Asymptotics of Integrals of. Hermite Polynomials

Transcription

1 Applied Mathematical Sciences, Vol. 4, 010, no. 61, Asymptotics of Integrals of Hermite Polynomials R. B. Paris Division of Complex Systems University of Abertay Dundee Dundee DD1 1HG, UK Abstract Integrals involving products of Hermite polynomials with the weight factor exp ( x ) over the interval (, ) are considered. A result of Azor, Gillis and Victor (SIAM J. Math. Anal. 1 (198) ] is derived by analytic arguments and extended to higher order products. An asymptotic expansion in the case of a product of four Hermite polynomials H n (x) asn is obtained by a discrete analogue of Laplace s method applied to sums. Keywords: Hermite polynomials, Moment integrals, Asymptotic expansion 1. Introduction Consider an even set of distinguishable elements divided into k subsets containing n r elements (r =1,,...,k) with k r=1 n r an even integer, s. In [], Azor, Gillis and Victor considered the combinatorial problem of the arrangement of these elements into s disjoint pairs. They showed that the number of possible pairings of different types is expressed in terms of integrals of the type I(n 1,n,...,n k )= k e x r=1 H nr (x) dx, (1.1) where H n (x) denotes a Hermite polynomial of degree n. The evaluation of the above integral in the case k = 4 when n 1 = n and n = n 4 was obtained by combinatorial arguments and expressed in terms of a terminating

2 044 R. B. Paris F generalised hypergeometric function. In the case n r = n (1 r 4) the asymptotics of this integral as n were obtained by means of an elaborate argument involving the use of a generating function derived from a contour integral containing a Legendre function. In this paper, we shall obtain the evaluation of (1.1) in the cases k =4, 5 and 6 by analytic arguments. In addition, we show how an asymptotic expansion in the case k = 4 and n r = n can be obtained as n by a discrete analogue of Laplace s method applied to sums. We assemble here the main properties of the Hermite polynomials that we shall require. The H n (x) are defined by H n (x) =( ) n e x and have the representation H n (x) = dn dx n (e x ), n =0, 1,,... ( ) n/ n! ( 1 n)! 1 F 1 ( 1 n; 1 ; x ) (n even), ( ) (n 1)/ n! ( 1 n 1 )!x 1F 1 ( 1 n + 1 ; ; x ) (n odd), (1.) where 1 F 1 denotes the confluent hypergeometric function. These polynomials satisfy the well-known orthogonality property e x H m (x)h n (x) dx = n πn! δ m,n, (1.) where δ m,n is the Kronecker delta. From [4, p. 88, 7.76], we have the integrals e x x ν H n (x) dx = ( )n n 1 π Γ( 1 ν + 1 )Γ(n + 1 ) 1F 1 ( n, 1 ν + 1 ; 1 ;1), e x x ν H n+1 (x) dx = ( )n n 1 π Γ( 1 ν + 1)Γ(n + ) 1F 1 ( n, 1 ν +1; ;1), the first integral holding for Re(ν) > 1 and the second integral for Re(ν) >. Making use of Vandermonde s theorem [5, p. 4] 1F 1 ( n, r + 1; 1;1)= ( r) n ( 1), n 1 There are misprints in the second of these evaluations in [4].

3 Asymptotics of integrals of Hermite polynomials 045 where (a) n =Γ(a+n)/Γ(a) is the Pochhammer symbol, we obtain the moment integrals for nonnegative integer r e x x r H n (x) dx =( ) n n π( 1 ) r( r) n, (1.4) e x x r+1 H n+1 (x) dx =( ) n n π( ) r( r) n. (1.5) Since ( r) n =( ) n r(r 1)...(r n + 1), we see that both these moment integrals vanish when r<n. Finally, with s = n 1 + n + n, we note the integral (1.1) corresponding to k = is given by [4, p. 88, 7.75] s πn 1!n!n! (s even), (s n e x 1 )!(s n )!(s n )! H n1 (x)h n (x)h n (x) dx = 0 (s odd); (1.6) this integral is also clearly zero when s is even if any one of the three indices is greater than the sum of the other two.. The case k =4 We consider the evaluation of the integral (1.1) when k = 4: this requires the sum of the indices n r to be an even integer for a nonzero value. Thus, the indices can be all even or odd, or two of them can be of different parity. With n + n 4 =p even, we can expand the product H n (x)h n4 (x) as a linear combination of H k (x) (0 k p) in the form where k π (k)!c k = p H n (x)h n4 (x) = c k H k (x), (.1) e x H n (x)h n4 (x)h k (x) dx by the orthogonality property (1.). Evaluation of this integral by (1.6) then yields the coefficients c k in the form c k = (n +n 4 )/ k n!n 4! ( n n 4 + k )! ( n 4 n + k )! ( n +n 4 k )!. (.)

