Three-Step and Four-Step Random Walk Integrals

Transcription

1 Three-Step and Four-Step Random Walk Integrals Jonathan M. Borwein, Armin Straub, and James Wan September, Abstract We investigate the moments of -step and 4-step uniform random walk in the plane. In particular, we further analyse a formula conjectured in [BNSW9] expressing 4-step moments in terms of -step moments. Diverse related results including hypergeometric and elliptic closed forms for W 4 (± are given and two new conjectures are recorded. Introduction and Preliminaries Continuing research commenced in [BNSW9], for complex s, we consider the n- dimensional integral s n W n (s := e πx ki dx ( [,] n which occurs in the theory of uniform random walk integrals in the plane, where at each step a unit-step is taken in a random direction. As such, the integral ( expresses the s-th moment of the distance to the origin after n steps. The study of such walks largely originated with Pearson more than a century ago [Pea95, Pea95b]. In his honor we call such integrals ramble integrals, as he posed such questions for a walker or rambler. As discussed in [BNSW9], and illustrated further herein, such ramble integrals are approachable by a mixture of analytic, combinatoric, algebraic and probabilistic methods. They provide interesting numeric and symbolic computation challenges. Indeed, nearly all of our results were discovered experimentally. CARMA, University of Newcastle, Australia. jonathan.borwein@newcastle.edu.au. Tulane University, New Orleans, USA. astraub@tulane.edu. CARMA, University of Newcastle, Australia. james.wan@newcastle.edu.au. k=

2 For n, the integral ( is well-defined and analytic for Re s >, and admits an interesting analytic continuation to the complex plane with poles at certain negative integers, see [BNSW9]. We shall also write W n for these continuations. In Figure we show the continuations of W and W 4 on the negative real axis. Observe the poles of W at negative even integers (but note that neither function has zeroes at negative odd integers even though the graphs may suggest otherwise (a W (b W 4 Figure : W, W 4 analytically continued to the real line. It is easy to determine that W (s =, and W (s = ( s s/. Furthermore, it is proven in [BNSW9] that, for k a nonnegative integer, in terms of the generalized hypergeometric function, we have W (k = Re F (, k, k, From here, the following expressions for W ( can be established: W ( = 4 ( (F,,, + 4 π = K (k + π K (k = ( / 6 π 4 Γ6 + 7 ( / 4 π 4 Γ6 = ( ( ( / π 4 β + / β 4. ( 4 F (,,, 4 ( (4 (5, (6

3 where K is the complete elliptic integral of the first kind, k := is the third singular value as in [BB87], and β(x := B(x, x is a central Beta-function value. More simply, but similarly, W ( = K (k = / π 6 π 4 Γ6 ( = 4π β (, (7 and, using the two-term recurrence for W (n given in [BNSW9], it follows that similar expressions can be given for W evaluated at odd integers. It is one of the goals of this paper to give similar evaluations for a 4-step walk. For s an even positive integer, the moments W n (s take explicit integer values. In fact, for integers k, ( k W n (k =. (8 a,..., a n a + +a n=k Based on the combinatorial properties of this evaluation, the following conjecture was made in [BNSW9]. Note that the case n = is easily resolved. Conjecture. For positive integers n and complex s, W n (s?[] = ( s/ W n (s j. (9 j j We investigate this conjecture in some detail in Section 4. For n =, in conjunction with ( this leads to a very efficient computation of W 4 at integers, yielding roughly a digit per term. Bessel integral representations We start with the result of Kluyver [Klu96], amplified in [Watson,.48] and exploited in [BNSW9], to the effect that the probability that an n-step walk ends up within a disc of radius α is P n (α = α From this, David Broadhurst [Bro9] obtains W n (s = s+ k Γ( + s Γ(k s J (αxj n (x dx. ( ( x k s k d J n (x dx ( x dx

