGRAL Haseo Ki Young One Kim Abstract. The zero-distribution of the Fourier integral Q(u)eP (u)+izu du, where P is a polynomial with leading term u (m 1) Q an arbitrary polynomial, is described. To this end, an asymptotic formula for the integral is established by applying the saddle point method. 1. Introduction Concerning the zeros of Fourier integrals, G. Pólya proved, among many other things, that all the zeros of the Fourier integral (1.1) e u +izu du (m = 1, 2, 3,... ) are real [11, 12]. If m = 1, it has no zeros at all, but if m > 1 it has infinitely many zeros. (See the remark after Theorem A below.) Recently, J. Kamimoto the authors proved that all the zeros of (1.1) are simple [7]. This is a special property of the polynomials u 2, u 4,.... There are other polynomials with the same property. Pólya proved that all the zeros of e u4m +au +bu 2 +izu du (m = 1, 2, 3,... ; a R; b 0) are real [12, p. 18], it is known that if m = 1, or if m 2 b > 0, then all the zeros are simple. (See Theorem 3.10 of [5], Theorem 1.1 of [8] Theorem 2.3 of [9].) N. G. de Bruijn proved that if P (u) is a polynomial with leading term u P (iu) has real zeros only, then e P (u)+izu du 1991 Mathematics Subject Classification. 30C15, 30D10, 30E15, 42A38. Key words phrases. Saddle point method, Zeros of Fourier integrals. The first author was supported by grant No. R from the Basic Research Program of the Korea Science Engineering Foundation, the second author was supported by SNU foundation in Typeset by AMS-TEX

2 has real zeros only [3, Theorem 20], it can be shown that all the zeros are simple. (See [7].) For general polynomials one cannot expect the same thing. For instance, all the zeros of e u +u+izu du (m = 2, 3,... ) are distributed on the line Im z = 1, because (1.1) has real zeros only. Nevertheless, it is natural to expect that if m is a positive integer P (u) is a polynomial with leading term u, then the zero-distribution of e P (u)+izu du is asymptotically equivalent to that of (1.1) in a suitable sense, because every polynomial is asymptotically equivalent to its leading term. The purpose of this paper is to show that this is in fact the case. The main result is this theorem. Theorem A. Let m be a positive integer, P (u) a polynomial whose leading term is u Q(u) an arbitrary polynomial which is not identically equal to zero. Let the entire function f(z) be defined by f(z) = Q(u)e P (u)+izu du (z C). Then the following hold: (1) The order of f(z) is 1. (2) For each ɛ > 0 all but a finite number of the zeros of f(z) lie in the set Im z ɛ Re z. (3) If P ( u) = P (u) for all u R, then for each ɛ > 0 all but a finite number of the zeros of f(z) lie in the strip Im z ɛ. (4) If P ( u) = P (u) Q( u) = Q(u) for all u R, then all but a finite number of the zeros of f(z) are real simple. The general properties of entire functions that are needed in our proof of the results can be found in [2]. If m = 1, then, by a direct calculation, one can show that f(z) has exactly d zeros, where d is the degree of Q(u) (see Section 2 of this paper); if m 2, then the first assertion implies that the order of f(z) is not an integer, hence Hadamard s factorization theorem implies that f(z) has infinitely many zeros. Suppose that P ( u) = P (u) Q( u) = Q(u) for all u R. If m 2, then it can be shown that the number of non-real zeros of f(z) does not exceed that of the polynomial Q(iu) (see [3, p. 224], [8, Section 3] [10, Section 2]); if m 3, or m = 2 Q(iu) has non-real zeros, then the last assertion of this theorem, de Bruijn s theorem [3, Theorem 13] (see also [9, Theorem 2.3]) Theorem 1.1 of [8] imply that there is a real constant Λ such that Q(u)e P (u)+λu2 +izu du 2