4 046 R. B. Paris Then the integral p I(n 1,...,n 4 ) = c k e x H n1 (x)h n (x)h k (x) dx p (n 1+n )/+k πn 1!n!(k)! = c k ( n1 n + k )! ( n n 1 + k )! ( n 1 +n k )! p ( n 1+n ) k ( n +n 4 ) k (k)! = A ( n 1 n + k)!( n n 1 + k)!( n n 4 + k)!( n 4 n + k)!, where A = σ/ π n 1! n! n! n 4! r ( n 1+n )!( n +n 4 )!, σ = n r r=1 and we have used the fact that ( a) k =( ) k a!/(a k)!. Since (k)! = k k!( 1 ) k by the duplication formula, the above sum can be expressed as a 5 F 4 generalised hypergeometric function to yield the final result I(n 1,...,n 4 )= σ/ πn 1! n! n! n 4! ( n 1+n )!( n +n 4 )!( n 1 n )!( n n 1 )!( n n 4 )!( n 4 n n 1+n, n +n 4, 1, 1, 1; 5 F 4 )!. (.) n 1 n +1, n n 1 +1, n n 4 +1, n 4 n +1; 4 In the case n 1 = n = m, n = n 4 = n, the 5 F 4 function reduces to a F function to yield the evaluation e x Hm (x)h n (x) dx =m+n πm!n! F m, n, 1; 4 (.4) 1, 1; as found in [] by means of combinatorial arguments. A similar reduction arises when n 1 = n = n = n, n 4 = m, where m and n have the same parity, to produce e x H m (x)hn(x) dx = (m+n)/ π(n!) m! ( m+n F (m, n), (.5) )! where m+n 1 n,, 1 F (m, n) := ( m n )!( n m )! F ; m n +1, n m +1; 4. (.6) We remark that the above hypergeometric functions have natural cut-offs. For example, the function F (m, n) can be written as k 1 m+n ( n) k ( ) k ( 1 F (m, n) = ) k k k=k 0 k!( m n + k)!( n m + k)!, (.7)

5 Asymptotics of integrals of Hermite polynomials 047 where k 0 = 1(m n), k 1 = min{n, 1 (m + n)}. From this, it is seen that when m>n we have k 0 >k 1, so that the sum (.7) is empty and the integral in (.5) is therefore zero; when n>mwe have k 0 <k 1, so that the integral (.5) is nonzero.. Examples of (1.1) when k =5and k =6 The same procedure used in Section, combined with the result in (.5), can be employed to evaluate cases of (1.1) corresponding to k = 5 and k =6. Consider the integral (1.1) with k = 5 and n r =n (1 r 5) given by I n = Then, since by (.1) and (.) H n (x) = n we have upon employing (.5) I n = n c k e x H 5 n(x) dx. (.1) c k H k (x), c k = n k ((n)!) (k!) (n k)!, = n π((n)!) n e x H n (x)h k(x) dx c k k (k)! F (k, n), (n + k)! where F (k, n) is defined in (.6). We therefore have the evaluation I n = 5n π ((n)!) 4 n ( 1 ) k k F (k, n) k!(n k)!(n + k)!. (.) In a similar manner we can evaluate (1.1) when k = 6 and n r = n (1 r 6), namely the integral J n = e x Hn 6 (x) dx. (.) If we first suppose that n is even, we have (upon replacing n by n) H n(x) = n c k H k (x),

