water wave part 9

the final part of Stationary phase

Method of steepest descents[1]

Method of steepest descents is applied to the approximate evaluation of integrals of the form \[ I=\int_A^B \chi(z) e^{t f(z)} d z \quad(1) \] where \(t\) is large, real and positive, and \(f(z)\) is analytic. We write \[ f(z)=\phi+i \psi \quad (2) \] separating its real and imaginary parts. \(\phi\) and \(\psi\) both satisfy Laplace's equation, and the integrand will be large where \(\phi\) is algebraically large. The transformation from \(x, y\) to \(\phi, \psi\) will be non-singular in a region containing no singularities or zeros of \(f^{\prime}(z)\). In such a region we can pass from \(A\) to \(B\) by a finite number of steps along lines of \(\phi\) or \(\psi\) constant. Put \(f(z)=\zeta\). Then \[ I=\int_{z=A}^B e^{t \zeta} \frac{\chi(z)}{f^{\prime}(z)} d \zeta=\int_{z=A}^B e^{t \zeta} g(\zeta) d \zeta \quad(3) \]

Suppose first that \(g(\zeta)\) is analytic in the region. Then \(g(\zeta)\) has a bounded derivative in the region and its real and imaginary parts separately will be of bounded variation on a finite path of \(\phi\) or \(\psi\) constant. We can then apply the inequalities derived for integrals in Abel's Lemma . If the path from \(A\) to \(B\) is one of \(\psi\) constant, and \(\phi_A>\phi_B\), the path from \(A\) to \(B\) is called one of steepest descent. Then \[ I=\int_{z=A}^B e^{t \zeta} e^{-t\left(\phi_{A}-\phi\right)} g(\zeta) d \phi \quad(4) \]

If the path is one of \(\phi\) constant \[ \begin{gathered} I=i \int_{z=A}^B e^{t \zeta_A} e^{-t\left(\psi-\psi_{A}\right)} g(\zeta) d \psi \quad(5)\\ \left|e^{-t \zeta_{A}} I\right|<\frac{2 \sqrt{2}\{|g(B)|+V(B)\}}{t}\quad(6) \end{gathered} \] where \(V(B)\) is the greater of the total variations of \(\Re g(\zeta)\) and \(\Im g(\zeta)\) on the path. In either case \[ I=O\left(e^{t \phi_A} / t\right)\quad(7) \] subject to \(\phi_A \geqslant \phi_B\). For infinite paths it is necessary to verify directly whether \(V(B)\) is finite. Since \(\phi_A-\phi\) is a real variable and \(g(\zeta)\) is analytic, we find for (4) by Watson's lemma, \[ I \sim-e^{t \zeta_A}\left\{\frac{g(A)}{t}+\sum_1^{\infty} \frac{(-1)^r}{t^{r+1}}\left(\frac{\partial^r}{\partial \zeta^r} g(\zeta)\right)_{z=A}\right\} \quad(8) \] Note

we have \(\zeta=\phi+i \psi\), as \(\psi=\) constant \(=\psi_A\) in path \(A \rightarrow B\) \[ \begin{aligned} I & =\int_{z=A}^B e^{t \zeta} g(\zeta) d \zeta \\ & =\int_{z=A}^B e^{t\left(\phi+i \psi_A\right)} e^{t\left(\phi_A+i \psi_A\right)} e^{-t\left(\phi_A+i \psi_A\right)} g(\zeta) d\left(\phi_{+i} \psi_A\right) \\ & =\int_{z=A}^B e^{t \zeta_A} e^{-t\left(\phi_A-\phi\right)} g(\zeta) d \phi \end{aligned} \] where \(\zeta_A=\phi_A+i \psi_A\) then we have equation (1).