4 valid as long as k > s > n/. Here and below J ν (z denotes the Bessel function of the first kind. Example (W n (±. In particular, from (, for n, we can write: W n ( = J n (x dx, W n ( = n n dx J (xj (x x. ( Equation ( enabled Broadhurst to verify Conjecture for n =,, 4, 5 and odd positive s < 5 to a precision of 5 digits. A different proof of ( is outlined in Remark below. In particular, for < s < n/, we have s Γ( s/ W n ( s = Γ(s/ x s J n (x dx, ( so that W n ( s essentially is the Mellin transform of (the analytic continuation of the nth power of the Bessel function J. Example. Using (, the evaluations W (s = and W (s = ( s s/ translate into x s J (x dx = s Γ(s/ Γ( s/, x s J (x dx = Γ(/ Γ(s/Γ(/ s/ Γ( s/ in the region where the left-hand side converges. The Mellin transforms of J and J 4 in terms of Meijer G-functions appear in the proofs of Theorems and. Remark. Here, we demonstrate how Ramanujan s master theorem may be applied to find the Bessel integral representation ( in a natural way; this and more applications of Ramanujan s master theorem will appear in [RMT]. For an alternative proof see [Bro9]. Ramanujan s master theorem [Har78] states that, under certain conditions on the analytic function ϕ, ( x ν ( k ϕ(kx k = Γ(νϕ( ν. (4 k! k= 4

5 Based on the evaluation (8, we have, as noted in [BNSW9], the generating function ( n W n (k ( xk (k! = ( x k = J (k! ( x n (5 k k for the even moments. Applying Ramanujan s master theorem (4 to ϕ(k := W n (k/k!, we find Γ(νϕ( ν = Upon a change of variables and setting s = ν, s Γ( s/ W n ( s = Γ(s/ x ν J n ( x dx. (6 x s J n (x dx. (7 This is the case k = of (. The general case follows from the fact that if F (s is the Mellin transform of f(x, then (s (s 4 (s kf (s k is the Mellin transform of ( d k x dx f(x.. Pole structure A very useful consequence of equation ( is the following proposition. Proposition (Poles. The structure of the poles of W n is as follows: (a (Reflection For n =, we have explicitly for k =,,,... that and the corresponding poles are simple. Res ( k (W = π W (k >, k (b For each integer n 5, the function W n (s has a simple pole at k for integers k < (n /4 with residue given by Res ( k (W n = ( k k (k! x k+ J n (x dx. (8 (c Moreover, for odd n 5, all poles of W n (s are simple as soon as the first (n / are. 5

6 In fact, we believe that for odd n, all poles of W n (s are simple as stated in Conjecture. For individual n this may be verified as in Example. Proof. (a For n = it was shown in [BNSW9] that Res (W = /( π. This also follows from (6 of Corollary. We remark that from [Watson, (4 p. 4] this is also the value of the conditional integral xj (x dx in accordance with (8. Letting r (k := Res ( k (W n, the explicit residue equation is r (k = ( k k + r (k (k r (k 9 (k, which has the asserted solution, when compared to the recursion for W (s: (s + 4 W (s + 4 (5s + s + 46W (s + + 9(s + W (s =. (9 We give another derivation in Example 7 in Section. (b For n 5 we note that the integral in (8 is absolutely convergent since J (x on the real axis and J (x /(πx cos(x π/4 (see [AS7, (9..]. Since the residue is as claimed by (7. ( k lim (s kγ( s/ = s k (k!, (c As shown in [BNSW9] W n, for odd n, satisfies a recursion of the form λ ( λ (n!! (s + j W n (s+c (sw n (s++ +(s + λ n W n (s+λ =, j= with polynomial coefficients of degree n where λ := (n + /. From this, on multiplying by (s+k(s+k (s k+λ one may derive a corresponding recursion for Res ( k (W n for k =,,... Inductively, this lets us establish that the poles are simple. The argument breaks down if one of the initial values is infinite as it is when 4 n. Example (Poles of W 5. We illustrate Proposition in the case n = 5. In particular, we demonstrate how to show that all poles are indeed simple. To this end, we start with the recursion: (s W 5 (s + 6 (5(s (s W 5 (s (s + 4 (59(s W 5 (s + = 5(s + 4 (s + W 5 (s. 6

