Asymptotics of Integrals of. Hermite Polynomials

Applied Mathematical Sciences, Vol. 4, 010, no. 61, 04-056 Asymptotics of Integrals of Hermite Polynomials R. B. Paris Division of Complex Systems University of Abertay Dundee Dundee DD1 1HG, UK R.Paris@abertay.ac.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) 879 890] 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

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 1 0 0 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].

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 )!. (.)

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)

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),

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 + 1 + 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)!,

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).

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),.... 5. 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.

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, 51840 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 + 1 115k + 100 41470k + (5.4) and Γ(k +1) = kγ(k) ( πk k+ 1 e k 1+ 1 1k + 1 88k 19 51840k +. (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 + 49 115k + 749 41470k + )

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 1 + 5 18 m + O(m ), c (m) = 8 1 64 m 1 + O(m ), c 4 (m) = m + 145 18 + O(m 1 ), c 6 (m) = 1 m + O(1), c 8 (m) = 77 16 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

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 1 5 1 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 = 17 144, b = 87 1960, We then finally obtain the expansion b 4 = 101 608, b 5 = 17181 61760,.... 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 = 59 104,.... 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.

054 R. B. Paris s n =50 n = 100 n = 00 0 5.000 10.500 10 1.50 10 1.58 10 5 6.97 10 6 1.568 10 6.558 10 7.161 10 8.98 10 9 4.59 10 9.801 10 10 1.79 10 11 4 1.956 10 10 5.91 10 1 1.87 10 1 5 1.06 10 11 1.55 10 1.87 10 15 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),

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 + 1 + 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 + 1 + 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) 879 890. [] C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers, McGraw-Hill, New York, 1978. [4] I. S. Gradshteyn and I. M. Rhyzhik, Table of Integrals, Series and Products, Academic Press, New York, 1980. [5] L. J. Slater, Generalised Hypergeometric Functions, Cambridge University Press, Cambridge, 1966.

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) 6 66. [7] E. T. Whittaker and G. N. Watson, Modern Analysis, Cambridge University Press, Cambridge 195. Received: May, 010