Similarly. if the path is one of \(\phi\) constant \[ I=i \int_{z=A}^B e^{t \zeta_A} e^{i t\left(\psi-\psi_A\right)} g(\zeta) d \psi \] we hare \(g(\zeta)=\frac{x(z)}{f^{\prime}(z)} \quad \phi, \psi\) contain no singularities or zeros of \(f^{\prime}(z)\) in the interval. it follow that \(\phi\) and \(\psi\) are continuous, \(\psi^{\prime}(z) \neq 0\) and \(\phi^{\prime}(z) \neq 0\) as well. then \(\psi(z)\) and \(\phi(z)\) are monotone functions, it says \(\phi_A>\phi_B\), then \(\phi(z)\) is monotone decreasing, which implies \(\phi(z)\) is non-increasing and non-negative (from \(\phi\) is a gebraically large) which is follow the Abel's Lemma.

Based on previous analysis, \(e^{t\left(\phi-\phi_A\right)}\) is monotone decreasing and \(\psi_A=\psi_B=\) constant, \(t\) is large and the limits of integration \(A, B\) correspond to \(\phi(A), \phi(B)\). \[ \begin{aligned} & \left|e^{t \zeta} I\right|=\left|\int_A^B e^{t\left(\phi-\phi_A\right)} g\left(\phi_{+} i \psi_A\right) d \phi\right| \\ & \left.=\left|\frac{e^{t\left(\phi-\phi_A\right)}}{t} g\left(\phi+i \psi_A\right)\right|_A^B-\int_A^B \frac{e^{t\left(\phi-\phi_A\right)}}{t} g^{\prime}\left(\phi_{+} i \psi_A\right) d \phi \right\rvert\, \\ &\begin{aligned} =\left|\frac{e^{t\left(\phi_B-\phi_A\right)}}{t} g\left(\phi_B+i \psi_B\right)\right.&-\frac{e^{t\left(\phi_A-\phi_A\right)}}{t} g\left(\phi_A+i \psi_A\right)\\ -&\left.\int_A^B \frac{e^{t\left(\phi-\phi_A\right)}}{t} g^{\prime}\left(\phi+i \psi_A\right) d \phi\right| \end{aligned} \end{aligned} \] as \(t\) is large and the first term above \(\sim \frac{1}{t e^t}\) which is much smaller than \(1 / t\) then it becomes (note that: \(g(A)\rightarrow g\left(\zeta_A\right)\)) \[ \begin{aligned} & \left|-\frac{g(A)}{t}-\int_A^B \frac{e^{t\left(\phi-\phi_A\right)}}{t} g^{\prime}\left(\phi+i \psi_A\right) d \phi\right| \\ & \leq \frac{1}{t}|g(A)|+\left|\int_A^B \frac{e^{t\left(\phi-\phi_A\right)}}{t} g^{\prime}\left(\phi+i \psi_A\right) d \phi\right|\\ &\leqslant \frac{1}{t}|g(A)|+\int_A^B \frac{e^{t\left(\phi-\phi_A\right)}}{t} \left| g^{\prime}\left(\phi+i \psi_A\right)\right| d \phi \end{aligned} \] Let's consider the second term individually, It's straightforward to obtain \(e^{t(\phi-\phi_A)}/t\) is monotone decreasing with respect to \(z\) then from the equation (3) we have \[ \int_A^B \frac{e^{t\left(\phi-\phi_A\right)}}{t}\left|g^{\prime}\left(\phi+i \psi_A\right)\right| d \phi \leqslant \frac{e^{t\left(\phi_A-\phi_A\right)}}{t} \int_A^B\left|g^{\prime}\left(\phi+i \psi_A\right)\right| d \phi \] Since \(V_A^x(q)=\int_A^x\left|g^{\prime}\left(\phi+i \psi_A\right)\right| d \phi\) then it's readily to obtain \[ \begin{aligned} \int_A^B \frac{e^{t\left(\phi-\phi_A\right)}}{t}\left|g^{\prime}\left(\phi+i \psi_A\right)\right| d \phi & <\frac{1}{t} \int_A^x\left|g^{\prime}\left(\phi+i \psi_A\right)\right| d \phi \\ & <\frac{1}{t} V_A^B(g)\rightarrow \left(\frac{1}{t} V(B)\right) \end{aligned} \]