6 048 R. B. Paris where by (1.) It then follows that J n = k π(k)!c k = n c k e x H n(x)h k (x) dx. (.4) e x H n (x)h k(x) dx = π n k (k)! c k. Evaluating the integral in (.4) by means of (.5), combined with a similar procedure for odd n, we then obtain the result J n = n π ((n)!) 4 n/ (n 1)/ ( 1 ) kk! k (( 1 n + k)!) F (k, n) (n even), ( ) kk! k (( 1 n k)!) F (k +1,n) (n odd). (.5) 4. Moment integrals In this section we evaluate the moment integrals involving a product of two Hermite polynomials defined by K(m, n, r) = where r is a nonnegative integer and m + n + r is even. From (1.) and (1.4) we find, for even m and n, K(m, n, r) =( ) n (n)! n! Upon noting that n e x x r H m (x)h n (x) dx, (4.1) ( n) k k!( 1 ) k = ( ) m+n m π (n)! n! n e x x k+r H m (x) dx ( n) k k!( 1 ) ( 1) k+r( k r) m. k ( 1 ) k+r =(1 ) r(r + 1 ) k, ( k r) m =( 1) m (r + k)! (r m + k)!,

7 Asymptotics of integrals of Hermite polynomials 049 we then obtain K(m, n, r) =( ) n m π (n)! ( 1 n! ) r n ( n) k (r + 1) k(r + k)! k!( 1). (4.) k(r m + k)! It is readily seen that when r<m n (m >n) all terms in the above sum vanish. Since m and n are interchangeable, it then follows that K(m, n, r) =0 when r< m n. An alternative representation is given by expressing the sum in (4.) as a F hypergeometric function of unit argument to obtain K(m, n, r) =( ) n m π (n)!r! n!(r m)! ( 1) r F n, r + 1,r+1; 1,r m +1; 1. (4.) A similar procedure can be applied to the remaining two cases of odd m, n and m, n of different parity with r odd to yield K(m+1, n+1, r) =( ) n m+1 (n + 1)!r! π n!(r m)! ( ) r F n, r +1,r+ ;,r m +1; 1 (4.4) and K(m +1, n, r +1)=( ) n m π (n)!r! n!(r m)! ( ) r F n, r +1,r+ ; 1,r m +1; 1. (4.5) It is found that K(m +1, n +1, r) in (4.4) similarly vanishes when r< m n, whereas K(m +1, n, r + 1) in (4.5) vanishes when r<m n (m >n) and r<n m 1(n>m+ 1). In the case m = n, we define k r (n) :=K(n, n, r) = e x x r H n(x) dx (4.6) for nonnegative integer n. If we apply Thomae s transformation [1, p. 14] a, b, c; Γ(d)Γ(e)Γ(s) d a, e a, s; F 1 = d, e; Γ(a)Γ(s + b)γ(s + c) F 1, s + b, s + c; where in this section s = d + e a b c denotes the parametric excess, to the hypergeometric functions appearing in (4.) and (4.4) we find that ( 1 ) r F r, r +1, 1n + 1; 1 k r (n) =( ) r n, 1; 1 (n even), πn! (4.7) ( ) r F r, r +1, 1 n +1; 1, ; 1 (n odd).

8 050 R. B. Paris Application of Sheppard s transformation for nonnegative integer r [1, p. 141] r, a, b; F 1 = (d a) r(e a) r r, a, 1 s; F d, e;; (d) r (e) r a r + d +1,a r e +1; 1 shows that the ratio of the two F functions appearing in (4.7) equals ( ) r/( 1 ) r. Hence we finally obtain the evaluation k r (n) =( ) r n r πn! r (n) (4.8) for nonnegative integers n and r, where the polynomial r (n) is given by r (n) = r ( 1 ) r F r, r +1, 1n + 1; 1, 1; 1. The first few polynomials r (n) are then found to be 0 (n) = 1, 1 (n) =n +1, (n) = (n +n +1), (n) = 5(4n +6n +8n +), 4 (n) = 5(n 4 +4n +10n +8n +), 5 (n) = 6(4n 5 +10n 4 +40n +50n +46n + 15), An asymptotic expansion for n The asymptotic behaviour of the integral (1.1) when one or more of the indices n r is large and k =, follows immediately from (1.) and (1.6), respectively. In [], Azor et al. considered the particular case of the next integral in the sequence corresponding k = 4 and n r = n (1 r n), namely I n = e x Hn 4 (x) dx, and derived by an elaborate argument an asymptotic estimate of this integral as n. They expressed I n as integral involving a Legendre function taken round a contour surrounding the origin in the complex plane and from this constructed a generating function to which they applied Darboux s method.

