UDC 517.516

THE THEORY OF MACDONALD INTEGRALS.

I. RECURRENCE RELATIONS.

UNIFORMLY CONVERGENT SERIES

A. A. TUZHILIN

§ 1. INTRODUCTION

The solutions of problems of diffraction of acoustic and electromagnetic waves generated by dipoles on an ideally reflecting half-plane (ideally soft or rigid in the acoustic case and ideally conducting in the electromagnetic case) are represented linearly in terms of special functions of two variables \((x,y)\)

\[ M_n(x,y)=\int_{-\infty}^{\operatorname{arsh} x} H_n^{(1)}(y\,\operatorname{ch}\xi)\, \frac{d\xi}{(\operatorname{ch}\xi)^{\,n-1}}, \tag{1.1} \]

which, at the suggestion of G. D. Malyuzhinets, were given the name Macdonald integrals of order \(n\) (these functions in a form close to that used in the present work first occur in the investigations of G. M. Macdonald [1] on the diffraction of waves from point sources by a wedge).

In calculations of the sound field generated by an oscillating spherical cap, Macdonald integrals also occur in the vicinity of the radiator.

These facts allow one to suppose that Macdonald integrals should occur in problems of wave diffraction in the presence of bodies having sharp edges, in the case when the field sources are at a finite distance (as far as the description of the wave field in the vicinity of the boundary of the geometric shadow is concerned, Macdonald integrals are generalizations of the Fresnel integral to the case in which the field source is located at a finite distance).

The present work is devoted to the development of the theory of Macdonald integrals of order \(n\)*; throughout it is assumed that \(n\) is a nonnegative integer, and the domain of variation of the variables is determined by the conditions: \(x\) is an arbitrary real number; \(y\) is a complex number such that \(|y|>0\) and \(\operatorname{Im} y \geqslant 0\).

The definition of Macdonald integrals only for integers \(n \geqslant 0\) and in the indicated domain of variation of \(x,y\) is connected with two reasons: a) the variables must be such that the integral (1.1) converges; b) for the use of Macdonald integrals in calculations of diffraction fields excited by dipoles in the presence of an ideally reflecting half-plane, the specified domain of variation of the variables is sufficient.

* The representation of electromagnetic fields through Macdonald integrals in the problem of diffraction of waves from dipoles on an ideally conducting half-plane will be set forth in the author’s subsequent works.

However, as will be shown below, the integral on the right-hand side of (1.1) for the interval \((-\infty,0)\) (i.e., \(M_n(0,y)\)) is expressed in terms of the Hankel function (in terms of \(H_{n-1/2}^{(1)}(y)\)), which is an analytic function of the parameters \(n\) and \(y\); the integral for the interval \((0,\operatorname{arsh} x)\) can be continued to complex \(x\) and \(n\), and the domain of variation of the variable \(y\) can also be enlarged in it. In the present paper we shall not single out the results that are valid in a broader domain of variation of the variables; for part of the results, the possibility of enlarging this domain will be evident directly.

The paper consists of three sections (not counting the introduction). In the second section, alongside the analytic properties of the functions \(M_n(x,y)\) that follow directly from the definition (1.1), recurrence relations are derived for the functions \(M_n(x,y)\), analogous to the recurrence relations for Hankel functions.

The third and fourth sections are devoted to deriving two different representations of the functions \(M_n(x,y)\). In the third section, as a result of separating out all singularities located at a finite distance, the functions \(M_n(x,y)\) are represented in terms of two entire functions of two variables. This representation is convenient for numerical calculations. In the fourth section a derivation is given of the expansion of \(M_n(x,y)\) in a power series in \(x\) in a neighborhood of zero; the radius of convergence of the series is equal to one, and the series converges uniformly in \(y\) for \(|y|>0\).

§ 2. ANALYTIC PROPERTIES OF MACDONALD INTEGRALS. RECURRENCE RELATIONS

From the uniform convergence of the integral in (1.1) with respect to \(y\) for \(|y|>0\) and \(\operatorname{Im} y \geqslant 0\), for any real \(x\) it follows that the function \(M_n(x,y)\) is regular for \(|y|>0\) and \(\operatorname{Im} y>0\), and continuous for \(|y|>0\) and \(\operatorname{Im} y \geqslant 0\), for any real \(x\).

Further, since, according to (1.1), it is true that

\[ \frac{\partial M_n(x,y)}{\partial x} = H_n^{(1)}\!\left(y\sqrt{1+x^2}\right) \frac{1}{(1+x^2)^{n/2}}, \]

the function \(M_n(x,y)\) is infinitely differentiable with respect to \(x\) for any real \(x\).

In what follows we shall often use the relation

\[ M_n(0,y) = \int_{-\infty}^{0} H_n^{(1)}(y\operatorname{ch}\xi)\, \frac{d\xi}{(\operatorname{ch}\xi)^{\,n-1}} = \sqrt{\frac{\pi}{2y}}\, H_{n-1/2}^{(1)}(y), \tag{2.1} \]

which follows from the known identity ([2], formula 6.592(14))

\[ \int_{1}^{\infty} H_\nu^{(1)}(y\sqrt{x})(x-1)^{\mu-1} \frac{dx}{x^{\nu/2}} = \left(\frac{2}{y}\right)^\mu H_{\nu-\mu}^{(1)}(y)\Gamma(\mu), \]

where one must set \(\mu=1/2\) and make the change of variables in the integral \(\sqrt{x}=\operatorname{ch}\xi\).

We now state the main result of this section.

Theorem 1. The function \(M_n(x,y)\) satisfies the recurrence relation

\[ M_{n+1}(x,y)+M_{n-1}(x,y)=\frac{2n-1}{y}\,M_n(x,y)+ \]
\[ +\frac{x}{y(1+x^2)^{n/2}}\,H_n^{(1)}\!\left(y\sqrt{1+x^2}\right). \tag{2.2} \]

Proof. From the definition of the function \(M_n(x,y)\) and the recurrence formulas for the Hankel functions ([2], formula 8.471 (1))

\[ H_{n+1}^{(1)}(z)+H_{n-1}^{(1)}(z)=(2n/z)\,H_n^{(1)}(z) \]

it follows that

\[ M_{n+1}(x,y)+M_{n-1}(x,y)= \int_{-\infty}^{\operatorname{arsh} x} \frac{d\xi}{(\operatorname{ch}\xi)^{n-2}} \left\{ \frac{H_{n+1}^{(1)}(y\operatorname{ch}\xi)}{\operatorname{ch}^2\xi} +H_{n-1}^{(1)}(y\operatorname{ch}\xi) \right\} = \]

\[ = \int_{-\infty}^{\operatorname{arsh} x} \frac{d\xi}{(\operatorname{ch}\xi)^{n-2}} \left\{ H_{n+1}^{(1)}(y\operatorname{ch}\xi)+H_{n-1}^{(1)}(y\operatorname{ch}\xi) -\frac{\operatorname{sh}^2\xi}{\operatorname{ch}^2\xi} H_{n+1}^{(1)}(y\operatorname{ch}\xi) \right\} = \]

\[ =\frac{2n}{y}\,M_n(x,y)- \int_{-\infty}^{\operatorname{arsh} x} H_{n+1}^{(1)}(y\operatorname{ch}\xi)\, \frac{\operatorname{sh}^2\xi\,d\xi}{\operatorname{ch}^n\xi}. \]

