This post mainly discusses analytic functions required for stationary phase method, as well as the asymptotic expansion and related mathematical lemma required in the expansion of analytic function into power series.

the method of stationary phase

Holomorphic function [1]

In mathematics, a holomorphic function is a complex-valued function of one or more complex variables that is complex differentiable in a neighbourhood of each point in a domain in complex coordinate space \(\mathbf{C}^n\). A holomorphic function is infinitely differentiable and locally equal to its own Taylor series (analytic).

The term complex analytic function is often used interchangeably with "holomorphic function", that can be written as a convergent power series in a neighbourhood of each point in its domain. That all holomorphic functions are complex analytic functions, and vice versa, is a major theorem in complex analysis.

Holomorphic functions are also sometimes referred to as regular functions.

In fact more is true: every holomorphic function is analytic, in the sense that it has a power series expansion near every point, and for this reason we also use the term analytic as a synonym for holomorphic.[2]

Asymptotic Expansions [3]


Nature of asymptotic expansions: [3, section 17] \[ I=\int_x^{\infty} e^{-t} t^{-n} d t \quad(1) \] where \(x\) and \(n\) are positive.

Poincare's definition:

The usual definition is that if \(f(z)\) is an analytic function, \(S_n(z)\) is the sum of the terms up to \(A_n z^{-n}\) of the series \[ S(z)=A_0+\frac{A_1}{z}+\ldots+\frac{A_n}{z^n}+\ldots,\quad(2) \] and if \(R_n(z)=f(z)-S_n(z)\), the series is called an asymptotic expansion of \(f(z)\) within a given interval of \(\arg z\) if for every \(n\) \[ \lim _{|z| \rightarrow \infty} z^n R_n(z)=0\quad(3) \]

We write \[ f(z) \sim S(z) \] Properties:

Asymptotic series can be multiplied unconditionally. For if \[ \begin{aligned} & S_n(z)=A_0+\frac{A_1}{z}+\ldots+\frac{A_n}{z^n}, \\ & T_n(z)=B_0+\frac{B_1}{z}+\ldots+\frac{B_n}{z^n}, \end{aligned}\quad(4) \] are asymptotic representations of \(f(z), g(z)\), we can choose \(z\) so that are arbitrarily small. \[ \left|z^n\left\{f(z)-S_n(z)\right\}\right|, \quad\left|z^n\left\{g(z)-T_n(z)\right\}\right|\quad(5) \] Also \[ S_n(z) T_n(z)=C_0+\frac{C_1}{z}+\ldots+\frac{C_n}{z^n}+o\left(z^{-n}\right)=U_n(z)+o\left(z^{-n}\right)\quad(6) \] whrere \[ C_m=A_0 B_m+A_1 B_{m-1}+\ldots+A_m B_0\quad(7) \] and \(z^n\left\{f(z) g(z)-U_n(z)\right\}\) is the sum of three terms that tend to zero with \(1 / z\) for all \(n\).

Asymptotic expansions are unique. For if we have for all \(|z|>R, \alpha \leqslant \arg z \leqslant \beta\) \[ \begin{aligned} &\lim _{z \rightarrow \infty} z^n\left\{f(z)-A_0-\frac{A_1}{z}-\ldots-\frac{A_n}{z^n}\right\}=0,\\ &\lim _{z \rightarrow \infty} z^n\left\{f(z)-B_0-\frac{B_1}{z}-\ldots-\frac{B_n}{z^n}\right\}=0, \end{aligned}\quad(8) \] we should have

\[ \quad \lim _{z \rightarrow \infty} z^n\left\{A_0-B_0+\frac{A_1-B_1}{z}+\ldots+\frac{A_n-B_n}{z^n}\right\}=0\quad(9) \]

and therefore \[ A_0=B_0, \quad A_1=B_1, \ldots, \quad A_n=B_n . \]

Appendix: Method for determining the convergence radius [4, section 4]



\[ \lim _{n \rightarrow \infty}\left|\frac{c_{n+1}}{c_n}\right|=\lambda\quad(10) \]