9 Asymptotics of integrals of Hermite polynomials 051 The asymptotics of I n as n were then deduced from the bahaviour of the generating function at its singularities on its circle of convergence. Here, we shall obtain an asymptotic expansion for I n by means of the discrete analogue of Laplace s method applied to sums. This method was employed by Stokes [6] in his determination of the leading behaviour of the generalised hypergeometric function p F q (x) for x +. An example of the application of this method can also be found in [, p. 04]. From (.4) together with ( n) k =( ) k n!/(n k)!, we find I n = n (n!) 4 n k Γ(k + 1 ) (k!) ((n k)!). (5.1) This sum consists of positive terms which are easily shown to possess a maximum for large n at k n m. Asn, the terms in the sum (5.1) peak sharply near the maximum term. For arbitrary ɛ>0 we then have S n := n k Γ(k + 1 ) (k!) ((n k)!) [m(1+ɛ)] k=[m(1 ɛ)] k Γ(k + 1 ) (k!) ((n k)!) (5.) with an error that is subdominant with respect to every power of 1/n as n. The terms retained in the sum on the right-hand side of (5.) can now be approximated by means of the well-known expansion for Γ(z) given by Γ(z) πz z 1 e z ( ) s γ s z s, z +, (5.) s=0 where the first few Stirling coefficients γ s are given by γ 0 =1,γ 1 = 1 1 γ = 1 88 = 19. Then, by the duplication formula for the gamma function, we have for large k πγ(k) Γ(k + 1) = k 1 Γ(k) s=0 πk k e k ( ) s γ s (k) s s=0 ( ) s γ s k s ( πk k e k 1 1 4k k k + (5.4) and Γ(k +1) = kγ(k) ( πk k+ 1 e k k k k +. (5.5) From (5.4) and (5.5) we therefore find k Γ(k + 1 ) (k!) k k k e k πk / ( 1 7 4k k k + )

10 05 R. B. Paris as k +. We now set k = m + t, m = n, τ = t/m, where t is small compared with m. We find from (5.) that (n k)! = ( 1 n t)! π( 1 n t) 1 n t+ 1 e 1 n+t s=0 ( ) s s γ s m s (1 τ) s. Some routine algebra then shows that the terms in the sum on the right-hand side of (5.) can be written as where k Γ(k + 1) (k!) ((n k)!) ( ( ) n n e n t 4π n n)/ /m G(τ,m)= (1 + τ) / 1 τ ( 1 7(1+τ) 1 exp [mτ m(1 + τ) log(1 + τ)] exp [m(1 τ) log(1 τ)] 4m ( 1+ (1 τ) 1 6m + 49(1+τ) 115m + (1 τ) 7m This produces the expansion for large n in the form + 749(1+τ) 41470m + ) 19(1 τ) 6480m + ). G(τ,m), k Γ(k + 1 ) (k!) ((n k)!) ( ( ) n n e n t 4π n n)/ /m c s (m)τ s, s=0 where, omitting the odd coefficients (as these will not be required), c 0 (m) =1 5 8 m m + O(m ), c (m) = m 1 + O(m ), c 4 (m) = m O(m 1 ), c 6 (m) = 1 m + O(1), c 8 (m) = m + O(1),.... We now extend the range of summation in (5.) to ± to obtain ( ( ) S n en n n 4π n n)/ e t /m (c 0 (m)+c 1 (m)τ + c (m)τ + ). t= (5.6) The sums in (5.6) may be evaluated using the Poisson-Jacobi transformation [7, p. 14] ( ) π e an = 1+ e π n /a, Re(a) > 0, n= a n=1