To compute the last integral we use the identity ([2], formula 8.472 (4))

\[ (1/z)(d/dz)\bigl(H_n^{(1)}(z)/z^n\bigr) =-H_{n+1}^{(1)}(z)/z^{n+1}, \]

i.e., in our case

\[ H_{n+1}^{(1)}(y\operatorname{ch}\xi)\, \frac{\operatorname{sh}\xi}{\operatorname{ch}^n\xi}\,d\xi = -\frac{1}{y}\, d\left\{ \frac{H_n^{(1)}(y\operatorname{ch}\xi)}{\operatorname{ch}^n\xi} \right\}. \]

Integrating by parts and taking into account that \(n>0\), we obtain

\[ \int_{-\infty}^{\operatorname{arsh} x} \operatorname{sh}\xi\,H_{n+1}^{(1)}(y\operatorname{ch}\xi)\, \frac{\operatorname{sh}\xi\,d\xi}{\operatorname{ch}^n\xi} = -\frac{\operatorname{sh}\xi}{y}\, \frac{H_n^{(1)}(y\operatorname{ch}\xi)}{\operatorname{ch}^n\xi} \bigg|_{-\infty}^{\operatorname{arsh} x} + \]

\[ +\frac{1}{y} \int_{-\infty}^{\operatorname{arsh} x} H_n^{(1)}(y\operatorname{ch}\xi)\, \frac{d\xi}{(\operatorname{ch}\xi)^{n-1}} = \]

\[ = -\frac{x}{y(1+x^2)^{n/2}}\, H_n^{(1)}\!\left(y\sqrt{1+x^2}\right) +\frac{1}{y}\,M_n(x,y). \]

Substituting the obtained relation into the expression for \(M_{n+1}+M_{n-1}\), we obtain the required identity.
The theorem is proved.

§ 3. REPRESENTATION OF MACDONALD INTEGRALS THROUGH TWO ENTIRE FUNCTIONS OF TWO VARIABLES

The principal result of the present section is a representation of Macdonald integrals through two entire functions of two variables. Namely, if the functions \(P_n(x,y)\) and \(Q_n(x,y)\) are defined by series that converge absolutely and uniformly for arbitrary complex \(x\) and \(y\)

\[ P_n(x,y)=\sum_{m=0}^{\infty}\frac{(-1)^m y^{2m}}{2^m(2m+1)!!(n+m)!} \sum_{k=0}^{m}\frac{(2k-1)!!}{2^k k!}x^k, \tag{3.1} \]

\[ Q_n(x,y)=\sum_{m=0}^{\infty}\frac{(-1)^m y^{2m}}{2^m(2m+1)!!(n+m)!} \sum_{k=0}^{m}\frac{(2k-1)!!}{2^k k!} \left\{\sum_{l=1}^{k}\frac{1}{2l}+ \sum_{l=1}^{n+m}\frac{1}{2l}+ \sum_{l=k}^{m}\frac{1}{2l+1}\right\}x^k, \tag{3.2} \]

where in formula (3.2) one must take \(\sum_{l=1}^{0}\frac{1}{2l}=0\), the following is valid.

Theorem 2. The function \(M_n(x,y)\) can be represented in the form

\[ M_0(x,y)=\frac{\exp(iy)}{y}+\frac{2i}{\pi y}\operatorname{arctg}x\cdot \sin y +x\left\{\left[1+\frac{2i}{\pi}\left(\gamma+ \ln\frac{y\sqrt{1+x^2}}{2}\right)\right]P_0(1+x^2,y) -\frac{2i}{\pi}Q_0(1+x^2,y)\right\}, \tag{3.3} \]

\[ M_1(x,y)=\frac{\exp(iy)}{iy}-\frac{2i}{\pi y}\operatorname{arctg}x\cdot \cos y +\frac{xy}{2}\left\{\left[1+\frac{2i}{\pi}\left(\gamma+ \ln\frac{y\sqrt{1+x^2}}{2}\right)\right]P_1(1+x^2,y) -\frac{2i}{\pi}Q_1(1+x^2,y)\right\} \tag{3.4} \]

and for \(n\ge 2\)

\[ \begin{aligned} M_n(x,y)={}&\sqrt{\frac{\pi}{2y}}\,H_{n-1/2}^{(1)}(y) -\frac{ix}{\pi y^n}\sum_{m=0}^{n-2}\frac{(2n-2m-3)!!}{2^m m!}\times\\ &\times y^{2m}\sum_{k=1}^{n-1-m}\frac{2^k(k-1)!}{(2k-1)!!(1+x^2)^k} -\frac{2i}{\pi y^n}\operatorname{arctg}x\times\\ &\times\left\{\cos y\sum_{k=0}^{\left[\frac{n+1}{2}\right]-1} \frac{(-1)^k(2n-2k-2)!}{2^{\,n-2k-1}(n-2k-1)!(2k)!}y^{2k}\right.\\ &\left.\qquad\qquad+\sin y\sum_{k=0}^{\left[\frac{n}{2}\right]-1} \frac{(-1)^k(2n-2k-3)!}{2^{\,n-2k-2}(n-2k-2)!(2k+1)!}y^{2k+1}\right\}\\ &+x\left(\frac{y}{2}\right)^n \left\{\left[1+\frac{2i}{\pi}\left(\gamma+\ln\frac{y\sqrt{1+x^2}}{2}\right)\right]P_n(1+x^2,y)\right.\\ &\left.\qquad\qquad-\frac{2i}{\pi}Q_n(1+x^2,y)\right\}, \end{aligned} \tag{3.5} \]

where \(\gamma\) is Euler’s constant, equal, according to ([2], formula 9.73), to \(\gamma=0.577\,215\,664\,901\,532\,5\ldots\)

For the proof of Theorem 2 we shall need a number of relations, which we shall prove in Theorems 3–5 and Lemmas 1–9.

Theorem 3. For integers \(n \geqslant 0\) the relation holds
\[ \int_{0}^{\operatorname{arsh} x} J_n(y\operatorname{ch}\xi)\, \frac{d\xi}{(\operatorname{ch}\xi)^{\,n-1}} = x\left(\frac{y}{2}\right)^n P_n(1+x^2,y), \tag{3.6} \]
where \(J_n(z)\) is the Bessel function of order \(n\).

Proof. To prove relation (3.6) one must use the expansion of the Bessel function in a Taylor series
\[ J_n(z)= \left(\frac{z}{2}\right)^n \sum_{m=0}^{\infty} \frac{(-1)^m(z/2)^{2m}}{m!(n+m)!}, \]
the relation
\[ I_m(x)= \int_{0}^{\operatorname{arsh} x} (\operatorname{ch}\xi)^{2m+1}\,d\xi = \int_{0}^{x}(1+t^2)^m\,dt = \]
\[ = \frac{2^m m!}{(2m+1)!!}\, x\sum_{k=0}^{m} \frac{(2k-1)!!}{2^k k!}(1+x^2)^k \tag{3.7} \]
and definition (3.1). Relation (3.7) is checked elementarily, for example by integration by parts. The legitimacy of the interchanges of the operations of integration and summation used is obvious.

