MATHEMATICS

A. V. LYUTOTSKII

ON THE ASYMPTOTIC BEHAVIOR OF ANALYTIC FUNCTIONS

(Presented by Academician S. L. Sobolev on 25 IV 1958)

The paper indicates a certain dependence between the asymptotic behavior

\[ f(z)=\sum_{s=1}^{\infty}(-1)^{s-1}a_s z^s = a_1z-a_2z^2+a_3z^3-\cdots \tag{1} \]

as \(z\to\infty\) along the positive real axis and the analytic properties of the function \(a_s=a(s)\), which is considered chiefly as a rational function of the complex variable \(s\). For this case \(f(z)\) can be analytically continued to the whole half-axis \(x\geqslant0\). If in the half-plane \(\operatorname{Re}(s)\geqslant0\) the function \(a_s\) has no poles, then \(f(x)\) tends to a finite limit. If \(a_s\) has poles in this half-plane, then \(f(x)\) grows without bound, and the order of growth depends on the abscissa of the rightmost pole. Knowing the poles of \(a_s\), one can indicate not only the order of growth, but also an exact asymptotic formula for \(f(x)\). This dependence, and moreover in the very same form, turns out to be valid not only for rational functions \(a_s\). Here we restrict ourselves to additional indications only for more or less special classes of functions for which the properties mentioned hold.

Definition. The function \(a_s\) is said to possess property A if it is defined at the integer points \(0,1,2,\ldots\), the series (1) converges in some neighborhood of the point \(z=0\), \(f(z)\) is analytically continuable to the whole half-axis \(x\geqslant0\), and tends to \(a_0\) as \(x\to\infty\).

Lemma. If the \(n\)-th difference \(\Delta^n a_s\) at the point \(0\) for the function \(a_s\) tends to zero as \(n\to\infty\), then \(a_s\) and \(a_s/\Gamma(s+1)\) possess property A.

Theorem 1. If \(a_s\) is a rational function having no poles in the half-plane \(\operatorname{Re}(s)\geqslant0\), then \(a_s\) and \(a_s/\Gamma(s+1)\) both possess property A.

Proof. We shall show that \(\Delta^n a_0\to0\) as \(n\to\infty\). For a polynomial the assertion is obvious. In the presence of poles it is sufficient to consider the case \(a_s=(s-p)^{-k}\), where \(\operatorname{Re}(p)<0\). Denote \(p=-q\), \(\operatorname{Re}(q)>0\); setting \(F_n(x,q)=x^{k-1}e^{-qx}(1-e^{-x})^n\), we have for \(n\geqslant0\)

\[ a_n=\frac{1}{(k-1)!}\int_0^\infty x^{k-1}e^{-(q+n)x}\,dx, \qquad \Delta^n a_0=\frac{(-1)^n}{(k-1)!}\int_0^\infty F_n(x,q)\,dx. \]

Take arbitrary \(x_0>0\); let

\[ I_1=\int_0^{x_0} F_n(x,q)\,dx, \qquad I_2=\int_{x_0}^{\infty} F_n(x,q)\,dx. \]

Then

\[ |\Delta^n a_0|\leqslant \frac{1}{(k-1)!}\bigl(|I_1|+|I_2|\bigr). \]

Let an arbitrary \(\varepsilon>0\) be given. Since, for any \(n\),

\[ |I_2|\leqslant \int_{x_0}^{\infty} F_n(x,\operatorname{Re}(q))\,dx < \int_{x_0}^{\infty} x^{k-1}e^{-x\operatorname{Re}(q)}\,dx, \]

\(x_0\) can be chosen so large that, for all \(n\), the inequality \(|I_2|<\varepsilon\) will hold. In the integral \(I_1\) we have \(0\leqslant x\leqslant x_0\), \(0\leqslant 1-e^{-x}\leqslant 1-e^{-x_0}=\lambda<1\); consequently,

\[ |I_1|\leqslant \lambda^n \int_0^{x_0} x^{k-1}e^{-x\operatorname{Re}(q)}\,dx, \]

where the second factor on the right-hand side does not depend on \(n\). Hence, for all sufficiently large \(n\), \(|I_1|<\varepsilon\), and then

\[ |\Delta^n a_0|<\frac{2\varepsilon}{(k-1)!}. \]

Theorem 2. Every entire function \(a_s\) of order \(1\) and type \(\sigma<\pi\) has property A.