11 Asymptotics of integrals of Hermite polynomials 05 so that for a 0+ we have (neglecting exponentially small terms of order e π /a ) ( ) s π d n s e an ( ) s n= da a =Γ(s + 1) 1 a s for s =0, 1,,.... Since odd powers of τ yield zero contribution to the sum in (5.6), we then find from (5.1) and (5.6) I n 4n π n n e n πn 6n (n!) 4 s=0 c s (m)γ(s + 1) (m) s. π Evaluation of this sum with the above values of c s (m)(0 s ) produces the value n 1 + O(n ). Continuation of this process with the help of Mathematica then yields the expansion I n ( 4n π 6n (n!) s=0 )( ) b s n s ( ) s γ s n s, where we have removed a factor of (n!) with the help of (5.) and s=0 b 0 =1, b 1 = 5 1, b = , b = , We then finally obtain the expansion b 4 = , b 5 = ,.... I n 4n π 6n (n!) a s n s (n ), (5.7) s=0 where a 0 =1, a 1 = 1 4, a = 1 16, a = 1, a 4 = 7 56, a 5 = ,.... The first three terms of this expansion were obtained in [], where the third coefficient was incorrectly given as a =. To demonstrate the validity of this 16 expansion we present in Table 1 the absolute relative error in the computation of I n in (5.1) by means of (5.7) for different values of n and truncation index s. An obvious misprint in [, Eq. (58)] has the factor (n!) in the denominator.

12 054 R. B. Paris s n =50 n = 100 n = Table 1: Values of the absolute relative error in the computation of I n by the asymptotic expansion (5.7) for different values of n and truncation index s. The integral (1.1) with n 1 = n = m, n = n 4 = n is much more straightforward to estimate asymptotically as n when m is finite. From (.5) and (.7), we have when n>m I m,n := e x H m(x)h n(x) dx = m+n (m!n!) m Γ(k + 1 )k (k!) (m k)!(n k)! and, for large n, the maximum term in the sum corresponds to k = m. With k = m j, we can rewrite the above sum as I m,n = m+n Γ(m + 1)n! n m d j (m), m j=0 (n m +1) j where d j (m) = ( )j j (j!) ( 1 m) j m. j This is in the form of an inverse factorial expansion in n which is suitable for computation as n. The behaviour of the integrals with n 1 = n = n = m, n 4 = n and n 1 = m, n = n = n 4 = n (where m and n have the same parity) as n can be obtained from (.5) and (.7). We have e x H m (x)h n(x) dx =0 (n>m), and e x H m (x)h n(x) dx = (m+n)/ π (n!) m! ( m+n )! F (m, n),

13 Asymptotics of integrals of Hermite polynomials 055 where F (m, n) is defined in (.6). If we suppose m and n are both even and define (a, j) :=(a + j)!(a j)!, we can express F (m, n) in the form F (m, n) = n m/ n! ( 1m + 1n)! π j= m/ j Γ( 1 n j) ( 1 n, j)( 1 m, j)(n 1 m + j)!; a similar result applies for m, n odd. A difficulty arises in the estimation of F (m, n) asn since, for finite m, the discrete analogue of Laplace s method cannot be employed. Since ( 1n, j +1)/( 1 n, j) 1asn, the ratio of consecutive terms in this sum is controlled by e j := j Γ( 1 n j)/(n 1 m + j)!. It is then easy to see that e j+1/e j asn, so that there is no clear maximum term in the sum. Finally, we can consider the integral e x H n(x)h n+m(x) dx = m+n (m + n)!n! n k Γ(k + 1 ) (k!) (n k)!(m + n k)! by (.5) and (.7). As n, the discrete analogue of Laplace s method can be used to show that that this integral possesses the leading behaviour 4n π 6n+m (m + n)! n m (n ). References [1] G. E. Andrews, R. A. Askey and R. Roy, Special Functions, Cambridge University Press, Cambridge, 000. [] R. Azor, J. Gillis and J. D. Victor, Combinatorial applications of Hermite polynomials, SIAM J. Math. Anal. 1 (198) [] C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers, McGraw-Hill, New York, [4] I. S. Gradshteyn and I. M. Rhyzhik, Table of Integrals, Series and Products, Academic Press, New York, [5] L. J. Slater, Generalised Hypergeometric Functions, Cambridge University Press, Cambridge, 1966.

14 056 R. B. Paris [6] G. G. Stokes, Note on the determination of arbitrary constants which appear as multipliers of semi-convergent series, Proc. Camb. Phil. Soc. 6 (1889) [7] E. T. Whittaker and G. N. Watson, Modern Analysis, Cambridge University Press, Cambridge 195. Received: May, 010