The power series is convergent as \(\lambda< 1\), and the radius of convergence is \(R=1/\lambda\). E.g.

\[ \lim _{n \rightarrow \infty} \frac{\left|c_{n+1}\right|\left|z-z_0\right|^{n+1}}{\left|c_n\right|\left|z-z_0\right|^n}=\lim _{n \rightarrow \infty}\left|\frac{c_{n+1}}{c_n}\right||z-z_0|\quad(11) \]

Then the power series converges within the circle \(|z-z_0|<R\)


Watson's lemma:

Consider the integral along the real axis \[ I=\int_0^z e^{-a z} z^m f(z) d z\quad(12) \] where \(f(z)\) is analytic on the path and not zero at \(z=0 ; Z\) is independent of \(a\), and may be infinite; \(a\) is large, real and positive; and \(I\) exists for some \(a\), say \(\alpha\). Hence \(m>-1\). Then within the circle of convergence of the series expansion of \(f(z)\) \[ f(z)=a_0+a_1 z+\ldots+a_{n-1} z^{n-1}+R_n(z) \quad(13) \] where \(R_n(z) / z^n\) tends to a finite limit as \(z \rightarrow 0\). Take a fixed \(A\) in \((0, Z)\) within the circle of convergence; then \[ I=\left[\int_0^A+\int_A^Z\right] e^{-a z} z^m f(z) d z \quad(14) \]

In \((0, A)\) the function \[ g(z)=\left\{f(z)-\left(a_0+a_1 z+\ldots+a_{n-1} z^{n-1}\right)\right\} z^{-n}\quad(15) \] is bounded. Let the upper bound of its modulus be \(M\). Then

\[ \begin{aligned} I_{A}= & \int_0^A e^{-a z} z^m f(z) d z \\ = & \int_0^A e^{-a z} z^m\left(a_0+a_1 z+\ldots+a_{n-1} z^{n-1}\right) d z \\ & +\int_0^A e^{-a z} z^m \theta M z^n d z \end{aligned}\quad(16) \] where \(|\theta|<1\).


we have \[ g(z)=\left\{f(z)-\left(a_0+a_1 z+\cdots+a_{n-1} z^{n-1}\right)\right\} z^{-n}\quad(17) \] and \[ |g(z)|<M \] then \(f(z)\) can be written as \[ \begin{aligned} f(z) & =a_0+a_1 z+\cdots+a_{n-1} z^{n-1}+g(z) z^n \\ & \leqslant a_0+a_1 z+\cdots+a_{n-1} z^{n-1}+|M| z^n \end{aligned}\quad(18) \] or we use a new parameter \(\theta\) to determine it. sing \(|M|>\theta M \quad(M\) is a complex number, \(|\theta|<\mid)\) then rewrite it \[ f(z)=a_0+a_1 z+\cdots+a_{n-1} z^{n-1}+\theta M z^n\quad(19) \] end

Put \(z=A(1+u)\); then \((1+u)^{m+r} \leqslant e^{(m+r) u}(m+r>0), \leqslant 1(m+r \leqslant 0)\), \[ \begin{aligned} \int_{A}^{\infty} e^{-a z} z^{m+r} d z&<\int_0^{\infty} A^{m+r+1} e^{-A a-a A u} e^{(m+r) u} d u\\ &<A^{m+r+1} e^{-A a} /(A a-m-r) \quad\text{where } (m+r>0) \\ \int_{A}^{\infty} e^{-a z} z^{m+r} d z&<A^{m+r} e^{-a A} / a \quad\text{where } (m+r \leqslant 0) \end{aligned}\quad(20) \]


consider \(z \in[A, \infty) \quad z=A(1+u)(u>0)\) based on \(e^u \geq u+1\) in \(R\) it's deduced that \(0<(u+1)^{m+r} \leq\left(e^u\right)^{m+r} \quad(m+r>0)\) then \[ 0<(u+1)^{m+r} \leqslant 1 \quad(m+v \leqslant 0) \]

