Summing Sneddon–Bessel series explicitly

We sum in a closed form the Sneddon–Bessel series ∑m=1∞Jα(xjm,ν)Jβ(yjm,ν)jm,ν2n+α+β−2ν+2Jν+1(jm,ν)2,$$ \sum \limits_{m&amp;#x0003D;1}&amp;#x0005E;{\infty}\frac{J_{\alpha}\left(x{j}_{m,\nu}\right){J}_{\beta}\left(y{j}_{m,\nu}\right)}{j_{m,\nu}&amp;#x0005E;{2n&amp;#x0002B;\alpha &amp;#x0002B;\beta -2\nu &amp;#x0002B;2}{J}_{\nu &amp;#x0002B;1}{\left({j}_{m,\nu}\right)}&amp;#x0005E;2}, $$ where 0


Introduction
Sneddon considered in [8, § 2.2] the following Bessel series in two variables: where 0 < x, 0 < y, x+y < 2 and {j m,ν } m≥0 are the zeros of the Bessel function J ν of order ν.
The purpose of this paper is to compute explicitly these Sneddon-Bessel series for q n = 2n + α + β − 2ν + 2, n ∈ Z, under mild conditions on the parameters α, β and ν.
The problem of the explicit summation of Bessel series is a classical topic, but there is no doubt that it remains of interest today ([1, § 6.8], [4]) and active research is still being done (specially in applied mathematics, mathematical physics and engineering).
Define now the polynomial δ α,β,ν n (x, y) (of degree 2n), n ≥ 0, by the generating function This generating function allows the explicit computation of the polynomials δ α,β,ν n (x, y) recursively from the Taylor coefficients of the functions Φ α , Φ β and Φ ν .
The content of the paper is as follows.In Section 3 we use the calculus of residues to find a partial fraction expansion of functions of the form f (z) Φν (z) 2 , where f is an entire function satisfying a suitable bound in C (see Theorem 1 for details).
Theorem 1 leads us to a partial differential equation for the Sneddon-Bessel series S α,β,ν qn (x, y), n ≥ 0, which we solve in Section 4 to find (1.5).Then, the identity (1.6) can be easily deduced from the case n = 0 by differentiation.
Using our identity (1.5) for the Sneddon-Bessel series we find in Section 5 some extensions of the Kneser-Sommerfeld expansion (1.7), among which is the following: if Re ν < Re β + 1 and 0 ≤ y ≤ x ≤ 1 (as usual Y ν denotes the Bessel function of the second kind).

Preliminaries
The zeros of the function Φ ν (w) defined by (1.2), that is, the zeros of the even function J ν (w)/w ν , are simple and can be ordered as a double sequence {j m,ν } m∈Z\{0} with j −m,ν = −j m,ν and 0 ≤ Re j m,ν ≤ Re j m+1,ν for m ≥ 1 ([9, § 15.41, p. 497]).Although these zeros depend on ν, we will often omit this dependence to avoid unnecessary complications in the notation.The imaginary part of these zeros is bounded and, when m is a sufficiently large integer, there is exactly one zero in the strip mπ It follows from the estimate 6,Eq. 10.7.8], see also [9, § 7.21(1), p. 199]) that where the limit z → ∞ is to be taken inside a sector | arg(z)| ≤ π − δ.Thus, for some constants c and C not depending on m.In terms of Φ ν+1 , for some constants c and C not depending on m.
Bessel functions satisfy the bound for |z| large enough, with a constant C depending only on ν.To be precise, for |z| > ε > 0 and ν on a compact set K, there is a constant C depending only on ε and K, as follows from [6, Eq. 10.4.4 and § 10.17(iv)].
For µ and η satisfying Re µ > Re η > −1, consider the integral transform T µ,η given by (with a small abuse of notation, we will often write T µ,η (f (x)) if it does not cause confusion).
Sonin's formula for the Bessel functions ([9, 12.11(1), p. 373]) can be written as x η , valid for Re µ > Re η > −1.For 2 Re η + r + 2 > 0, we also have x r , as follows from the identity The identities (2.4) and (2.5) can be extended for Re η < −1 as follows.For complex numbers µ, η and a positive integer h satisfying Re η > − h 2 − 1, Re µ > Re η + h, consider the integral transform T µ,η,h given by (2.6) To be precise, η should not be half a negative integer (in case it is we will manage somehow).
It is then easy to check that