3 has non-real zeros if only if λ < Λ. See also [10]. The last assertion of Theorem A is a direct consequence of the third one the following theorem [9, Theorem 2.2] which solved a conjecture of de Bruijn [3, pp. 199, 205]. Theorem B. Let F (u) be a complex-valued function defined on the real line; suppose that F (u) is integrable, ( (1.2) F (u) = O e u b) ( u, u R) for some constant b > 2, F ( u) = F (u) for all u R. Suppose also that for each ɛ > 0 all but a finite number of the zeros of the Fourier integral of F (u) lie in the strip Im z ɛ. Then for each λ > 0 all but a finite number of the zeros of the Fourier integral of e λu2 F (u) are real simple. Proof that (3) implies (4). Suppose P ( u) = P (u) Q( u) = Q(u) for all u R. Let λ be an arbitrary positive constant put F (u) = Q(u)e P (u) λu2. We may assume, without loss of generality, that m 2. Then the function F (u) satisfies (1.2) with b = 1 > 2, it is clear that F ( u) = F (u) for all u R. Since P λ (u) = P (u) λu 2 is a polynomial with leading term u P λ ( u) = P λ (u) for all u R, (3) implies that for each ɛ > 0 all but a finite number of the zeros of the Fourier integral of F (u) lie in the strip Im z ɛ. Hence, by Theorem B, all but a finite number of the zeros of the Fourier integral of e λu2 F (u) = Q(u)e P (u) are real simple. The other assertions of Theorem A will be proved in Section 2. They are consequences of Theorem C stated below which describes the asymptotic behavior of the function f(z). In order to state the theorem, we need some notation. Let the polynomial P (u) be given by P (u) = u + a 1 u a 1 u + a 0. Suppose that Re z 0 write z = re iθ with r 0 π/2 θ π/2. We put (1.3) R = ( r ) 1 1 ζ = r 1 1 e iθ 1. Thus z = ζ 1, we have P (u) + iz = u 1 + ( 1)a 1 u a 1 + iζ 1. There is a positive constant r 1 such that if r > r 1, then the equation P (u) + iz = 0 has exactly 1 (distinct simple) roots in the complex u-plane. Suppose that r > r 1 let u j, j [ m + 1, m 1] Z, denote the 1 roots of the equation P (u) + iz = 0. By taking r 1 sufficiently large, we may assume that these roots are given by (1.4) u j = () 1 1 e πi 1( j) ζ(1 + A j1 ζ 1 + A j2 ζ 2 + ), 3

4 where the A jk s are constants independent of z, the series converge absolutely uniformly for r > r 1. Each u j is an analytic function of z, which is defined for r > r 1 θ π/2. It is clear that for each j we have (1.5) u j = Re 1( i π +θ+2jπ) ( O(R 1 ) ) (1.6) P (u j ) + izu j = ( 1)u j ( 1 + O ( R 1 )) = ( 1)R e i( π 2 +θ+ 1( 1 π +θ+2jπ)) ( O ( R 1)) = ( 1)() 1 e m+2j 1 πi ζ ( 1 + O( ζ 1 ) ) for r θ π/2. Theorem C. Suppose m 2. Let Q(u) be a monic polynomial of degree d f(z) be as in Theorem A. Then, with A = i d π m( 1) ( ) 1 m+d Re iθ 1 B = i d e πi 4m 2 (1 m+d), we have f(z) = A [Be P (u 0)+izu 0 ( ( 1 + O R 1 )) + Be P (u m 1)+izu m 1 ( ( 1 + O R 1 ))] where R is defined in (1.3). ( θ π/2, r ), Several authors applied the saddle point method to obtain asymptotic formulas for the integral (1.1). See, for instance, [1, 4, 6, 13]. Theorem C is also proved by an application of the saddle point method (Section 3). In fact, we will prove a more precise formula (see (3.1) in Section 3). 2. Proof of (1), (2) (3) of Theorem A If m = 1, that is, if P (u) = u 2 + au + b for some constants a b, then we have f(z) = πe b Q( id) exp ( (z ia) 2 /4 ), D = d/dz, hence f(z) has exactly d zeros, where d is the degree of the polynomial Q(u), it is clear that f(z) is of order 2. This proves the theorem in the case when m = 1. From here on, we assume that m 2. It is enough to prove the assertions in the right half plane Re z 0. If we put ( π K(θ, j) = ( 1) cos 2 + θ ( π 2 + θ + 2jπ ) ),

