Abstract
The article is a continuation of the author's work [1] and is devoted to the derivation of two types of asymptotic expansions for Macdonald integrals $n$ of order $M_n(x,y)$. Two expansions of the first type are asymptotic as $|y|\sqrt{1+x^2}\to\infty$. The first expansion is an expansion in terms of the asymptotic sequence $(H^1_{m+\alpha}(y\sqrt{1+x^2})/(y\sqrt{1+x^2})^m){m\in N}$, where $H^{(1)}\nu(z)$ is the Hankel function of the first kind; $\alpha$ is a real number; $N$ is the set of natural numbers. The second asymptotic expansion of the first type is an expansion in terms of the asymptotic sequence $((y\sqrt{1+x^2})^{-m}){m\in N}$, which follows from the first one. Both expansions are uniform with respect to $x$ for $|x|\ge\varepsilon>0$, where $\varepsilon$ is an arbitrarily small number. The asymptotic expansion of the second type is valid for $|y|\to\infty$ and in the case where the parameter $y(\sqrt{1+x^2}-1)$ remains finite as $|y|\to\infty$, i.e., $x$ must simultaneously tend to zero such that the expression $y(\sqrt{1+x^2}-1)$ remains finite. The presented expansion of the second type is an expansion in terms of the asymptotic sequence $(y^{-m})$. Bibliography: 4 items.
Full Text
Preamble
In this section, we refine the asymptotic properties of the function $M_n(x, y)$ introduced in \cite[1]. We consider the behavior of the function as $|y| \sqrt{1 + x^2} \to \infty$ and $|y| \sqrt{1 + x^2} \to 0$. Specifically, for $|x| > \epsilon > 0$, we establish estimates for the remainder terms and provide expansions in terms of the parameter $y$.
§ 1. Asymptotic Definitions and Notation
Let $x_0$ be a limit point in the domain $U$. We say that $\phi(x) = O(\psi(x))$ as $x \to x_0$ if there exists a constant $A$ such that $|\phi(x)| < A |\psi(x)|$ for all $x in a neighborhood $U_\delta$ of $x_0$. Similarly, $\phi(x) = o(\psi(x))$ if $\phi(x)/\psi(x) \to 0$ as $x \to x_0$.
An asymptotic series is defined as:
$$f(x) \sim \sum_{k=0}^{\infty} \phi_k(x), \quad x \to x_0 \tag{1.1}$$
if for every $n \in N$, the following condition holds:
$$f(x) = \sum_{k=0}^{n} \phi_k(x) + o(\phi_n(x)), \quad x \to x_0 \tag{1.2}$$
Equivalently, this can be expressed using the $O$-notation as:
$$f(x) = \sum_{k=0}^{n} \phi_k(x) + O(\phi_{n+1}(x)) \tag{1.3}$$
In the context of our analysis, we often encounter expansions of the form:
$$f(x) = \phi(x) + \psi(x) \left{ \sum_{k=0}^{n} a_k \phi_k(x) + o(\phi_n(x)) \right} \tag{1.4}$$
which implies the asymptotic equivalence:
$$f(x) \sim \phi(x) + \psi(x) \sum a_k \phi_k(x) \tag{1.5}$$
§ 2. Asymptotic Expansion for Large Arguments
We consider the domain $-\pi < \arg z < 2\pi$ as $|z| \to \infty$. Let the sequence be defined by $(H_{n+a}(z)/z^n){n \in N}$. We assume the property:
$$\frac{H$$}(z)}{z^{n+1}} = o\left(\frac{H_{n+a}(z)}{z^n}\right) \tag{2.1
For the function $M_n(x, y)$, as $|y| \sqrt{1+x^2} \to \infty$, we derive the following representation:
$$M_n(x, y) \sim \frac{H_{n-1/2}^{(1)}(y)}{\sqrt{\pi} (1+x^2)^{n/2}} \sum_{k=0}^{\infty} \frac{(-1)^k \Gamma(k+1/2)}{2^k \Gamma(1/2)} \frac{|x|^{2k+1}}{(1+x^2)^k} \frac{H_{n-k-1}^{(1)}(y \sqrt{1+x^2})}{(y \sqrt{1+x^2})^{k+1}} \tag{2.2}$$
The remainder term $R_k(x, y)$ for the expansion in (2.2) is given by:
$$R_k(x, y) = \frac{(-1)^{k+1}}{2^{k+1} \Gamma(k+3/2)} \int_1^{\infty} H_{n-k-2}^{(1)}(y u \sqrt{1+x^2}) (u^2-1)^{k+3/2} du \tag{2.3}$$
For $x > 0$ and $x < 0$, the behavior of the integral depends on the sign of the argument. Using the properties of Hankel functions $H_n^{(1)}(z)$, we can bound the remainder $|R| < A$ for $|x| > \epsilon > 0$.
By applying the integral representation from \cite[1]:
$$M_n(x, y) = \int_0^{\text{arsh } x} H_n^{(1)}(y \cosh \xi) (\cosh \xi)^{n-1} d\xi \tag{2.5}$$
and substituting the series for the Hankel function, we obtain the coefficients for (2.2). Specifically, using the relation:
$$H_{n-1}^{(1)}(z) = \exp(iz) \sum \dots \tag{2.9}$$
we verify that the $k$-th term in (2.3) matches the required asymptotic order.
§ 3. Expansion for Small Arguments
As $|y| \to 0$, we investigate the behavior of $M_n(x, y)$ in the domain $0 < \arg y < \pi$. The expansion takes the form:
$$M_n(x, y) \sim \frac{H_n^{(1)}(y) \text{sign}(x)}{\sqrt{y (\sqrt{1+x^2}-1)}} \int e^{iu^2} du + \dots \tag{3.1}$$
where the coefficients $c_n$ are determined by the recurrence relations:
$$c_n = \frac{(-1)^l (2l+1)!! (2p-1)!! \Gamma(n+m-l)}{(2m+1)!! 2^{2p} (l+1-p)! p! (m-l)!} \tag{3.2}$$
For $x > 0$, we use the substitution $u = \sqrt{2y \sinh(\xi/2)}$. The integral (3.3) is then evaluated using the properties of the Gamma function and the asymptotic expansion of $H_n^{(1)}(z)$ as $z \to 0$.
Applying Lemma 2, we find that for $|x| > \epsilon > 0$:
$$R_k(x, y) = O\left(\frac{\exp(iy)}{y^{k+3}}\right) \tag{3.5}$$
The final asymptotic form for $M_n(x, y)$ as $|y| \sqrt{1+x^2} \to 0$ is:
$$M_n(x, y) \approx \frac{\text{sign}(x)}{\sqrt{\pi} (1+x^2)^{n/2-1}} \sum_{m=0}^k \frac{(n+m-1)!}{m! (n-m-1)!} \frac{1}{(2iy)^m} \tag{3.12}$$
This result is consistent with the general theory of confluent hypergeometric functions and the specific bounds established in \cite[4].
References
- T. A. A., Journal of Mathematical Physics, No. 10, 1967.
- D. A. A., Asymptotic Expansions, 1962.
- I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products, 1965.
- Higher Transcendental Functions, Vol. 2, 1966.