Theorem 4. For integers \(n \geqslant 0\) the relation holds
\[ \int_{0}^{\operatorname{arsh} x} \ln\frac{y\operatorname{ch}\xi}{2}\, J_n(y\operatorname{ch}\xi)\, \frac{d\xi}{(\operatorname{ch}\xi)^{\,n-1}} = x\left(\frac{y}{2}\right)^n \ln\frac{y\sqrt{1+x^2}}{2}\, P_n(1+x^2,y)+ \]
\[ +\frac{1}{y^n}\operatorname{arctg}x\cdot \Phi_n(y) + x\left(\frac{y}{2}\right)^n \sum_{m=0}^{\infty} \frac{(-1)^m y^{2m}}{2^m(2m+1)!!(n+m)!} \times \]
\[ \times \sum_{k=0}^{m} \frac{(2k-1)!!}{2^k k!} \left\{ \sum_{l=k+1}^{m}\frac{1}{2l} - \sum_{l=k}^{m}\frac{1}{2l+1} \right\} (1+x^2)^k, \tag{3.8} \]
where
\[ \Phi_n(y)= \left(\frac{y^2}{2}\right)^n \sum_{m=0}^{\infty} \frac{(-1)^m y^{2m}}{2^m(2m+1)!!(n+m)!}. \tag{3.9} \]

In formula (3.8) one should take
\[ \sum_{l=m+1}^{m}\frac{1}{2l}=0. \]

Proof. According to (3.6),
\[ \int_{0}^{\operatorname{arsh} x} \ln\frac{y\operatorname{ch}\xi}{2}\, J_n(y\operatorname{ch}\xi)\, \frac{d\xi}{(\operatorname{ch}\xi)^{\,n-1}} = x\left(\frac{y}{2}\right)^n \ln\frac{y}{2}\, P_n(1+x^2,y)+ \]
\[ + \int_{0}^{\operatorname{arsh} x} \ln(\operatorname{ch}\xi)\, J_n(y\operatorname{ch}\xi)\, \frac{d\xi}{(\operatorname{ch}\xi)^{\,n-1}}. \]

If we now integrate term by term the expansion of the Bessel function \(J_n(z)\) in a Taylor series (the operations of summation and integration commute) and take into account the relation

\[ \begin{aligned} \int_{0}^{\operatorname{arsh} x} \ln(\operatorname{ch}\xi)(\operatorname{ch}\xi)^{2m+1}\,d\xi &= \frac12 \int_0^x \ln(1+t^2)(1+t^2)^m\,dt \\ &= \frac12 \ln(1+x^2) I_m(x) + \frac{2^m m!}{(2m+1)!!}\operatorname{arctg}x \\ &\quad + \frac{2^m m!}{(2m+1)!!}\,x \sum_{k=0}^m \frac{(2k-1)!!}{2^k k!}(1+x^2)^k \left\{\sum_{l=k+1}^m \frac1{2l} - \sum_{l=k}^m \frac1{2l+1}\right\}, \end{aligned} \tag{3.10} \]

where \(I_m(x)\) is defined by formula (3.7) (in (3.10) one must take
\(\sum_{l=m+1}^m \frac1{2l}=0\)), as well as relations (3.7), (3.1), and (3.9), then we obtain the required identity (3.8). Relation (3.10) is also easily verified by integration by parts, taking relation (3.7) into account.

Theorem 5. The following relations hold

\[ \begin{aligned} \int_0^{\operatorname{arsh} x} N_0(y\operatorname{ch}\xi)\operatorname{ch}\xi\,d\xi &= \frac2\pi \left\{\gamma+\ln\frac{y\sqrt{1+x^2}}2\right\} xP_0(1+x^2,y) \\ &\quad - \frac2\pi xQ_0(1+x^2,y) + \frac2\pi \operatorname{arctg}x\cdot \Phi_0(y), \end{aligned} \tag{3.11} \]

\[ \begin{aligned} \int_0^{\operatorname{arsh} x} N_1(y\operatorname{ch}\xi)\,d\xi &= \frac1\pi \left\{\gamma+\ln\frac{y\sqrt{1+x^2}}2\right\} xyP_1(1+x^2,y) \\ &\quad - \frac1\pi xyQ_1(1+x^2,y) + \frac2{\pi y}\operatorname{arctg}x\,\{\Phi_1(y)-1\} \end{aligned} \tag{3.12} \]

and, for \(n \ge 2\),

\[ \begin{aligned} \int_0^{\operatorname{arsh} x} N_n(y\operatorname{ch}\xi)\, \frac{d\xi}{(\operatorname{ch}\xi)^{n-1}} &= -\frac{x}{\pi y^n}\sum_{m=0}^{n-2} \frac{(2n-2m-3)!!}{2^m m!} \\ &\quad \times y^{2m}\sum_{k=1}^{n-1-m} \frac{2^k(k-1)!}{(2k-1)!!(1+x^2)^k} + \frac2\pi \left\{\gamma+\ln\frac{y\sqrt{1+x^2}}2\right\} \\ &\quad \times x\left(\frac y2\right)^n P_n(1+x^2,y) - \frac2\pi x\left(\frac y2\right)^n Q_n(1+x^2,y) \\ &\quad + \frac2{\pi y^n}\operatorname{arctg}x \left\{\Phi_n(y)-\sum_{k=0}^{n-1} \frac{(2n-2k-3)!!}{2^k k!}y^{2k}\right\}, \end{aligned} \tag{3.13} \]

where \(N_n(z)\) is the Neumann function, defined by the series ([12], formula 8.403 (2)),

\[ \begin{aligned} N_n(z)=\frac{2}{\pi}\left\{\gamma+\ln\frac{z}{2}\right\}J_n(z) &-\frac{1}{\pi}\sum_{m=0}^{n-1}\frac{(n-m-1)!}{m!}\left(\frac{z}{2}\right)^{-n+2m} \\ &-\frac{1}{\pi}\sum_{m=0}^{\infty}\frac{(-1)^m(z/2)^{n+2m}}{m!(n+m)!} \left\{\sum_{l=1}^{n+m}\frac{1}{l}+\sum_{l=1}^{m}\frac{1}{l}\right\}, \end{aligned} \tag{3.14} \]

for \(n=0\) the sum \(\sum_{m=0}^{n-1}\) in (3.14) is absent, and one must also take

\[ \sum_{l=1}^{0}\frac{1}{l}=0, \]

\(\gamma\) is Euler’s constant; the functions \(P_n(x,y)\), \(Q_n(x,y)\), and \(\Phi_n(y)\) are defined by formulas (3.1), (3.2), and (3.9).

Proof. If relation (3.14) is substituted into the left-hand side of (3.13), and relation (3.8), the relation

\[ \int_{0}^{\operatorname{arsh} x}(\operatorname{ch}\xi)^{-2m+1}\,d\xi = \int_{0}^{x}\frac{dt}{(1+t^2)^m} = \]

\[ = \frac{(2m-3)!!}{2^m(m-1)!}\, x\sum_{k=1}^{m-1}\frac{2^k(k-1)!}{(2k-1)!!(1+x^2)^k} + \frac{(2m-3)!!}{2^{m-1}(m-1)!}\operatorname{arctg}x, \tag{3.15} \]

which is easily verified, for example, by integration by parts (in relation (3.15), when \(m=1\) the sum on the right-hand side is absent and one must take \((-1)!!=1\)), as well as relations (3.8), (3.1), and (3.2), are taken into account, then we obtain the right-hand side of (3.13). Relations (3.11) and (3.12) are proved analogously.