7 From here, lim (s + s W 5 (s = 4 5 (85W 5( W 5 ( + 6W 5 (4 = which shows that the first pole is indeed simple as is also guaranteed by Proposition b. Similarly, lim (s + s 4 4 W 5 (s = 4 5 (5W 5( W 5 ( = showing that the second pole is simple as well. It follows from Proposition c that all poles of W 5 are simple. More specifically, let r 5 (k := Res ( k (W 5. With initial values r 5 ( =, r 5 ( and r 5 (, we derive that r 5 (k + = k4 r 5 (k (5 + 8 k + 6 k + 7 k + 5 k 4 r 5 (k + 5(k + (k + + ( k + 59 k r 5 (k + 5(k +. Example 4 (Poles of W 4. Let r 4 (k := lim s k (s + k W 4 (s, then the recursion for W 4 (s (s + 4 W 4 (s + 4 4(s + (5s + s + 48W 4 (s (s + W 4 (s = gives us r 4 (k + = (k + (5k + 5k + (k + r 4 (k + 64 k (k + r 4(k. We also compute that π = r 4( = lim (s + s W 4 (s = + 4W 4( W 4(. 8 The first equality is obtainable from (7. Further, L Hôpital s rule shows that the residue at s = is lim s d ds ((s + W 4 (s = 9 + 8W 4( W 4( + 4W 4 ( W 4 ( 6 7

8 9 with a numerical value of which we were able to identify as log(. π 9 Similarly, the second residue is found to be (4 log( with similar formulae 8π for the other residues. We finally record a remarkable identity related to the pole of W 4 at that was established in [Watson, ( p. 45]. It is J 4 ν (xx ν dx = π Γ(νΓ(ν Γ(νΓ(ν + / ( for Re ν >. Hence J 4 ν (xx ν dx /ν 4π / as ν.. Meijer G-function representations We recall that the Meijer G-function introduced in 96 by the Dutch mathematician Cornelis Simon Meijer ( is defined, for parameter vectors a and b [AAR99], by ( ( a G m,n p,q b x = G m,n a,..., a p p,q b,..., b q x ( = m k= Γ(b k t n k= Γ( a k + t πi q L k=m+ Γ( b k + t p k=n+ Γ(a k t xt dt. ( In the case x < and p = q the contour L is a loop that starts at infinity on a line parallel to the positive real axis, encircles the poles of the Γ(b k t once in the negative sense and returns to infinity on another line parallel to the positive real axis; with a similar contour when x >. Moreover G p,q m,n is analytic in each parameter; in consequence so are the compositions arising below. Our main tool below is the following special case of Parseval s formula giving the Mellin transform of a product. Theorem (Mellin transform. Let G(s and H(s be the Mellin transforms of g(x and h(x respectively. Then x s g(xh(x dx = πi δ+i δ i for any real number δ in the common region of analyticity. This leads to: G(zH(s z dz ( 8

9 Theorem (Meijer form for W. For all complex s ( Γ( + s/,, W (s = Γ(/Γ( s/ G,, /, s/, s/. (4 4 Proof. We apply Theorem to J = J J for s in a vertical strip. Using Example we then obtain x s J (x dx = πi δ+i δ i s s z Γ(z/Γ(/ z/ Γ(s/ z/ Γ(/ Γ( z/ Γ( s/ + z/ dz δ/+i = Γ(/ πi δ/ i ( s = Γ(/ G,, /, s/, s/ t Γ(tΓ(/ tγ(s/ t 4 Γ( t Γ( s/ + t dt,, 4 where < δ <. The claim follows from (7 by analytic continuation. Similarly we obtain: Theorem (Meijer form for W 4. For all complex s with Re s > ( W 4 (s = s Γ( + s/, ( s/,, π Γ( s/ G, 4,4 /, s/, s/, s/. (5 Proof. We now apply Theorem to J 4 = J J, again for s in a vertical strip. Using once more Example, we obtain x s J 4 (x dx = δ+i Γ(z/Γ(/ z/ Γ(s/ z/γ(/ s/ + z/ dz πi δ i 4π Γ( z/ Γ( s/ + z/ = (, ( + s/,, π G, 4,4 /, s/, s/, s/ where < δ <. The claim again follows from (7. We illustrate with graphs of W, W 4 in the complex plane in Figure. Note the poles and removable singularities. These graphs were produced employing the Meijer forms in their hypergeometric form as presented in the next section. In the case n = 4, the functional equation is employed for s with Re s. 9