Partial fraction decomposition of Bessel functions
In this section, we use the calculus of residues to find a partial fraction expansion of functions of the form f (z) Φν (z) 2 , where f is an entire function with some growth control.
Theorem 1.Let f be an entire function satisfying . .} and n be a nonnegative integer such that Then , where the convergence is uniform in bounded subsets of C \ {j m : m ∈ Z \ {0}}.
Proof.Let us fix t ∈ C \ {j m : m ∈ Z \ {0}} and consider the holomorphic function It has a pole at t of order n + 1, and a double pole at each j m , m ∈ Z \ {0}.The residue at t is, therefore, Let us consider separately the last term: Now, the identities so that, going back to (3.1) and using (3.2) again, the residue at j m is Thus, if D = {z ∈ C : |z| = A} is a large circle of radius A > |t| with the only condition, at the moment, that none of the points j m lie in D, the calculus of residues gives . Now, the value of A can be chosen arbitrarily large and such that there exists some constant c > 0, independent of A, satisfying for w ∈ D, where C is a constant, independent of A, but possibly different at each occurrence.The natural parametrization of D then gives Taking this bound into (3.3) and letting A be arbitrarily large proves the theorem.
Evaluating at t = 0 the identity of Theorem 1, gives , under the assumption that We then define the double Bessel numbers δ f,ν n by These are the Taylor coefficients of f (t) Φν (t) 2 at t = 0, in other words, in a neighbourhood of 0.
Under the stronger condition termwise differentation in (4.1) is allowed.In particular, we obtain (and the same for the other partial derivative).

The case n ≥ 0
Let us assume firstly that n is a nonnegative integer (later on we will address the case when n is negative).The function f (z) = Φ α (xz)Φ β (yz) meets the conditions of Theorem 1 with N = − Re α − Re β − 1 and κ = x + y, and the condition N + 2 Re ν < 2n of Theorem 1 is therefore (4.2).Thus, (3.4) becomes where the function is a polynomial in x 2 and y 2 (that is, even powers of x and y) which could be computed recursively from the Taylor coefficients of the functions Φ α , Φ β and Φ ν involved.Notice that Let us write (4.9) and assume also, for simplicity, that ν = 0, 1, . . ., n.Then, it is easy to see that the solution to (4.6) is if (4.4) holds, where φ n,α,β,ν is a one variable function to be determined.In case ν ∈ {0, 1, . . ., n}, some logarithmic terms appear also.Before going on, let us focus on the dependence of these functions and constants on the parameter α and β.It is apparent from (4.7) and (4.9) that each A α,β,ν 2j,2k,n is a rational function of α, β and ν.If n, ν, x, y and β (respectively, α) are fixed, then the function Φ α (xj m ) is holomorphic on α ∈ C \ {−1, −2, . . .}, and so is ξ n,α,β,ν (x, y) (resp., β) under the condition (4.2) (the series involved converge uniformly on α-compacts, as follows from (1.2) and (2.2)).The same applies therefore to φ n,α,β,ν (y/x).This analytic dependence on α (resp., β) will eventually allow us to extend some identities by analytic continuation.
Thus, formula (4.10), which in principle requires (4.4) to hold, extends to the whole range (4.2) in this way: firstly, (4.5) can be written as on the whole range (4.2); using now (4.10) on the right-hand side gives an expression for ξ n,α,β,ν (x, y) with holomorphic coefficients, which by analytic continuation must equal the coefficients in (4.10).
Let us consider now the case Re ν < −1.Take a positive integer h such that Re ν > −h/2 − 1 and α, β satisfying Re α, Re β > Re ν + h.Let us assume for the moment that ν is not half a negative integer; using the integral transform T α,ν,h defined by (2.6) acting on x and T β,ν,h acting on y, we get from (2.7): The function ξ 0,ν,ν,ν (x, y) is given by (4.12).Therefore, ξ 0,ν,ν,ν is analytic in ν and so is the function on the right hand side of the above identity (on the region Re ν > −h/2 − 1).For ν > −1, integrating by parts we deduce that this function is equal to Proceeding as before, we deduce that for ν > −1 and 0 < y ≤ x < 1, the function (4.14) is equal to the function on the right hand side of (4.13), which is also analytic in ν.This proves the identity (4.13) also for Re ν > −h/2 − 1 and Re α, Re β > Re ν + h.Using again an argument of analyticity on the variables α and β, we prove that (4.13) holds indeed for 2 Re ν < 1 + Re α + Re β (and also for x + y = 2 if 2 Re ν < Re α + Re β).The requirement that ν is not half a negative integer can be suppressed by continuity.
By the way, this means that which, together with (4.11), allows to find the functions φ n,α,β,ν and, therefore, ξ n,α,β,ν for every positive integer n.
For the sake of completeness, we display in full extension the cases n = 0, 1.