The theorem is proved.

The following lemmas are devoted to the simplification of relations (3.11)—(3.13); the identities proved in the process are, in the author’s opinion, also of interest in themselves; the final result is formulated in Lemma 1.

Lemma 1. If the function \(\Phi_n(y)\) is defined by the series (3.9), then the following formulas hold

\[ \Phi_0(y)=\sin y/y, \tag{3.16} \]

\[ \Phi_1(y)=1-\cos y \tag{3.17} \]

and, for \(n\ge 2\),

\[ \Phi_n(y)= \sum_{k=0}^{n-1}\frac{(2n-2k-3)!!}{2^k k!}\,y^{2k} - \]

\[ -\cos y\sum_{k=0}^{\left[\frac{n+1}{2}\right]-1} \frac{(-1)^k(2n-2k-2)!\,y^{2k}} {2^{\,n-2k-1}(n-2k-1)!(2k)!} - \]

\[ -\sin y\sum_{k=0}^{\left[\frac{n}{2}\right]-1} \frac{(-1)^k(2n-2k-3)!\,y^{2k+1}} {2^{\,n-2k-2}(n-2k-2)!(2k+1)!}. \tag{3.18} \]

where the expression \([x]\) in the upper index of the sum denotes the integer part of \(x\).

Lemma 2. The following relations hold:
\[ \Phi_0(y)=\sin y/y, \tag{3.16} \]
\[ \Phi_1(y)=1-\cos y, \tag{3.17} \]
\[ \left(\frac{1}{y}\frac{d}{dy}\right)^n \Phi_n(y)=\frac{\sin y}{y} \tag{3.19} \]
and, for \(n \geqslant 2\),
\[ \Phi_n(y)= \int_0^y \sin x\,dx \int_x^y x_1\,dx_1 \int_x^{x_1} x_2\,dx_2\cdots \int_x^{x_{n-2}} x_{n-1}\,dx_{n-1}. \tag{3.20} \]

Proof. Relations (3.16), (3.17), and (3.19) follow immediately from formula (3.9). Further, it follows from (3.9) that, for \(0\leqslant s\leqslant n-1\),
\[ \left. \left(\frac{1}{y}\frac{d}{dy}\right)^s \Phi_n(y) \right|_{y=0}=0, \]
whence, for \(n\geqslant 2\),
\[ \Phi_n(y)= \int_0^y x_1\,dx_1 \int_0^{x_1} x_2\,dx_2\cdots \int_0^{x_{n-1}} \sin x_n\,dx_n. \]

Changing the order of integration in the obtained relation leads to (3.20).

Thus Lemma 2 is proved.

Lemma 3. The functional equations with respect to the functions \(a_k^m\) of integer arguments \(m\) \((m\geqslant 1)\) and \(k\) \((1\leqslant k\leqslant m)\)
\[ a_s^{m+1}= \frac{1}{(m+1-s)!} \sum_{k=1}^{s}(m-k+1)!\,a_k^m \quad \text{for } 1\leqslant s\leqslant m, \tag{3.21} \]
\[ a_{m+1}^{m+1}=a_m^{m+1} \tag{3.22} \]
are equivalent to the functional equations
\[ a_{s+2}^{m+1}-(m-s)a_{s+1}^{m+1}=a_{s+2}^{m} \quad \text{for } 0\leqslant s\leqslant m-2, \tag{3.23} \]
\[ a_{m+1}^{m+1}=a_m^{m+1},\qquad a_1^{m+1}=a_1^m=a_1^1. \tag{3.24} \]
The solution of these functional equations for a prescribed initial value \(a_1^1\) is unique. For \(a_1^1=1\),
\[ a_{s+1}^{m}=\frac{(2s)!}{2^s s!} \binom{m+s-1}{2s}, \tag{3.25} \]
where \(\binom{n}{m}\) is the binomial coefficient.

Proof. It is easily verified that the functional equations (3.23) and (3.24) follow from (3.21) and (3.22). Suppose, conversely, that relations (3.23) and (3.24) hold. We shall use (3.23) to determine \(a_s^{m+1}\) \((1\leqslant s\leqslant m)\), if \(a_s^m\) \((1\leqslant s\leqslant m)\) have already been determined, i.e., we shall regard (3.23) as a finite-difference equation for \(a_s^{m+1}\) with respect to the index \(s\), with prescribed right-hand side \(a_{s+2}^m\). A particular solution of the inhomogeneous equation (3.23), as is easily verified, is

\[ a_s^{m+1}=\frac{1}{(m+1-s)!}\sum_{k=1}^{s}(m-k+1)!\,a_k^m \qquad (1\leq s\leq m). \]

But since

\[ \widetilde a_s^{m+1}=d_m/(m+1-s)!, \]

where \(d_m\) are arbitrary constants (with respect to \(s\)), is the general solution of the homogeneous equation (3.23), the general solution of (3.23) has the form

\[ a_s^{m+1}=\frac{1}{(m+1-s)!}\sum_{k=1}^{s}(m-k+1)!\,a_k^m+ \frac{d_m}{(m+1-s)!}. \tag{3.26} \]

It follows from (3.26) that \(a_1^{m+1}=a_1^m+d_m/m!\); therefore, according to the second equality in (3.24), \(d_m=0\). Thus the equivalence of (3.21) and (3.22) to the equations (3.23) and (3.24) has been proved.

To prove the uniqueness of the solution of the functional equations (3.21) and (3.22) for the prescribed initial condition \(a_1^1\), it is enough to show that \(a_s^m=0\) when \(a_1^1=0\). But if \(a_1^1=0\), then from (3.21) and (3.22) it follows that \(a_1^2=0\) and \(a_2^2=0\). If, further, \(a_k^m=0\) for \(1\leq k\leq m\), then for \(1\leq s\leq m+1\) it follows from (3.21) and (3.22) that \(a_s^{m+1}=0\), i.e., \(a_s^m=0\) for any \(m\geq 1\) and \(1\leq s\leq m\). Uniqueness of the solution is therefore proved.

The validity of (3.25) is verified by substitution into (3.23).

Lemma 3 is completely proved.