Therefore \[ \int_A^{\infty} e^{-a z} z^{m+r} d z=\int_0^{\infty} e^{-a A(1+u)}[A(1+u)]^{m+r} A d u\quad(21) \] end

\[ I_A=\sum_{r=0}^{n-1} a_r \frac{(m+r) !}{a^{m+r+1}}+O\left(e^{-a A}\right)+\theta \frac{M(m+n) !}{a^{m+n}+1}\quad(22) \]


the proof of equation(22) \[ \begin{aligned} I_A= & \int_0^A e^{-a z} z^m\left(a_0+a_1 z+\cdots+a_{n-1} z^{n-1}\right) d z \\ & +\int_0^A e^{-a z} z^{m+n} \theta M d z \end{aligned}\quad(23) \] rewrite \(I_A=I_A^1+I_A^2\) where \[ \begin{aligned} & I_A^1=\int_0^A e^{-a z} z^m \sum_{r=0}^{n-1} a_r z^r d z \\ & I_A^2=\int_0^A e^{-a z} z^{m+n} \theta M d z \end{aligned}\quad(24) \] we start from \(I_A^1\). It follows that \[ \small\begin{aligned} & I_A^1=\int_0^A e^{-a z}\left(a_0 z^m+a_1 z^{m+1}+\cdots+a_{n-1} z^{m+n-1}\right) d z \\ & =\int_0^A e^{-a z} a_0 z^m d z+\int_0^A a_1 z^{m+1} e^{-a z} d z+\cdots+\int_0^A a_{n-1} z^{m+n-1} e^{-a z} d z \end{aligned}\quad(25) \] calculate an arbitrary term from the above equations it follows that \[ \left.I_A^1\right|_r=\int_0^A a_r z^{m+r} e^{-a z} d z\quad(26) \] integrate it by parts we have \[ \left.I_A^1\right|_r=a_r\left\{\left.\left(\frac{e^{-a z}}{-a} z^{m+r}\right)\right|_0 ^A-\left(-\frac{1}{a}\right)(m+r) \int_0^A e^{-a z} z^{m+r-1} d z\right\}\quad(27) \] Calculation of the first term in the above equation: \[ \left.\frac{e^{-a z}}{-a} z^{m+r}\right|_0 ^A=\frac{e^{-a A}}{-a} A^{m+r}\quad(28) \]

Since \(a\) is a larger number then equation \((28) \rightarrow O\left(e^{-a A}\right)\), repeat performing integration by parts for \(m+r\) times we have \[ \begin{aligned} & I_A^1=a_r\left(\frac{1}{a}\right)^{m+r}(m+r) \cdot(m+r-1) \cdots 1 \int_0^A e^{-a z} d z \\ & =\left.a_r\left(\frac{1}{a}\right)^{m+r}(m+r) !\left(\frac{e^{-a z}}{-a}\right)\right|_0 ^A \\ & =a_r\left(\frac{1}{a}\right)^{m+r}(m+r) !\left[O\left(e^{-a A}\right)+\frac{1}{a}\right] \\ & =a_r \frac{(m+r) !}{a^{m+r+1}}+O\left(e^{-a A}\right) \end{aligned}\quad(29) \] Therefore \[ I_A^1=\sum_{r=1}^{n-1} a_r \frac{(m+r) !}{a^{m+r+1}}+O(e^{-a A})\quad(30) \]

Similarly \[ \begin{aligned} I_A^2 & =\int_0^A e^{-a z} \theta M z^{m+n} d z \\ & =\theta M \frac{(m+n) !}{a^{m+n+1}}+O\left(e^{-a A}\right) \end{aligned}\quad(31) \] The above derivation yields the equation \[ I_A=\sum_{r=0}^{n-1} a_r \frac{(m+r) !}{a^{m+r+1}}+O\left(e^{-a A}\right)+\theta M \frac{(m+n) !}{a^{m+n+1}}\quad(32) \] end

Appendix: Abel's lemma [3]