5 then (1.6) implies that Re (P (u j ) + izu j ) = R ( K(θ, j) + O(R 1 ) ) ( θ π/2, r ). Since K(θ, m 1) < K(θ, 0) for π/2 θ < 0 K(θ, 0) < K(θ, m 1) for 0 < θ π/2, the first the second assertions are immediate consequences of Theorem C. To prove (3), suppose that P ( u) = P (u) for all real u. We will show the existence of positive constants β, C 1, C 2 C 3 such that 0 < β < 1 ( ) (2.1) Re (P (u m 1 ) + izu m 1 ) Re (P (u 0 ) + izu 0 ) C 1 y x 1 1 C2 C 3 (x > 1, y βx), where x = Re z y = Im z. It is clear that this inequality, Theorem C (2) imply (3): Since β > 0, (2) implies that all but a finite number of the zeros of f(z) lie in the set {z : Im z β Re z }; (2.1) together with Theorem C imply that for every ɛ > 0 there is a positive constant x 1 such that f(z) does not vanish in the set {z : Re z x 1, ɛ < Im z βre z}. We next prove inequality (2.1). Since P ( u) = P (u) for all real u, the coefficients a 1, a 3,..., a 1 are purely imaginary the coefficients a 2, a 4,..., a 0 are real. If z is real, then the roots of the equation P (u) + iz = 0 are symmetrically located with respect to the imaginary axis in the complex u-plane. In particular ū 0 = u m 1 for real z, we have P (u 0 ) + izu 0 = P ( ū 0 ) + iz( ū 0 ) = P (u m 1 ) + izu m 1 (z R). Hence, by (1.4), we have (2.2) P (u 0 ) + izu 0 = B k ζ k P (u m 1 ) + izu m 1 = k=0 k=0 B k ζ k for some constants B 0, B 1, B 2,..., the series converge absolutely uniformly for r > r 1. Thus (2.3) ( ) Re (P (u m 1 ) + izu m 1 ) Re (P (u 0 ) + izu 0 ) = Re Bk B k ζ k k=0 = 2 ( Im B 0 Im ζ + Im B 1 Im ζ Im B 1 Im ζ ) + O(1) for r > r 1. From (1.6) (2.2), we see that Im B 0 > 0. For a constant a, we define the function h a by ( ) a h a (s) = ( 1) n s 2n ( s < 1). 2n + 1 n=1 5

6 If z = x + iy, with x > 0 x < y < x, then Im ζ k = yx 1 k 1 ( k ( y ) ) 1 + h (k = 0, 1, 2,..., 1). k 1 x There is a constant β such that 0 < β < 1 s β h (s) 1 < 1. Suppose z = x + iy, x > 1 βx y βx. Then we have ( Im ζ k k y 1 + sup Im ζ > 1 1 y x 1 1 s β h k 1 ) (s) (k = 1, 2,..., 1). From these inequalities (2.3), we obtain the desired result. 3. Proof of Theorem C Suppose r > r 1 π/2 θ π/2. Put z = re iθ, J = {j Z : j m 1} J + = {0, 1,..., m 1}. Let ρ 1 be a positive constant such that P (u), Q(u) 0 for u > ρ 1, let D = {ρe iφ : ρ > ρ 1, π/2 < φ < 3π/2}. There is a unique analytic function V (u) defined in D such that P (u)v (u) 2 = 2 (u D) V (u) = 1 m( 1) u 1 m ( 1 + O ( u 1)) (u D, u ). From (1.5), we may assume that u j D for j J +. We put v j = ( 1) j V (u j ) ϕ j (z) = πv j Q(u j )e P (u j)+izu j (j J + ). Each v j is an analytic function of z satisfying v j = 1 R 1 m e m( 1) P (u j ) 2 1((1 m)θ+( i 1 m = v 2 j 2 +j)π) ( 1 + O(R 1 ) ) 6 ( θ π/2, r ).

7 Each ϕ j is an analytic function of z does not vanish in the region r > r 1, θ π/2. Since m 2, we have < 2. Hence (1.6) implies that 1 ln ϕ j (z) = O ( r 2) ( θ π/2, r ). We will prove that (3.1) f(z) = ϕ 0 (z) ( 1 + O(R ) ) + ϕ m 1 (z) ( 1 + O(R ) ) ( θ π/2, r ). A straightforward calculation shows that this implies Theorem C. Let j J + be arbitrary. There is a curve (continuous piecewise smooth function) γ j : R C such that γ j (0) = u j (3.2) P (γ j (s)) + izγ j (s) = P (u j ) + izu j s (s R). We must have We also have or equivalently Im (P (γ j (s)) + izγ j (s)) = Im (P (u j ) + izu j ) (s R) lim γ j(s) = lim γ j(s) =. s s lim s 0 P (u j )(γ j (s) u j ) 2 P (u j )(γ j (s) u j ) 2 = 1, (γ j (s) u j ) 2 lim s 0 γ j (s) u j 2 = v2 j v j 2. We may assume, by replacing γ j (s) with γ j ( s) if necessary, that γ j (s) u j (3.3) lim s 0+ γ j (s) u j = v j v j. If the values Im (P (u j ) + izu j ), j J, are all different, then the curves γ j, j J +, are uniquely determined by (3.2) (3.3); if θ (2k+1)π 4m for all k [ m, m 1] Z, then Im (P (u j ) + izu j ), j J, are all different, whenever r becomes sufficiently large. Let α be a constant such that 0 < α < π 4m, let R = R 1 R 2 R 3, where R 1 = {re iθ : r > r 1, θ α}, R 2 = {re iθ : r > r 1, π 2 θ π 2 + α} R 3 = {re iθ : r > r 1, π 2 α θ π 2 }. 7