Proof. From the inequality \(|a_s|<Me^{\sigma|s|}\) we conclude for the series (1) that its radius of convergence is \(R\geqslant e^{-\sigma}\). Denote by \(\psi(s)\) the function associated with \(a_s\) by Borel. It is known that \(\psi(s)\) is regular outside the circle \(|s|=\sigma\), and \(a_s\) is represented in the whole plane by means of it by the integral

\[ a_s=\frac{1}{2\pi i}\int_C e^{s\zeta}\psi(\zeta)\,d\zeta, \tag{2} \]

where the contour \(C\) may be taken to be the circle \(|\zeta|=r\), \(\sigma<r<\pi\).

From (1) and (2) it follows that there is an \(x_0>0\) such that, for \(0\leqslant x\leqslant x_0\), one can write

\[ f(x)=a_0-\frac{1}{2\pi i}\int_C \psi(\zeta)\,\frac{d\zeta}{1+xe^\zeta}. \tag{3} \]

For fixed \(x>0\), the poles of the expression \((1-xe^\zeta)^{-1}\) will be
\(\zeta=-\ln x+(2k+1)\pi i\) \((k=0,\pm1,\pm2,\ldots)\). They all lie on a straight line perpendicular to the real axis, and as \(x\) increases from \(0\) to \(\infty\), the abscissa of these poles will change from \(\infty\) to \(-\infty\). In doing so, the poles move parallel to the real axis and on their path do not meet the contour \(C\). It follows that the right-hand side of (3) is a regular function of \(x\) not only for \(0\leqslant x\leqslant x_0\), but also on the entire half-axis \(x\geqslant 0\). Directly from (3) it follows that \(f(x)\to a_0\) as \(x\to\infty\).

In this theorem the condition \(\sigma<\pi\) cannot be replaced by a weaker one, since for the function \(a_s=\frac{1}{s}\sin\pi s\) \((\sigma=\pi)\) the assertion of the theorem is already false. But low growth is not in itself a necessary condition for the validity of property A. Thus, for example, the function \(1/\Gamma(s+1)\) obviously has this property, and yet its type \(\sigma=\infty\). In view of the special role of the factor \(1/\Gamma(s+1)\) in the composition of the coefficients of a power series, we give one theorem whose proof is analogous to the preceding one.

Theorem 3. Let \(a_s\) be an entire function of order \(1\) and type \(\sigma<\pi/2\). Then not only \(a_s\), but also \(a_s/\Gamma(s+1)\), has property A.

The class of functions \(a_s\) for which the validity of property A can be asserted admits further extensions. We state without proof the following “multiplication” theorem.

Theorem 4. If \(\varphi(s)\) has property A, then \(a_s=\varphi(s)\mu(s)\) has the same property, where \(\mu(s)\) is either any rational function having no poles in the half-plane \(\operatorname{Re}(s)\geqslant 0\) and at infinity, or the Dirichlet series \(\sum_{k=1}^{\infty} b_k k^{-s}\), absolutely convergent for \(\operatorname{Re}(s)\geqslant 0\), or \(\Gamma(\beta)\Gamma(s+\alpha)/\Gamma(s+\beta)\Gamma(\alpha)\), where \(\operatorname{Re}(\beta)>\operatorname{Re}(\alpha)>0\).

Let us now consider the case where \(a_s\) has poles in the right half-plane. We shall write \(f(x)\sim\varphi(x)\) if \(f(x)=\varphi(x)(1+\eta)\), \(\eta\to 0\) as \(x\to\infty\).