Abel's lemma

If \(\left\{v_r\right\}\) is a non-increasing sequence of non-negative quantities, and if the sums \[ s_p=a_1+a_2+\ldots a_p\quad(33) \] satisfy the inequalities \(h \leqslant s_p \leqslant H\) for all \(p\), then \(h v_1 \leqslant \sum_1^n a_p v_p \leqslant H v_1\) for all \(n\). We have \[ \begin{aligned} a_1 & =s_1, a_2=s_2-s_1, \ldots, a_n=s_n-s_{n-1} \\ \boldsymbol{S}_n & =\sum_1^n a_p v_p=s_1 v_1+\sum_{p=2}^n\left(s_p-s_{p-1}\right) v_p \\ & =s_1\left(v_1-v_2\right)+\ldots+s_{n-1}\left(v_{n-1}-v_n\right)+s_n v_n . \end{aligned}\quad(34) \] Since all \(v_p-v_{p+1} \geqslant 0\) and \(v_n \geqslant 0\), the last sum will not be decreased if all the \(s_p\) are replaced by \(H\); and therefore \(\boldsymbol{S}_n \leqslant H v_1\). Similarly the sum will not be increased if all the \(s_p\) are replaced by \(h\); hence \(\boldsymbol{S}_n \geqslant h v_1\).

Abel's lemma for integrals

If \(v(x)\) is non-negative, bounded in \(a \leqslant x \leqslant b\) and non-increasing with \(x\), and if \(h, H\) are the lower and upper bounds of \[ F(\xi)=\int_a^{\xi} f(x) d x\quad(35) \] for \(a \leqslant \xi \leqslant b\), then \[ h v(a) \leqslant \int_a^b f(x) v(x) d x \leqslant H v(a)\quad(36) \] Please refer to the equation (33)-(34) for the proof of equation (36), also the following is another way.

put \[ \begin{gathered} I=\int_a^b f(x) v(x) d x=\int_{x=a}^b v(x) d F(x) \\ =v(b) F(b)-\int_{x=a}^b F(x) d v(x) . \end{gathered}\quad(37) \] This is valid because \(F(x)\) is an integral and therefore continuous, and \(v(x)\) is of bounded variation. Then, since \(v(x)\) is nowhere increasing, \(I\) will not be decreased if \(F(x)\) is replaced everywhere by its upper bound, or increased if it is replaced by its lower bound; then \[ h\left\{v(b)-\int_{x=a}^b d v(x)\right\} \leqslant I \leqslant H\left\{v(b)-\int_{x=a}^b d v(x)\right\}\quad(38) \] that is, \[ h v(a) \leqslant I \leqslant H v(a) \quad(39) \]


For \(a=\beta\) let the upper bound of \(\left|\int_A^X e^{-\beta z} z^m f(z) d z\right|\) for \(X\) in \((A, Z)\) be \(N\). Then from Abel's lemma for integrals (since \(e^{-(a-\beta) z}\) is a positive decreasing function of \(z\) ) \[ \left|\int_{A}^Z e^{-a z} z^m f(z) d z\right| \leqslant 2 e^{-(a-\beta) A} N \quad(40) \]


put \[ F(x)=\int_A^x e^{-\beta z} z^m f(z) d z\quad(41) \] and we have \[ |F(x)| \leq N\quad(42) \] put \(v(z)=e^{-(\alpha-\beta) z}\), \(v(z)\) is strictly positive and non-increasing hence \[ \begin{aligned} & \left|\int_A^z e^{-\beta z} z^m f(z) \cdot e^{-(a-\beta) z} d z\right| \\ = & \left|\int_A^z e^{-a z} z^m f(z) d z\right| \\ \leqslant & e^{-(\alpha-\beta) A} N \end{aligned}\quad(43) \] end

and therefore \[ \left|I-\sum_{r=0}^{n-1} a_r \frac{(m+r) !}{a^{m+r+1}}\right|<\frac{M(m+n) !}{a^{m+n+1}}+K e^{-a A}\quad(44) \] where \(K\) is independent of \(a\). If we multiply by \(a^{m+n}\) and make \(a \rightarrow \infty\) the right side tends to 0 . Hence \[ \int_0^Z e^{-a z} z^m f(z) d z \sim \sum_{r=0}^{\infty} \frac{(m+r) ! a_r}{a^{m+r+1}}\quad(45) \]