8 We may assume, by taking r 1 sufficiently large, that for every z R the curves γ j, j J +, are uniquely determined. Using an elementary argument, one can prove that if r 1 is sufficiently large, then the following hold: (1) If z R 1, lim s γ j (s) γ j (s) = ejπi/m lim s γ j (s) γ j (s) = e(j+1)πi/m (j J + ). (2) If z R 2, (3) If z R 3, lim s γ 0 (s) γ 0 (s) γ 0 (s) = 1 lim s γ 0 (s) = 1. lim s γ m 1 (s) γ m 1 (s) γ m 1 (s) = 1 lim s γ m 1 (s) = 1. From here on, we assume the above three statements. Let ɛ be a positive constant such that 0 < ɛ < π 4m, put S j = { ρe iφ : ρ > 0, If z R 1, then there is a positive constant s 1 such that jπ m ɛ φ jπ } m + ɛ. s s 1 γ j (s) S j s s 1 γ j (s) S j+1 hold for every j J + ; since the leading term of P (u) is u 0 < ɛ < π 4m, there are positive constants A B such that Q(u)e P (u)+izu Ae B u (u S 0 S 1 S m ). Hence, by Cauchy s theorem, (3.4) f(z) = m 1 j=0 γ j Q(u)e P (u)+izu du (z R 1 ). Similarly, we have (3.5) f(z) = Q(u)e P (u)+izu du γ 0 (z R 2 ), (3.6) f(z) = Q(u)e P (u)+izu du (z R 3 ). γ m 1 Now, we need a lemma whose proof will be given after the proof of (3.1). 8

9 Lemma. For arbitrary j J + we have γ j Q(u)e P (u)+izu du = ϕ j (z) ( 1 + O ( R )) (z R, r ). From (3.5), (3.6) the lemma, we have { ϕ0 (z) ( 1 + O(R ) ) (θ = π/2, r ), (3.7) f(z) = ϕ m 1 (z) ( 1 + O(R ) ) (θ = π/2, r ). If we put then (1.6) implies that ( π K(θ, j) = ( 1) cos 2 + θ + 1 ( π ) ) θ + 2jπ, Re (P (u j ) + izu j ) = R ( K(θ, j) + O(R 1 ) ) ( θ π/2, r ). We have K(θ, j) < min {K(θ, 0), K(θ, m 1)} for θ α 1 j m 2. Hence, by (3.4) the lemma, (3.8) f(z) = ϕ 0 (z) ( 1 + O(R ) ) + ϕ m 1 (z) ( 1 + O(R ) ) (z R 1, r ). Now (3.1) will follow, once we show that (3.9) f(z) = ϕ 0 (z) ( 1 + O(R ) ) ( π/2 θ α, r ) (3.10) f(z) = ϕ m 1 (z) ( 1 + O(R ) ) (α θ π/2, r ), because K(θ, 0) > K(θ, m 1) for π/2 θ < 0 K(θ, m 1) > K(θ, 0) for 0 < θ π/2. We prove (3.9) only: (3.10) is proved by the same way. We have ζ = () 1/( 1) R. Hence (3.9) is equivalent to the assertion that the analytic function ( ) f(z) h 0 (z) = ζ ϕ 0 (z) 1 is bounded in the region R 4 = {re iθ : r > r 1, π/2 θ α}. From (3.7), h 0 (z) is bounded on the ray θ = π/2, r > r 1. Since K( α, 0) > K( α, m 1), (3.8) implies that f(z) = ϕ 0 (z) ( 1 + O(R ) ) (θ = α, r ). Hence h 0 (z) is bounded on the ray θ = α, r > r 1. Therefore h 0 (z) is bounded on the boundary of the region R 4. It is known that the order of the entire function f(z) is at most (< 2). (For a proof, see [12, pp. 9 10].) Since 1 ln ϕ 0 (z) = O(r 2 ) ( θ π/2, r ), 9