Lemma. Suppose \(0<\operatorname{Re}(p)<1\) or \(p=0\). If \(\psi(x)\) has the properties: 1) in a neighborhood of \(x=0\) it has a convergent expansion of the form \(\psi(x)=b_1x+\cdots\); 2) it admits an analytic continuation to the half-axis \(x\geqslant 0\) and
\[ \psi(x)\sim Ax^p(\ln x)^\alpha, \]
where \(A\) is an arbitrary complex number not equal to zero, and \(\alpha\geqslant 0\) is an integer, then the function
\[ \chi(x)=x^p\int_0^x t^{-1-p}\psi(t)\,dt \]
has the same properties 1), 2), and
\[ \chi(x)\sim Ax^p\frac{(\ln x)^{\alpha+1}}{\alpha+1}. \]

Theorem 5. Let \(a_s\) be a rational function having poles in the half-plane \(\operatorname{Re}(s)>0\), and suppose that none of these poles has an integral abscissa, except possibly the point \(s=0\), which may be a pole. Denote by \(p\) the rightmost pole. Assume that on the line \(\operatorname{Re}(s)=p\) the function \(a_s\) has no other poles.

Then the series (1) has an analytic continuation to the half-axis \(x\geqslant 0\), and
\[ f(x)\sim \begin{cases} c_{-m}\dfrac{(\ln x)^m}{m!}, & \text{in the case } p=0,\\[1.2em] c_{-m}\dfrac{\pi}{\sin p\pi}\dfrac{(\ln x)^{m-1}}{(m-1)!}\,x^p, & \text{in the case } \operatorname{Re}(p)>0, \end{cases} \tag{4} \]
where \(m\) is the multiplicity of the pole \(s=p\), and \(c_{-m}\) is the coefficient of \((s-p)^{-m}\) in the meromorphic part of \(a_s\) at this pole.

Proof. First consider the case \(0<\operatorname{Re}(p)<1\). Let
\[ \varphi_k(x)=\frac{x}{(1-p)^k}-\frac{x^2}{(2-p)^k}+\cdots,\qquad k=0,1,2,\ldots, \]
a sequence of series with common radius of convergence \(R=1\).

For \(0\leqslant x<1\),
\[ \varphi_{k+1}(x)=x^p\int_0^x \frac{\varphi_k(t)}{t^{p+1}}\,dt,\qquad k=0,1,2,\ldots. \]

For \(k=0\),
\[ \varphi_1(x)=x^p\int_0^x \frac{dt}{t^p(1+t)}\sim \frac{\pi}{\sin p\pi}\,x^p. \]

Using the lemma, by induction we prove the assertion of the theorem for any integer \(m\geqslant 1\). Passing from the first strip \(0<\operatorname{Re}(p)<1\) to the second \(1<\operatorname{Re}(p)<2\) is possible by considering \(b_s=a_{s+1}\), and so on. The case of the pole \(s=0\) follows directly from the lemma.

It turns out that the assertion of Theorem 5 is valid not only for rational functions \(a_s\). Thus, for example, the following holds:

Theorem 6. The asymptotic formulas (4) are valid for every function \(a_s\) of the form \(a_s=r(s)\varphi(s)\), where \(r(s)\) is a rational function satisfying the condition of Theorem 5, and \(\varphi(s)\) belongs to one of the following types: 1) \(1/\Gamma(s+1)\); 2) an entire function of first order of type \(\sigma<\pi\); 3) the sum of a Dirichlet series absolutely convergent in the half-plane \(\operatorname{Re}(s)\geqslant 0\); 4) \(\Gamma(\beta)\Gamma(s+\alpha)/\Gamma(s+\beta)\Gamma(\alpha)\), where \(\operatorname{Re}(\beta)>\operatorname{Re}(\alpha)>0\) (\(p\) and \(c_{-m}\) with respect to \(a_s\) are understood in the same sense as in Theorem 5).

The proof is omitted for lack of space.

Ivanovo State
Pedagogical Institute

Received
10 II 1958