10 (a W (b W 4 Figure : W via (4 and W 4 via (5 in the complex plane.. Hypergeometric representations By Slater s theorem [Mar8, p. 57], the Meijer G-function representations for W (s and W 4 (s given in Theorems and can be expanded in terms of generalized hypergeometric functions. For n =, 4 we obtain the following: Corollary (Hypergeometric forms. For s not an odd integer, we have W (s = ( πs ( ( tan s s+ s F,, ( ( s+, s s s+ + s F, s, s 4, s and, if also Re s >, we have W 4 (s = ( πs ( tan s s s ( 4F,,, s + s+, s+, s+ + ( s s 4 4F (, s, s, s,, s, (6. These lovely analytic continuations of W and W 4, first found in [Cra9] using Mathematica, can also be obtained by symbolic integration of ( in Mathematica. Example 5. From (6 and taking the limit using L Hôpital s rule, we have ( W ( = 6 π K log + ( n ( n k k= k. (8 π n 4 4n n= (7

11 In conjunction with (7 we obtain ( n ( n k ( k= k = 6 ( n 4 4n π π log K. (9 n= For comparison, (7 produces W 4 ( = 4 π ( n 4 n 4 4n n= k=n+ ( k+. k We see that while Corollary makes it easy to analyse the poles, the provably removable singularities at odd integers are much harder to resolve explicitly [Cra9]. For W 4 ( we proceed as follows: Theorem 4 (Hypergeometric form for W 4 (. W 4 ( = π ( F,,,,,, 4 6,,,,,. ( Proof. Using Theorem we write W 4 ( = π G, 4,4 4 (,,,,,,. Using the definition ( of the Meijer G-function as a contour-integral, we see that the corresponding integrand is Γ( t Γ(t Γ( + t Γ( t xt = Γ( t Γ(t 4 Γ( + t sin (πt π x t, ( where we have used Γ(tΓ( t = π. sin(πt We choose the contour of integration to enclose the poles of Γ( t. Note then that the presence of sin (πt does not interfere with the contour or the residues (for sin (πt = at half integers. Hence we may ignore sin (πt in the integrand altogether. Then the right-hand side of ( is the integrand of another Meijer G-function; thus we have shown that (,,, G, 4,4 = (,,, π G,4 4,4. (,,,,,,

12 The same argument shows that the factor of applies to all W π 4 (s when we change from G, 4,4 to G,4 4,4. Now, using the transformation ( ( a x α G m,n a + α p,q b x = G m,n p,q b + α x ( we deduce that W 4 ( = π G,4 4,4 (,,,,,,. Finally, we appeal to Bailey s identity [Bai, Formula (.4]: ( a, + a, b, c, d, e, f 7F 6, + a b, + a c, + a d, + a e, + a f a Γ( + a bγ( + a cγ( + a dγ( + a eγ( + a f = Γ( + aγ(bγ(cγ(dγ( + a b cγ( + a b dγ( + a c dγ( + a e f ( G,4 e + f a, b, c, d 4,4, + a b c d, e a, f a. (4 The claim follows upon setting all parameters to /. An attempt to analogously apply Bailey s identity for W 4 ( fails, since its Meijer G representation as obtained from Theorem does not meet the precise form required in the formula. Nevertheless, a combination of Nesterenko s theorem ([Nest] and Zudilin s theorem ([Zudilin] gives the following result: Theorem 5 (Hypergeometric form for W 4 (. W 4 ( = π ( F,,,,,, 4 6,,,,, π ( F,,,,,, 4 6,,,,,. (5 4 Proof. We first prove a result that will allow us to use Nesterenko s theorem, which converts the Meijer G form of W 4 ( to a triple integral. We need the following identities which can be readily verified: ( ( ( d z b G, a, a, a, a 4 4,4 dz b, b, b, b 4 z = z b G, a, a, a, a 4 4,4 b +, b, b, b 4 z (6 ( ( ( d z a G, a, a, a, a 4 4,4 dz b, b, b, b 4 z = z a G, a, a, a, a 4 4,4 b, b, b, b 4 z (7 4