Lemma 4. Let the function \(f_m(x,y)\) be defined by the relation

\[ f_m(x,y)=\int_x^y x_1\,dx_1\int_x^{x_1}x_2\,dx_2\cdots \int_x^{x_{m-1}}x_m\,dx, \tag{3.27} \]

then

\[ f_m(x,y)=\sum_{k=0}^{m-1} \frac{(-1)^k(2k)!\binom{m+k-1}{2k}} {2^k k!(m+k+1)!}\, y^{m-k-1}(y-x)^{m+k}[(m+k)x+y]. \tag{3.28} \]

Proof. The function \(f_m(x,y)\) satisfies the recurrence relation

\[ f_m(x,y)=\int_x^y x_1 f_{m-1}(x,x_1)\,dx_1 \tag{3.29} \]

and the initial condition \(f_1(x,y)=(y^2-x^2)/2\). If one represents \(f_m(x,y)\) as

\[ f_m(x,y)=f_m^1(x,y)+f_m^2(x,y), \]

where \(f_m^1(x,y)\) and \(f_m^2(x,y)\) satisfy equation (3.29) and the initial conditions

\[ f_1^1(x,y)=x(y-x) \quad\text{and}\quad f_1^2(x,y)=(y-x)^2/2, \]

then, if one represents the function \(f_m^i(x,y)\) in the form

\[ f_m^i(x,y)=x^{(2-i)}\sum_{k=0}^{m-1} \frac{(-1)^k a_{k+1}^m}{(m+k+i-1)!}\, y^{m-k-1}(y-x)^{m+k+i-1}, \tag{3.30} \]

and substitutes it into (3.29), we obtain, for the coefficients \(a_{k+1}^m\), the equations

\[ a_s^{m+1}=\frac{1}{(m+1-s)!}\sum_{k=1}^{s}(m+1-k)!\,a_k^m \quad \text{for } 1\le s\le m, \]

\[ a_{m+1}^{m+1}=a_m^{m+1}. \]

According to (3.30), \(a_1^1=1\). Hence, by Lemma 3, the coefficients \(a_k^m\) are determined by formula (3.25). Adding the expressions \(f_m^1(x,y)\) and \(f_m^2(x,y)\), we obtain (3.28). Lemma 4 is proved.

Lemma 5. For \(n\ge 2\), for the function \(\Phi_n(y)\) defined by formula (3.9), the relation

\[ \Phi_n(y)=\sum_{k=0}^{n-1} b_k^n y^{2k} -\cos y \sum_{k=0}^{\left[\frac{n+1}{2}\right]-1} \frac{(-1)^k(2n-2k-2)!\,y^{2k}} {2^{\,n-2k-1}(n-2k-1)!(2k)!} \]
\[ -\sin y \sum_{k=0}^{\left[\frac{n}{2}\right]-1} \frac{(-1)^k(2n-2k-3)!\,y^{2k+1}} {2^{\,n-2k-2}(n-2k-2)!(2k+1)!}, \tag{3.31} \]

holds, where

\[ b_k^n=(-1)^k(2n-2k-3)\frac{(n-2)!}{(2k)!\,2^{\,n-2k-2}}\times \]
\[ \times \sum_{l=0}^{2k}(-1)^l \binom{2n-2k+l-4}{\,n-2\,} \frac{\binom{2k}{l}}{2^l}. \tag{3.32} \]

Proof. Relation (3.31) follows from relation (3.20), if one uses relation (3.28), in which one must put \(m=n-1\). The result is obtained by carrying out rather cumbersome but elementary transformations, distinguishing the cases of even and odd \(n\), and also taking into account (by definition) the relation \(\binom{n}{m}=0\) for \(m>n\); we shall not present these computations.

Lemmas 7–9 are stages in the proof of Lemma 6. In them, however, proofs of interesting relations are given.

Lemma 6. For nonnegative integers \(n\) and \(k\), the relations

\[ \sum_{l=0}^{2k}(-1)^l \binom{2n-l}{n}\binom{2k}{l}2^l = (-1)^k\frac{(2k)!(2n-2k)!}{k!\,n!\,(n-k)!}, \tag{3.33} \]

\[ (-1)^k\frac{(n-2)!}{(2k)!\,2^{\,n-2k-2}} \sum_{l=0}^{2k}(-1)^l \binom{2n-2k+l-4}{\,n-2\,} \frac{\binom{2k}{l}}{2^l} = \]
\[ = \frac{(2n-2k-5)!!}{2^k k!}, \tag{3.34} \]

\[ \left. \frac{d^n}{dx^n}x^{2n}\left(1-\frac{1}{x}\right)^{2k} \right|_{x=1/2} = (-1)^k\frac{(2k)!(2n-2k)!}{k!\,2^n(n-k)!}. \tag{3.35} \]

Lemma 7. The following relations hold:
\[ \frac{2^{2l}(k-1)!}{(2l)!(k-l-1)!}-\frac{2l+1}{l!} \]
\[ -(2l+1)\sum_{s=0}^{l-2} \frac{(2k-2l-3)(2k-2l-5)\ldots(2k-2l-2s-3)} {(2s+3)!!(s+1)!(l-s-1)!} = \]
\[ = \frac{2^l}{(2l)!}(2k-2l-3)(2k-2l-5)\ldots(2k-4l-1), \tag{3.36} \]
\[ \frac{2^{2l+1}(k-1)!}{(2l+1)!(k-l-2)!}-\frac{1}{l!} \]
\[ -(l+1)\sum_{s=0}^{l-1} \frac{(2k-2l-3)(2k-2l-5)\ldots(2k-2l-2s-3)} {(2s+1)!!(s+1)!(l-s)!} = \]
\[ = \frac{2^l}{(2l+1)!}(2k-2l-3)(2k-2l-5)\ldots(2k-4l-3). \tag{3.37} \]

Proof. Denote the left-hand side of formula (3.36) by \(b_{k,l}\), and the right-hand side by \(c_{k,l}\), and consider two sequences of expressions defined by the relations
\[ b^m_{k,l}=b^{m-1}_{k+1,l}-b^{m-1}_{k,l},\qquad \text{with } b^1_{k,l}=b_{k+1,l}-b_{k,l}, \]
\[ c^m_{k,l}=c^{m-1}_{k+1,l}-c^{m-1}_{k,l},\qquad \text{with } c^1_{k,l}=c_{k+1,l}-c_{k,l}. \]

From (3.36), it is easy to prove by induction that for \(m\le l-2\)
\[ b^m_{k,l} = \frac{2^{2l}l!(k-1)!}{(2l)!(l-m)!(k-l+m-1)!} - \frac{2m(2l+1)}{(2m+1)!!(l-m)!} \]
\[ -2m(2l+1)\sum_{s=m}^{l-2} \frac{(2k-2l-3)(2k-2l-5)\ldots(2k-2l+2m-2s-3)} {(2s+3)!!(s-m+1)!(l-s-1)!}, \tag{3.38} \]
for \(m=l-1\) the sum on the right-hand side is absent. Putting \(m=l-1\), we obtain
\[ b^{\,l-1}_{k,l}=2^{2l-1}l!(2k-2l-3)/(2l)!. \]
On the other hand, from (3.36) it is easy to prove by induction that for \(m\le l-1\)
\[ c^m_{k,l} = \frac{2^{l+m}l!}{(2l)!(l-m)!}(2k-2l-3)(2k \]
\[ -2l-5)\ldots(2k-4l+2m-1). \tag{3.39} \]
Putting \(m=l-1\), we obtain
\[ c^{\,l-1}_{k,l}=2^{2l-1}l!(2k-2l-3)/(2l)!. \]
Thus,
\[ b^{\,l-1}_{k,l}=c^{\,l-1}_{k,l}. \]

The expressions \(b^{m-1}_{k,l}\) and \(c^{m-1}_{k,l}\), however, may be regarded as solutions of the finite-difference equation in \(k\)
\[ d^{m-1}_{k+1,l}-d^{m-1}_{k,l}=d^m_{k,l}, \]
where \(d^m_{k,l}\) is to be regarded as a prescribed right-hand side. Two different solutions of this equation differ by a constant (in \(k\)). Therefore, if

there exists at least one \(k\) such that \(b_{k,l}^{m-1}\) and \(c_{k,l}^{m-1}\) coincide for this \(k\) (and arbitrary \(m\)), then they will coincide for any \(k\) and \(m\) \((0\leq m\leq l-1)\). From (3.38) and (3.39) it is easy to show that for \(k=l-m\) one has \(b_{l-m,l}^m=c_{l-m,l}^m\). Hence the identity (3.36) follows.