The case n = 0
Identities (4.3) and (4.13) give valid for 2 Re ν < 1 + Re α + Re β, ν = 0, and 0 < y ≤ x, x + y < 2 (and also for x + y = 2 if 2 Re ν < Re α + Re β).Now, S α,β,0 q0 (x, y) can be obtained taking ν → 0 in the above formula: on one hand, after writing the hypergeometric function as a power series and looking for a hypergeometric representation of the resulting limit, it turns out that lim On the other hand, L'Hôpital's rule gives lim where is the harmonic number of order α (as usual γ denotes the Euler constant).Then, we conclude that

The case n < 0
Once we have determined the case n ≥ 0, let us consider now the case when n is a negative integer.From (4.5), we get An easy computation using (4.13) gives from where the identity (1.6) for n = −1 follows easily using (4.3).The identities (1.6) for an integer n < −1 can be proved similarly.The case ν ∈ {0, 1, . . ., n} follows by passing to the limit.

The one variable case
We will need later the following Sneddon-Bessel series in one variable: If we assume 2 Re ν < 2n + 1/2 + Re α, then the uniform convergence of the Sneddon-Bessel series (1.1) holds also for y = 0, hence the Sneddon-Bessel series (4.16) arises after dividing the Sneddon-Bessel series (1.1) by y β and taking y → 0. This can be done in the identities (1.5) and (1.6).To this end, we have to compute the sequence After some easy computations, using (1.3) and (1.4), we arrive at , n ≥ 0.

Extending the Kneser-Sommerfeld expansion
In this section we use the identities (4.10) to prove some extensions of the Kneser-Sommerfeld expansion (1.7).
For the sake of completeness, we first prove the Kneser-Sommerfeld expansion (1.7).In terms of the functions Φ ν defined by (1.2), the identity to be proved is Write ϕ(x, y, z) and ψ(x, y, z) for the left and right hand sides of (5.1), respectively.Using the geometric series, we can write where the function ξ n,ν,ν,ν (x, y) is defined by (4.1).Consider now the partial differential equation (5.2) On the one hand, using the partial differential equation (4.6) for ξ n (x, y) and (4.8), we get that ϕ(x, y, z) satisfies the partial differential equation (5.2).On the other hand, it is a matter of computation to check that the function ψ(x, y, z) satisfies the partial differential equation (5.2) as well.Hence, we deduce that ϕ(x, y, z) − ψ(x, y, z) = x −2ν ρ(y/x, z/x), for certain two-variable function ρ.But the definition of ϕ and ψ as both sides of (5.1) shows that ϕ(1, y, z) = ψ(1, y, z) = 0, so that ρ = 0 and the identity (5.1) holds.For Re β > Re ν > −1, the identity (1.8) follows by applying the integral transform T β,ν defined by (2.3) acting in the variable y to both sides of the Kneser-Sommerfeld expansion (1.7) and using Sonin's formula (2.4).With a standard argument of analyticity, the identity (1.8) extends to −1 < Re ν < Re β + 1.
The Kneser-Sommerfeld expansion has the following one variable version: valid for Re ν < 1/2 (it follows easily dividing the identity (1.7) by y ν and then taking limit as y → 0).Now, the well-known properties of the Bessel function of the second kind Y ν allow us to rewrite (5.3) as (5.4) (as usual, if ν = n is a nonnegative integer, the function on the right can be understood as the limit as ν → n).