13 ( Let a(z := G,,,, 4,4 z. Note that a( = πw,,, 4 ( by Theorem. Applying (6 to a(z and using the product rule, we get a( + a ( = c, where c := G, 4,4 where b := G, 4,4 (,,,,, (,,,,,,,. Applying (7 and ( to a(z, we obtain a ( = b. Appealing to equation (7, we see that b = c. Hence a( = 4c. Converting c to a G,4 4,4 as in (, which finally satisfies the conditions of Nesterenko s theorem, we obtain: W 4 ( = 4 π We now make a change of variable z = z. Writing x( y( z dx dy dz. ( xyz( x( yz (z = (z ( ( z = (z (z ( z splits the previous triple integral into two terms. Each term satisfies Zudilin s theorem and so can be written as a 7 F 6. We thence obtain the result as claimed. The following alternative relation was first predicted by the integer relation algorithm PSLQ in a computational hunt for results similar to that in Theorem 4: Theorem 6 (Alternative hypergeometric form for W 4 (. W 4 ( = 9π ( F,,,,,, ( 5 4 6,,,,, π 7 F,,,,,, 4 6,,,,,. (8 4 Proof. For notational convenience, let ( 7 A := π4 8 7 F,,,,,, 4 6,,,,,, 4 ( 7 B := π F,,,,,, 4 6,,,,,, 4 ( 5 C := π4 6 7 F,,,,,, 4 6,,,,,. 4 By (5, W 4 ( = (/π (A B, and the truth of (8 is equivalent to the evaluation W 4 ( = (/π (A C. Thus, we only need to show A + B C =. The triple integral for A encountered in the application of Zudilin s theorem is A = 8 x( y dx dy dz, ( xyz( z( x( yz 4

14 and can be reduced to a one dimensional integral: A = A := (K (k E (k k dk, Here, as usual, K (k := K ( k and E (k := E( k. Happily, we may apply a non-trivial action on the exponents of x, y, z and leave the value of the integral unchanged (see [Zudilin4], remark after lemma 8. We obtain: A = x( yz dx dy dz 8 xyz( x( y( z = A := K (ke (k dk. The like integral for B can also be reduced to a one dimensional integral, B = B := k K (k dk. But B also satisfies the conditions of Bailey s identity and Nesterenko s theorem, from which we are able to produce an alternative triple integral, and reduce it to: B = B := (K (k E (k ( E (k k K (k dk k. As for C, equation (64 details its evaluation, which we also record here: C = K (k dk. Now A + B C = A + A + B C =, because the integrand of the later expression is zero. Note that the theorem gives the identity among others. K (ke (k dk = ( k K (k dk, (9 4

15 Remark. Note that each of the 7 F 6 s involved in Theorems 4, 5 and 6 can also be easily written as a sum of two 6 F 5 s. Also note that the first 7 F 6 term in Theorem 6 satisfies the conditions of Bailey s identity (4 (with a = e = f =, b = c = d = : ( 7 7F,,,,,, 4 6,,,,, 4 = 6 π 4 G,4 4,4 (,,,,,,. (4 We can thence convert the right-hand side to a Meijer G form. On the other hand, W 4 ( = (,,, π G,4 4,4,,,. We thus obtain the non-trivial identity: ( G,4,,, ( 4,4,,, = 4 G,4,,, 4,4,,, + 8 G,4 4,4 (,,,,,, Corollary (Elliptic integral representation for W 4 (. We have W 4 ( = 6 π. (4 ( k K (k dk. (4 Proof. The conclusion of Theorem 6 implies (π /6 W 4 ( = C B = C B. Probabilistically inspired representations In this section, we build on the probabilistic approach taken in Section 6 of [BNSW9]. We may profitably view a (m+n-step walk as a composition of an m-step and n-step walk for m, n. Different decompositions make different structures apparent. We express the distance z of an (n+m-step walk conditioned on a given distance x of the first n steps as well as the distance y of the remaining m steps. Then, by the cosine rule, z = x + y + xy cos(θ, where θ is the outside angle of the triangle with sides of lengths x, y, and z: z y x θ 5