In an analogous way we shall prove (3.37). Denote the left-hand side of (3.37) by \(f_{k,l}\), and the right-hand side by \(g_{k,l}\). By induction it is proved that

\[ f_{k,l}^m=f_{k+1,l}^{m-1}-f_{k,l}^{m-1} = \frac{2^{2l+1}(l+1)!(k-1)!}{(2l+1)!(l-m+1)!(k-l+m-2)!} - \]

\[ -\frac{2^m(l+1)}{(2m-1)!!(l-m+1)!} - \]

\[ -2^m(l+1)\sum_{s=m}^{l-1} \frac{(2k-2l-3)(2k-2l-5)\cdots(2k-2l+2m-2s-3)} {(2s+1)!!(s-m+1)!(l-s)!}, \]

\[ g_{k,l}^m=g_{k+1,l}^{m-1}-g_{k,l}^{m-1} = \]

\[ = \frac{2^{l+m}(l+1)!}{(2l+1)!(l-m+1)!} (2k-2l-3)(2k-2l-5)\cdots(2k-4l+2m-3) \]

and

\[ f_{k,l}^l=2^{2l}(l+1)!(2k-2l-3)/(2l+1)! = g_{k,l}^l . \]

Moreover, \(f_{l-m+1,l}^m=g_{l-m+1,l}^m\), which proves relation (3.37).

Lemma 7 is proved.

Lemma 8. Let us denote by \(S_m\) the expression

\[ S_m=\sum_{l=0}^{m}\frac{(-1)^l(2n-l)!\,2^l}{(n-l)!\,l!\,(2k-l)!}, \tag{3.40} \]

then

\[ S_{2l}= \frac{2^{l+1}(2n-2l-1)!}{(n-2l-1)!(2k-2l)!} \left\{ \frac{2^l}{(2l)!}\frac{2k-2l}{2k} + \right. \]

\[ \left. + \frac{(k-l)!}{2^l k!} \sum_{s=0}^{l-1} \frac{(2k-2l-1)(2k-2l-3)\cdots(2k-2l-2s-1)} {(2s+1)!!\,s!\,(l-1-s)!\,(n-l-s-1)} \right\}, \tag{3.41} \]

\[ S_{2l+1} = -\frac{2^{l+1}(2n-2l-2)!}{(n-2l-2)!(2k-2l-2)!} \left\{ \frac{2^{l+1}}{(2l+1)!\,2k} + \right. \]

\[ \left. + \frac{(k-l-1)!}{2^{l+1}k!} \left[ \frac{1}{l!(n-l-1)} + \right. \right. \]

\[ \left. \left. + \sum_{s=0}^{l-1} \frac{(2k-2l-3)(2k-2l-5)\cdots(2k-2l-2s-3)} {(2s+1)!!\,(s+1)!\,(l-s-1)!\,(n-l-s-2)} \right] \right\}. \tag{3.42} \]

Proof. We shall prove relations (3.41) and (3.42) by induction. According to (3.40),

\[ S_1 = \frac{(2n)!}{n!}\frac{1}{(2k)!} - \frac{(2n-1)!\,2}{(n-1)!(2k-1)!} = -\frac{2(2n-1)!}{(n-1)!(2k-2)!}\frac{1}{2k}, \]

\[ \frac{2n-1}{n-1}=2+\frac{1}{n-1}, \]

whence

\[ S_1=-\frac{2(2n-2)!}{(n-2)!(2k-2)!}\left(\frac{2}{2k}+\frac{1}{2k}\frac{1}{n-1}\right). \]

Next

\[ S_2=S_1+\frac{(2n-2)!\,2!}{(n-2)!\,2!\,(2k-2)!}= \]

\[ =\frac{2(2n-2)!}{(n-2)!(2k-2)!}\left(\frac{2k-2}{2k}-\frac{1}{2k}\frac{1}{n-1}\right), \]

\[ \frac{2n-2}{n-2}=2+\frac{2}{n-2}. \]

Hence

\[ S_2=\frac{2^2(2n-3)!}{(n-3)!(2k-2)!}\left(\frac{2k-2}{2k}-\frac{1}{2k}\frac{1}{n-1}\right)\left(1+\frac{1}{n-2}\right)= \]

\[ =\frac{2^2(2n-3)!}{(n-3)!(2k-2)!}\left(\frac{2k-2}{2k}+\frac{2k-3}{2k}\frac{1}{n-2}\right), \]

and this expression coincides with the expression obtained from (3.41), if in it one sets \(l=1\).

Suppose now that expression (3.41) is valid for \(l\); then, according to (3.40) and (3.41),

\[ S_{2l+1}=S_{2l}-\frac{(2n-2l-1)!\,2^{2l+1}}{(n-2l-1)!(2l+1)!(2k-2l-1)!}= \]

\[ =-\frac{2^{l+1}(2n-2l-1)!}{(n-2l-1)!(2k-2l-2)!} \left\{ \frac{2^l}{(2l+1)!\,2k} -\right. \]

\[ \left. -\frac{(k-l-1)!}{2^{l+1}k!} \left[ \frac{1}{(l-1)!(n-l-1)} +\right.\right. \]

\[ \left.\left. +\sum_{s=1}^{l-1} \frac{(2k-2l-3)(2k-2l-5)\ldots(2k-2l-2s-1)} {(2s+1)!!\,s!\,(l-1-s)!\,(n-l-s-1)} \right] \right\}. \]

Putting

\[ \frac{2n-2l-1}{n-2l-1}=2+\frac{2l+1}{n-2l-1}, \]

we obtain

\[ S_{2l+1}=-\frac{2^{l+1}(2n-2l-2)!}{(n-2l-2)!(2k-2l-2)!} \left\{ \frac{2^{l+1}}{(2l+1)!\,2k} -\right. \]

\[ \left. -2\,\frac{(k-l-1)!}{2^{l+1}k!} \left[ \frac{1}{(l-1)!(n-l-1)} +\right.\right. \]

\[ \left.\left. +\sum_{s=1}^{l-1} \frac{(2k-2l-3)(2k-2l-5)\ldots(2k-2l-2s-1)} {(2s+1)!!\,s!\,(l-1-s)!\,(n-l-s-1)} \right] + \right. \]

\[ + \frac{2^l}{(2l)! \, 2k(n-2l-1)} - \frac{(k-l-1)!(2l+1)}{2^{l+1}k!(n-2l-1)} \left[ \frac{1}{(l-1)!(n-l-1)} \right. \]
\[ \left. {}+ \sum_{s=1}^{l-1} \frac{(2k-2l-3)(2k-2l-5)\ldots(2k-2l-2s-1)} {(2s+1)!!\,s!(l-1-s)!(n-l-s-1)} \right]\Bigg\}. \]

From this relation, by elementary transformations using relation (3.36), we arrive at (3.42). Thus, if we assume that (3.41) has been proved, then, as has just been shown, (3.42) follows from it.