10 it follows that ( h 0 (z) = O e Cr2) (z R 4, r ) for some positive constant C. It is obvious that the two rays θ = π/2, r > r 1 θ = α, r > r 1 make an angle less than π/2. Therefore the Phragmén-Lindelöf theorem [2, p.4] implies that the function h 0 (z) is bounded throughout the region R 4. This proves (3.9). Proof of the Lemma. Let j J + be arbitrary. Put s j = inf {s R : γ j (s) u j R 1 3 s + j = sup {s R : γ j (s) u j R 1 3 } }. We also put α j = γ j (s j ) u j β j = γ j (s + j ) u j. Thus Q(u)e P (u)+izu du = I 1 + I 2 + I 3, γ j where I 1 = I 2 = I 3 = uj +β j u j +α j + s + j s j Q(u)e P (u)+izu du, Q(γ j (s))e P (γ j(s))+izγ j (s) γ j (s)ds, Q(γ j (s))e P (γ j(s))+izγ j (s) γ j (s)ds. If we put u = u j + w, then Q(u)e P (u)+izu = Q(u j )e P (u j)+izu j (1 + F (w)) ( 1 + G(w) + G(w) 2 H(w) ) e P (u j )w 2 /2, where F (w) = deg Q n=1 Q (n) (u j ) n!q(u j ) wn, G(w) = we have P (u j )/2 = v 2 j. Hence n=3 P (n) (u j ) w n H(w) = eg(w) 1 G(w) n! G(w) 2 ; βj I 1 = Q(u j )e P (u j)+izu j (1 + F (w)) ( 1 + G(w) + G(w) 2 H(w) ) e v α j 10 2 j w 2 dw,

11 a straightforward calculation shows that βj α j Therefore (1 + F (w)) ( 1 + G(w) + G(w) 2 H(w) ) e v 2 j w 2 dw = ( ( πv j 1 + O R )) From (3.2), we have I 1 = ϕ j (z) ( 1 + O ( R )) Since s + j > 0, u j γ j (s + j ) = R1 3 it follows that ( θ π/2, r ). I 2 = e P (u j)+izu j Q(γ j (s))γ j (s)e s ds. s + j s + j = P (u j) + izu j ( P (γ j (s + j )) + izγ j(s + j )) = n=2 P (n) (u j ) (γ j (s + j n! ) u j) n, s + j m( 1)R/3 (z R, r ). We may assume, by taking r 1 sufficiently large, that Then we have γ j (s) = In particular, γ j (s) = u j + α j + γ j (s) u k 1 (z R, s R, k J \ {j}). 1 P (γ j (s)) + iz = 1 s s + j k J γ j (s) u k 1 R 3 1 γ j (s)ds u j + α j + s s+ j Now, it is clear that we can find positive constants A B such that Q(γ j (s))γ j (s)e s ds (z R). Ae RB Therefore we have s + j the same argument gives This proves the lemma. I 2 = ϕ j (z) O ( R ) I 3 = ϕ j (z) O ( R ) (z R, r ), (z R, r ). ( θ π/2, r ). (z R, s s + j ). R 3 1 (z R, s s + j ). Acknowledgment. The authors thank the referee for helpful comments. 11

12 References [1] N. G. Bakhoom, Asymptotic expansions of the function F k (x) = 0 e uk +xu du, Proc. London Math. Soc. 35 (1933), [2] R. P. Boas, Entire Functions, Academic Press, New York, [3] N. G. de Bruijn, The roots of trigonometric integrals, Duke Math. J. 17 (1950), [4] W. R. Burwell, Asymptotic expansions of generalized hypergeometric functions, Proc. London Math. Soc. 22 (1923), [5] T. Craven G. Csordas, Differential operators of infinite order the distribution of zeros of entire functions, J. Math. Anal. Appl. 186 (1994), [6] J. Kamimoto, On an integral of Hardy Littlewood, Kyushu J. Math. 52 (1998), [7] J. Kamimoto, H. Ki Y.-O. Kim, On the multiplicities of the zeros of Laguerre-Pólya functions, Proc. Amer. Math. Soc. 128 (2000), [8] H. Ki Y.-O. Kim, A generalization of Newman s result on the zeros of Fourier transforms, Comput. Methods Funct. Theory 2 (2002), [9] H. Ki Y.-O. Kim, De Bruijn s question on the zeros of Fourier transforms, J. Anal. Math. 91 (2003), [10] C. M. Newman, Fourier transforms with only real zeros, Proc. Amer. Math. Soc. 61 (1976), [11] G. Pólya, On the zeros of an integral function represented by Fourier s integral, Messenger of Math. 52 (1923), [12], Über trigonometrische Integrale mit nur reellen Nullstellen, J. Reine Angew. Math. 158 (1927), [13] D. Senouf, Asymptotic numerical approximations of the zeros of Fourier integrals, SIAM J. Math. Anal. 27 (1996), Department of Mathematics, Yonsei University, Seoul , Korea address: haseo@yonsei.ac.kr Department of Mathematical Sciences, Seoul National University, Seoul , Korea address: kimyo@math.snu.ac.kr 12