Ultimately, we arrive at the form of \(\left|e^{-t \zeta_A} I\right|\) as \(\psi\) is constant \[ \left|e^{-t\zeta_A} I\right|=\frac{1}{t}[|g(A)|+V(B)] \]

Similarly. for \(\phi=\) constant \[ \begin{aligned} & \left|e^{-t \xi_A} I\right|=\left|\int_{z=A}^B e^{i t\left(\psi-\psi_A\right)} g\left(\phi_A+i \psi\right) d \psi\right| \\ & \leq \int_{z=A}^B\left|e^{i t\left(\psi-\psi_A\right)}\right|\left|g\left(\phi_A+i \psi\right)\right| d \psi\\ &\leqslant \int_{z= A}^B\left|g\left(\phi_A+i \psi\right)\right| d \psi \end{aligned} \] ... That appears to be incorrect. (；´д｀)ゞ

the method of stationary phase as\(\varphi^{\prime\prime}(x)=0\)[2]

consider the equations as following \[ I(k)=\int_a^b \psi(x) e^{\imath k \varphi(x)} d x \quad(16) \] Suppose now that \(\varphi(x)\) has one stationary value at \(x=\alpha\) in the segment \(\alpha-\varepsilon_1 \leq x \leq \alpha+\varepsilon_1, \varepsilon_1>0\), i.e., \(\varphi^{\prime}(x)\) vanishes only at \(x=\alpha\) in this interval. Suppose, in addition, that the third derivative \(\varphi^{\prime \prime}(x)\) does not vanish at \(x=\alpha\). We shall show that a positive number \(\varepsilon \leq \varepsilon_1\) exists such that

\[ I(k)=\int_{\alpha-\varepsilon}^{\alpha+\varepsilon} \psi(x) e^{i k p(x)} d x\quad(17) \] we begin by introducing new variables as follows: \[ \quad x=\alpha+u, \quad \varphi(x)=\varphi(\alpha)+w(u)\quad(18) \]

where \(\varphi^{\prime}(a)=\varphi^{\prime \prime}(a)=0, \varphi^{\prime \prime \prime}(a)>0, a\) is the stationary point. Consider first the integral \(I\left(k, \varepsilon_1\right)\) : \[ I\left(k, \varepsilon_1\right)=e^{i k \varphi(\alpha)} \int_{-\varepsilon_1}^{\varepsilon_1} e^{i k w(u)} \psi(\alpha+u) d u=e^{2 k \varphi(\alpha)} J \quad(19) \]

To solve this. here we introduce a new integration variable \(t\), defined as follow \[ \begin{aligned} t^3&=w(u), \quad-\varepsilon_1 \leq u \leq \varepsilon_1\\ w(u)&=\varphi(a+u)-\varphi(a)=a_3 u^3+a_4 u^4+\cdots \end{aligned}\quad(20) \] Note

A:taylor expand of \(h(t)\) if \(h^{\prime}\left(t_0\right)=h^{\prime \prime}\left(t_0\right)=0, h^{\prime \prime \prime}\left(t_0\right)>0\), we have \[ h(t)=h\left(t_0\right)+\frac{1}{6} h^{\prime \prime \prime}\left(t_0\right)\left(t-t_0\right)^3+\cdots\quad(21) \] we define a new variable \(s\) \[ \begin{aligned} & h(t)-h\left(t_0\right)=s^3 \\ & \Rightarrow s^3=\frac{1}{6} h^{\prime \prime \prime}\left(t_0\right)\left(t-t_0\right)^3+O\left(s^4\right) \\ &\begin{aligned} t-t_0&=\left[\frac{6 s^3}{h^{\prime \prime \prime}\left(t_0\right)}\right]^{\frac{1}{3}}+O\left(s^4\right)\\ & =\left[\frac{6}{h^{\prime \prime \prime}\left(t_0\right)}\right]^{\frac{1}{3}} s+O\left(s^2\right) \end{aligned}\\ \end{aligned}\quad(22) \] Q:why do we start from $u^3$
A:method 1