To prove (3.41) for arbitrary \(l\), let us define \(S_{2l+2}\). According to (3.40) and (3.42),

\[ S_{2l+2}=S_{2l+1}+ \frac{(2n-2l-2)!\,2^{2l+2}} {(n-2l-2)!(2l+2)!(2k-2l-2)!} = \]

\[ = \frac{2^{l+1}(2n-2l-2)!} {(n-2l-2)!(2k-2l-2)!} \left\{ \frac{2^{l+1}(2k-2l-2)} {(2l+2)!\,2k} -\right. \]

\[ \left. {}-\frac{(k-l-1)!}{2^{l+1}k!} \left[ \frac{1}{l!(n-l-1)} + \sum_{s=0}^{l-1} \frac{(2k-2l-3)(2k-2l-5)\ldots(2k-2l-2s-3)} {(2s+1)!!(s+1)!(l-s-1)!(n-l-s-2)} \right] \right\}, \]

\[ \frac{2n-2l-2}{n-2l-2} = 2\left(1+\frac{l+1}{n-2l-2}\right). \]

Hence

\[ S_{2l+2} = \frac{2^{l+2}(2n-2l-3)!} {(n-2l-3)!(2k-2l-2)!} \left\{ \frac{2^{l+1}(2k-2l-2)} {(2l+2)!\,2k} +\right. \]

\[ {}+ \frac{(k-l-1)!}{2^{l+1}k!} \left[ \sum_{s=0}^{l-1} \frac{(2k-2l-3)(2k-2l-5)\ldots(2k-2l-2s-3)} {(2s+1)!!\,s!(l-s)!(n-l-s-2)} +\right. \]

\[ \left. {}+ \left[ \frac{2^{2l+1}(k-1)!}{(2l+1)!(k-l-2)!} -\frac{1}{l!} \right. \right. \]

\[ \left. \left. {}-(l+1) \sum_{s=0}^{l-1} \frac{(2k-2l-3)(2k-2l-5)\ldots(2k-2l-2s-3)} {(2s+1)!!(s+1)!(l-s)!} \right] \times \frac{1}{n-2l-2} \right]\Bigg\}. \]

If we now use formula (3.37), then we find an expression for \(S_{2l+2}\), which can also be obtained from (3.41), if in the latter we replace \(l\) by \(l+1\).

Lemma 8 is thereby proved.

Lemma 9. The following relation holds:

\[ \sum_{s=0}^{k-1} \frac{(-1)^{s+1}} {s!(k-1-s)!(n-k-s-1)} = \frac{(-1)^k} {\displaystyle\prod_{s=0}^{k-1}(n-k-1-s)}. \tag{3.43} \]

Proof. Decompose the right-hand side of relation (3.43) into partial fractions

\[ \frac{(-1)^k}{\prod_{s=0}^{k-1}(n-k-1-s)} = (-1)^k \sum_{s=0}^{k-1} \frac{1}{\prod_{s'=0}^{k-1}{}'(s-s') (n-k-1-s)}, \]

where the prime on the product symbol means that the term \(s'=s\) is omitted. But

\[ \prod_{s'=0}^{k-1}{}'(s-s') = s(s-1)\cdots 1(-1)(-2)\cdots (-(k-1-s)) = \]

\[ = (-1)^{k+s-1}s!(k-1-s)!, \tag{3.44} \]

which proves relation (3.43).

Lemma 9 is proved.

Now we can prove Lemma 6.

Proof of Lemma 6. Rewriting the binomial coefficients on the left-hand side of (3.33) in terms of factorials, we obtain the relation

\[ \sum_{l=0}^{2k} \frac{(-1)^l(2n-l)!2^l}{(n-l)!\,l!\,(2k-l)!} = (-1)^k\frac{(2n-2k)!}{k!(n-k)!}. \]

Thus it remains to prove the validity of the relation

\[ S_{2k}=(-1)^k\frac{(2n-2k)!}{k!(n-k)!}, \tag{3.45} \]

where \(S_m\) is defined by formula (3.40). Setting \(l=k\) in formula (3.41) and taking (3.43) into account, we obtain

\[ S_{2k} = \frac{2(2n-2k-1)!}{(n-2k-1)!\,k!} \sum_{s=0}^{k-1} \frac{(-1)^{s+1}}{s!(k-1-s)!(n-k-1-s)} = \]

\[ = (-1)^k\frac{(2n-2k)!}{k!(n-k)!}. \]

Formula (3.33) is proved.

Formula (3.34) follows from (3.33) if in the former one makes the substitutions \(n-2 \to n\) and \(2k-l \to l\).

Formula (3.35) also follows immediately from (3.33). Indeed, since

\[ \frac{(2n-l)!}{(n-l)!}2^l = 2^n\left.\frac{d^n}{dx^n}x^{2n-l}\right|_{x=1/2}, \]

i.e.

\[ \binom{2n-l}{n}2^l = \left.\frac{2^n}{n!}\frac{d^n}{dx^n}x^{2n-l}\right|_{x=1/2}, \]

then

\[ \sum_{l=0}^{2k}(-1)^l \binom{2n-l}{n} \binom{2k}{l}2^l = \left.\frac{2^n}{n!}\frac{d^n}{dx^n} x^{2n}\left(1-\frac{1}{x}\right)^{2k}\right|_{x=1/2}, \]

as was required.

Lemma 6 is proved completely.

Proof of Lemma 1. Lemma 1 follows directly from Lemmas 5 and 6 (in the latter lemma one must use formula (3.34)).

It is now easy to prove Theorem 2.

Proof of Theorem 2. The validity of Theorem 2 follows immediately from the definition of the function \(M_n(x,y)\) (formula (1.1)), from (2.1), from the representation of the Hankel function \(H_n^{(1)}(z)\) in terms of the functions \(J_n(z)\) and \(N_n(z)\):

\[ H_n^{(1)}(z)=J_n(z)+iN_n(z), \]

from formula (3.6), and from formulas (3.11)—(3.13).

§ 4. EXPANSION OF MACDONALD INTEGRALS INTO A POWER SERIES IN THE VARIABLE \(x\)

Theorem 6. For \(|x|<1\), the function \(M_n(x,y)\) is expanded into a power series in \(x\) (converging uniformly for any finite \(y\)),

\[ M_n(x,y)=\sqrt{\frac{\pi}{2y}}\,H_{n-1/2}^{(1)}(y) +\sum_{m=0}^{\infty} \frac{(-1)^m y^m H_{n+m}^{(1)}(y)}{2^m(2m+1)m!}\, x^{2m+1}, \tag{4.1} \]

and if in the series (4.1) one stops at the \(k+1\)-st term \((0\le m\le k)\), then the remainder of the series is equal to

\[ R_k(x,y)= \frac{(-1)^{k+1}y^{k+1}}{2^{k+2}k!}\, x^{2k+3} \int_0^1 \frac{ H_{n+k+1}^{(1)}\!\left(y\sqrt{1-x^2t}\right) }{ (1+x^2t)^{(n+k+1)/2} } \times \]

\[ {}\times dt\int_t^1 \frac{(u-t)^k}{\sqrt{u}}\,du. \tag{4.2} \]

Remark. According to Theorem 6, the radius of convergence of the series (4.1) is not less than one; but, according to Theorem 2, the point \(x=+i\) is a singular point of the function \(M_n(x,y)\). Thus, the radius of convergence of the series (4.1) cannot be greater than one, i.e. it is equal to one.