16 It follows that for s >, the s-th moment of an (n + m-step walk conditioned on the distance x of the first n steps and the distance y of the remaining m steps is g s (x, y := π ( z s dθ = x y s F, s π 4xy. (4 (x y Here we appealed to symmetry to restrict the angle to θ [, π. We then evaluated the integral in hypergeometric form which, for instance, can be done with the help of Mathematica or Maple. Remark (Alternate forms for g s. Using Kummer s quadratic transformation [AAR99], we obtain ( g s (x, y = Re y s s F, s x (44 y for general positive x, y. This provides an analytic continuation of s g s (x, y. In particular, we have g (x, y = π Re ( x y K (45 y and, with E the complete elliptic integral of the second kind, we have g (x, y = ( ( } x {E π Re y ( x x K. (46 y y y This later form has various re-expressions. Denote by p n (x the density of the distance x for an n-step walk. Since W n+m (s is the s-th moment of the distance of an (n + m-step walk, we obtain W n+m (s = n m g s (x, y p n (xp m (y dy dx, (47 for s. Since for the -step walk we have p (x = δ (x, this generalizes the corresponding formula given for W n+ (s in [BNSW9]. In (47, if n =, then we may take p (x = δ (x, and regard the limits of integration as from ɛ and +ɛ, ɛ. Then g s = y s as the hypergeometric collapses to, and we recover the basic form W m (s = m y s p m (y dy. (48 6

17 It is also easily shown that the probability density for a -step walk is given by p (x = π 4 x for x and otherwise. The density p (x for x can be expressed by ( x ( (x + p (x = Re π K ( x, (49 6x see, e.g., [Pea96]. To make (49 more accessible we need the following cubic identity. Proposition. For all x we have ( ( 6x K = x ( x ( + x + x K 6x. ( x( + x Proof. Both sides satisfy the differential equation 4x (x+ f(x+(x (x+ ((x 9x 9x+9f (x+x(x x 9x+9f (x =, and both of their function values and derivative values agree at the origin. We can use Proposition to deduce the more symmetrical formula p (x = π Re x ( ( AGM + x( x, (5 ( x( + x since AGM(, k = π denotes the Gaussian arithmetic-geometric mean. K(k We also apply Jacobi s imaginary transform ([BB87], p7, Re K(x = K ( x x for x > to express p (x as a real function over [, ] and [, ]. This leads to ( ( ( x(+x p (x W ( = x dx = 4 6x K ( x(+x π dx+ K 6x dx. ( x( + x π x Now in the last integral, we make the change of variable x t, and it transforms exactly into the second last integral. +t Therefore, W ( = 7 p (x x dx. (5

18 To make sense of what has happened more abstractly, let σ(x := x + x, λ(x := ( + x ( x. (5 6x Then for < x < we have σ (x = x and λ(xλ(σ(x =. In consequence σ is an involution that sends [, ] to [, ] and p (x = Example 6 (Series for p and W (. We know that 4x ( x(x + p (σ(x. (5 W (k = k ( ( k j j j j= is the sum of squares of trinomials (see (8 and [BNSW9]. Using Proposition, we may now apply equation (84 in [BBBG8, Section 5.] to obtain p (x = x π k= ( x k W (k, (54 with radius of convergence. For < x <, on using (5 we obtain 8x ( k x p (x = π W (k. (55 (x + + x From (54 and (5 we deduce that as a type of reflection formula. k= W ( = 4 π k= We can use ( to deduce for all n 4 that p n (α = α W (k 9 k (k + J (t n J (αtt dt. (56 Alternatively, setting φ n (r := p n (r/(πr, we have that for n ([Hug95] φ n (r = π π ( φ n r + r cos t dt. (57 8

19 The densities p and p 4 are shown in Figure. Note that p has a singularity at as follows from (49. We remark that the derivative of p 4 has singularities at and 4. We also record that p 4 ( while p + 4 ( =. This can be proven by using the large-order asymptotic expansion for J ν to estimate p 4(s for s near as a combination of Fresnel integrals. Further, computed from (56, the densities p 5, p 6, p 7, and p 8 are shown in Figure 4 to illustrate that the case of small n is very different from the case of large n. In particular, it should be noted that, for n 7, the density p n is already quite well approximated by the limiting x n e x /n. Also notice that, as the pictures suggest, p n is continuously differentiable for n 6 while p 5 is not. This, as well as similar statements about higher-order differentiability, may be proven using dominated convergence and ( Figure : The densities p (L and p 4 (R. Example 7 (Poles of W. From here we may efficiently recover the explicit form for the residues of W given in Proposition a. Fix integers N > k > and < α <. Use the series p (x = j a jx j+ in (54 to write W (s α p (xx s dx α a j x j++s dx = j=n = α N j= N j= a j x j++s dx a j α j+s j + s, (58 and observe that both sides are holomorphic and so (58 holds in a neighborhood of s = k. Since only the first term on the left has a pole at k we may deduce that 9

20 (a p 5 (b p (c p 7 (d p 8 Figure 4: Densities p n with the limiting behaviour superimposed.

21 Res ( k (W = a k. Equivalently, Res ( k (W = π W (k, k which exposes an elegant reflection property. Remark 4 (W 5. Using (47 we may express W 5 (s and W 6 (s as double integrals, for example, W 5 ( = 4 ( ( ( ( x x (x + π 4 y 4 y Re ( x K Re K dy dx. y 6x We also have an expression based on taking two -step walks and a -step walk: W 5 ( = 8 π 4 = 8 π 4 π π π π Re K( x + y + xy cos z dz dx dy (4 x (4 y ( Re K sin a + sin b + sin a sin b cos c dc da db, but we have been unable to make further progress with these forms.. Elliptic integral representations From (47, we derive W 4 (s = s+ π = s+ π π/ π/ g s (x, y dx dy ( x ( y g s (sin u, sin v du dv. where s >. In particular, for s =, again using Jacobi s imaginary transformation, we have: W 4 ( = 4 π Re = 8 π K(x/y y dx dy (59 ( x ( y K(t dy dt ( t y ( y = 8 K (k dk. (6 π

22 The corresponding integral at s = is W 4 ( = (k + (K(k E(k π k ( k E dk. k + Also, the following Fourier series [BBBG8, Eqn (7] allows one to apply the Parseval-Bessel formula from which we may obtain W 4 ( = 4 π n,m K (sin θ = n= ( ( n( ( m n!m! Γ( + n sin ((4n + θ, (6 Γ(n + (4(m n + (4(m + n +, (6 in terms of the rising factorial (a n := a(a + (a + (a + n. Starting instead with Nesterenko s theorem [Nest] we have the following: W 4 ( = dxdydz. (6 π [,] xyz( x( y( z( x( yz (Such integrals are related to Beukers integrals, which were used in the elementary derivation of the irrationality of ζ(. Upon computing the dx integral, followed by the change of variable k = yz, we have: W 4 ( = π = π K( yz yz( y( z dy dz (64 K( k dy dk k y( y(y k = 4 K (k dk. (65 π Compare this with the corresponding (59. In particular, appealing to Theorem 4 we derive the closed forms: ( π ( 4 5 K(k dk = K (k dk = 7F,,,,,, 4 6,,,,,. (66 4

23 Recalling Corollary and equation (9 we also deduce that W 4 ( = 96 π E (kk (k dk 8 W 4 (. (67 If we make a trigonometric change of variables in (64, we obtain W 4 ( = 4 π π/ π/ K ( sin x sin y dx dy. (68 We may rewrite the integrand as a sum, and then interchange integration and summation to arrive at a slowly convergent representation of the same general form as in Conjecture : W 4 ( = ( / n= n F (,, n,. (69 Remark 5 (Relation to Watson integrals. From the evaluation (7 we note that W ( equals twice the second of three triple integrals considered by Watson in [Watson9]: W ( = π π π π dudvdw cos v cos w cos w cos u cos u cos v. (7 This is derived in [BBG5] and various related extensions are to be found in [BBBG8]. It is not clear how to generalize this to W 4 (. Watson s second integral (7 also gives the alternative representation: ( W ( = π 5/ G,,,,,, 4. (7 The equivalence of this and the Meijer G representation coming from Theorem can be established similarly to the proof of Theorem 4 upon using the Meijer G transformation ( ( a G m,n b p,q b x = G n,m q,p a. (7 x Remark 6 (Probability of return to the unit disk. By a simple geometric argument, there is a chance of returning to the unit disk in a -step walk. Similarly, for a