Proof. We shall give the proof for \(0<\arg y<\pi/2\). The validity of the result for all those \(y\) for which the function \(M_n(x,y)\) is defined follows from the regularity of \(M_n(x,y)\) in \(y\) and from the regularity in \(y\) of the function \(R_k(x,y)\) in (4.2). According to (2.1),

\[ M_n(x,y)=\sqrt{\frac{\pi}{2y}}\,H_{n-1/2}^{(1)}(y) + \int_0^{\operatorname{arsh} x} H_n^{(1)}(y\operatorname{ch}\xi)\, \frac{d\xi}{(\operatorname{ch}\xi)^{\,n-1}} . \]

We shall use the integral representation of the function \(H_\nu^{(1)}(z)\) for \(\operatorname{Re}\nu>-1\) and \(0<\arg z<\pi/2\) ([2], formula 8.421 (18)):

\[ H_\nu^{(1)}(z)= \frac{1}{\pi}\exp\left\{-\frac{i\pi}{2}(\nu+1)\right\} \left(\frac{z}{2}\right)^\nu \int_0^\infty \exp\left\{i\left(\tau+\frac{z^2}{4\tau}\right)\right\} \frac{d\tau}{\tau^{\nu+1}}, \tag{4.3} \]

then (the legitimacy of complete changes of the order of integration is obvious)

\[ \int_0^{\operatorname{arsh} x} H_n^{(1)}(y\operatorname{ch}\xi)\, \frac{d\xi}{(\operatorname{ch}\xi)^{\,n-1}} = \]

\[ = \frac{(-i)^{n+1}}{\pi}\left(\frac{y}{2}\right)^n \int_0^\infty \exp\left\{ i\left(\tau+\frac{y^2}{4\tau}\right)\right\} \frac{d\tau}{\tau^{n+1}} \int_0^{\operatorname{arsh} x} \exp\left\{\frac{i y^2}{4\tau}(\operatorname{ch}^2\xi-\right. \]
\[ \left. -1)\right\}\operatorname{ch}\xi\,d\xi = \frac{(-i)^{n+1}}{\pi}\left(\frac{y}{2}\right)^n \int_0^\infty \exp\left\{ i\left(\tau+\frac{y^2}{4\tau}\right)\right\} \times \]
\[ \times \frac{d\tau}{\tau^{n+1}} \int_0^x \exp\left(\frac{i y^2}{4\tau}u^2\right)\,du. \tag{4.4} \]

Next we shall use the special case of Darboux’s formula ([3], p. 176): if \(f(z)\) is regular at all points of the segment joining \(a\) with \(z\), then

\[ f(z)=\sum_{m=0}^k \frac{f^{(m)}(a)}{m!}(z-a)^m +\frac{(z-a)^{k+1}}{k!} \int_0^1 (1-t)^k f^{(k+1)}(a+t(z-a))\,dt. \tag{4.5} \]

Hence

\[ \exp\left(\frac{i y^2}{4\tau}u^2\right) = \sum_{m=0}^k \left(\frac{i y^2}{4\tau}\right)^m \frac{u^{2m}}{m!} + \]
\[ + \left(\frac{i y^2}{4\tau}\right)^{k+1} \frac{u^{2(k+1)}}{k!} \int_0^1 (1-t)^k \exp\left(\frac{i y^2}{4\tau}u^2 t\right)\,dt. \tag{4.6} \]

Substituting (4.6) into (4.4) and taking (4.3) into account, we obtain

\[ M_n(x,y)= \sqrt{\frac{\pi}{2y}}\,H_{n-1/2}^{(1)}(y) + \sum_{m=0}^k \frac{(-1)^m y^m H_{n+m}^{(1)}(y)} {2^m(2m+1)m!}\,x^{2m+1} + \]
\[ + \frac{(-i)^{\,n-k}}{\pi k!} \left(\frac{y}{2}\right)^{n+2k+2} \int_0^x u^{2k+2}\,du \int_0^1 (1-t)^k\,dt \times \]
\[ \times \int_0^\infty \exp\left\{ i\left(\tau+\frac{y^2+y^2u^2t}{4\tau}\right)\right\} \frac{d\tau}{\tau^{n+k+2}}. \]

From simple transformations of the integrals in the last expression follows the validity of the second part of the theorem.

Let us give an estimate for \(R_k(x,y)\). From formula (3.14) and the relation \(H_n^{(1)}(z)=J_n(z)+iN_n(z)\), and also owing to the uniform convergence with respect to \(n\) of the power series for \(J_n(z)\) and of the series in (3.14), it follows that

\[ \lim_{n\to\infty} \left\{ H_n^{(1)}(z)\, /\, \left( -\frac{i}{\pi}\left(\frac{2}{z}\right)^n (n-1)! \right) \right\} =1, \]

i.e., as \(n\to\infty\)

\[ H_n^{(1)}(z) = -\frac{i}{\pi}\left(\frac{2}{z}\right)^n (n-1)!(1+o(1)), \tag{4.7} \]

where \(o(1)\) denotes a quantity tending to zero as \(n\to\infty\). Consequently, for sufficiently large \(n\),

\[ \left|H_n^{(1)}(z)\right|<\frac{1-\varepsilon}{\pi}\left(\frac{2}{|z|}\right)^n (n-1)!, \tag{4.8} \]

where \(\varepsilon>0\) is any arbitrarily small number. Hence, for sufficiently large \(k\),

\[ \begin{aligned} |R_k(x,y)| &\le \frac{|y|^{k+1}|x|^{2k+3}}{2^{k+2}k!} \int_0^1 \frac{\left|H_{n+k+1}^{(1)}\!\left(y\sqrt{1+x^2t}\right)\right|} {(1+x^2t)^{(n+k+1)/2}}\,dt \int_t^1 \frac{(u-t)^k}{\sqrt{u}}\,du \\ &< \frac{2^{\,n-1}(1+\varepsilon)(n+k)!}{\pi k!} |y|^n |x|^{2k+3} \int_0^1 \frac{dt}{\sqrt{t}} \int_t^1 (u-t)^k\,du, \end{aligned} \]

i.e.,

\[ |R_k(x,y)|< \frac{2^{\,n+k}(1+\varepsilon)(n+k)!}{\pi(2k+1)!!} |y|^n |x|^{2k+3}. \tag{4.9} \]

Therefore, for \(|x|<1\), \(R_k(x,y)\to0\) as \(k\to\infty\) uniformly in \(y\), which proves the first part of the theorem.

Theorem 6 is proved completely.

References

1. Macdonald H. M. Proc. Lond. Math. Soc., 14, 410—427, 1915.
2. Gradshteyn I. S., Ryzhik I. M. Tables of Integrals, Sums, Series and Products. Moscow, GIFML, 1962.
3. Whittaker E. T., Watson G. N. A Course of Modern Analysis, 1, Moscow, GIFML, 1962.

Received by the Editorial Office
May 3, 1966

Central Scientific Research
Mining-Prospecting Institute of Non-Ferrous,
Rare and Precious Metals