24 -step walk, if the second step makes an angle θ with the first step, then the third step can only vary over a range of θ to return to the unit disk (it can be parallel to the first step, to the second step, or anywhere in between. Thus the probability of returning to the unit disk in three steps is π θ dθ = 4π π 4 = p (x dx. Appealing to (54 we deduce that k= W (k 9 k (k + = π 4. In fact, as Kluyver showed [Klu96], the probability of an n-step walk ending in the unit disk is /(n +. This is easily obtained by setting α = in (. 4 Partial resolution of Conjecture We may now investigate Conjecture which is restated below for convenience. Conjecture. For positive integers n and complex s, W n (s?[] = j ( s/ W n (s j. (7 j We can resolve this conjecture modulo a conjectured technical estimate given in Conjecture. The proof outline below certainly explains Conjecture by identifying the terms of the infinite sum as natural residues. Proof. Using (7 we write W n as a Bessel integral s Γ( s/ W n ( s = Γ(s/ x s J n (x dx. Then we apply Theorem to J n = J n J for s in a vertical strip. Since, again by (7, we have x s J n (x dx = s Γ(s/ Γ( s/ W n( s 4

25 we obtain s Γ( s/ W n ( s = Γ(s/ = Γ( s/ Γ(s/ πi δ+i δ i x s J n (x J (x dx Γ(z/Γ(s/ z/ Γ( z/γ( s/ + z/ W n ( z dz where < δ <. Observe that the integrand has poles at z = s, s +, s + 4,... coming from Γ(s/ z/ as well as (irrelevant for current purposes poles at z =,, 4,... coming from Γ(z/. On the other hand, the term W n ( z has at most simple poles at z =, 4, 6,... which are cancelled by the corresponding zeros of Γ( z/. This asserted pole structure of W n was shown in Proposition for n =, 4 and that argument can be extended to 7, 9,..., but this will not resolve the general case. (See Conjecture below. Next, we determine the residue of the integrand at z = s+j. Since Γ(s/ z/ has a residue of ( j /j! at z = s + j, the residue of the integrand is ( j Γ(s/ + j (j! Γ( s/ j W n ( (j + s = Γ(s/ ( s/ W n ( s j. Γ( s/ j Thus it follows that ( s/ W n ( s j, (74 W n ( s = j j which is what we want to prove, provided that we can close the contour in the right half-plane so as to show that Γ(z/Γ(s/ z/ lim inf α Γ( z/γ( s/ + z/ W n ( z dz =. (75 γ α Here, γ α is a right half-circle of radius r α around δ. This follows from Conjecture below. To make this proof rigorous we therefore need to show that the next two conjectures hold. Conjecture (Poles of W n. For each n all poles of W n are simple. 5

26 Conjecture (Growth of W n. For given s, the maximum modulus of Γ(z/Γ(s/ z/ Γ( z/γ( s/ + z/ W n ( z over the half-circle γ αm, is achieved on the real axis at a point a m ; and the value W n (a m tends to zero for properly chosen r αm. Remark 7 (Other approaches to Conjecture. We restrict ourself to the core case with n =. One can prove that both sides of the needed identity satisfy the recursion for W 4. Hence, it suffices to show that the conjecture is correct for s = ±. Working entirely formally with ( and ignoring the restriction on s we have: ( / ( / W ( j = j Γ( j x j J (x dx j j = = j= j J (x = W 4 (, Γ( + j ( / Γ( j ( x j dx j Γ( + j j= J 4 (x dx on appealing to Example, since ( / Γ( j j Γ( + = J (x jxj j= for x >. There is a corresponding manipulation for s =, but we cannot make them rigorous. However, in [BSW], we prove the conjecture for n = and s an integer. Conclusion In addition to the two new conjectures made explicitly above, it would be fascinating to obtain closed forms for any of the residues in Proposition with n 5. It would likewise be very informative to obtain a closed form for W 5 (±. Acknowledgements We are grateful to Wadim Zudilin for much useful discussion, as well as for pointing out [Bai], [Nest], and [Zudilin], which have been crucial in obtaining the closed forms of W 4 (±. 6

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