we have \[ \small w(0)=\varphi(a)=w^{\prime}(0)=\varphi^{\prime}(a)=w^{\prime}(0)=\varphi^{\prime \prime}(a), w^{\prime \prime \prime}(0)=\varphi^{\prime \prime \prime}(a) \neq 0\quad(23) \]

Then \(a_0=a_1=a_2=0, w(u)=a_3 u^3+a_4 u^4+\cdots=t^3\) Subsequently. the determination of \(a_3\) proceeds, take derivate with respect to \(t\) in both sides \[ w= \sum_{n=3}^{\infty} a_n u^n\left(=t^3\right)\quad(24) \] then we have \[ \begin{aligned} \frac{d^3 w}{d t^3}= & u^{\prime \prime \prime} \sum_{n=3}^{\infty} n a_n u^{n-1}+u^{\prime \prime} \sum_{n=3}^{\infty} n(n-1) a_n u^{n-2} \\ &\begin{aligned}+u^{\prime \prime} \sum_{n=3}^{\infty} n(n-1) a_n u^{n-2}+u^{\prime} \sum_{n=3}^{\infty} n(n-1)(n-2) &a_n u^{n-3} \\ & (=6) \end{aligned}\\ \end{aligned}\quad(25) \] given \(t=0, u=0\) then \[ 6 u^{\prime} a_3=6\quad(26) \] hence \(a_{3} \neq 0\)

Based on equation (24) \[ t=u \sqrt[3]{a_3+a_4 u+\cdots}\quad(27) \] Let \(\xi(u)=\sqrt[3]{a_3+a_4 u+\cdots}\), take deviate with respect to \(t\) in both sides \(\left(u^{\prime}=\frac{d u}{d t}\right)\) \[ 1=u^{\prime} \xi(u)+\xi_t(u) u\quad(28) \]

Since \(t=0, u=0\) therefore \(u^{\prime}(0) \sqrt[3]{a_3} \neq 0 \rightarrow u^{\prime}(0) \neq 0, a_3 \neq 0\), finally we get the invert series to \(u\) as power series of \(t\).

since \(t^3=w(u)\) we may write \[ t=u\left(a_3+a_4 u+\cdots\right)^{\frac{1}{3}}\quad(29) \] Since \(a_3 \neq 0\), we may express the inverts the series to obtain \(u\) as power series in \(t\) as follows: \[ u=c_1 t+c_2 t^2+c_3 t^3+\cdots\quad(30) \] Note

A:method 1

By comparing equation (20) (21) (22), it readily to obtain that \(t\) in equation (21) correspond to \(s\) in equation (20) and \(u\) in equation (20) correspond to \(t\) in equation (22), consequently \[ u=\left[\frac{6}{\varphi^{\prime \prime \prime}(a)}\right]^{\frac{1}{3}} t+O\left(t^2\right)\quad(31) \]

since \(u=0\), as \(t=0\) then \(c_0=0\), back to equation \[ \varphi(a+u)-\varphi(a)=\omega(u)=t^3\quad(32) \]

taking the third derivative With respect to \(t\) in this equation in both sides \[ \begin{aligned} & \frac{d^2}{d t^2}\left(\frac{d \varphi(a+u)}{d u} \frac{d u}{d t}\right)=\frac{d^2}{d t^2}\left(3 t^2\right) \\ & \frac{d}{d t^2}\left[\varphi^{\prime}\left(c_1+2 c_2 t+3 c_3 t^2+\cdots\right)\right]=\frac{d^2}{d t^2}\left(3 t^2\right) \\ & \small \frac{d}{d t}\left[\frac{d \varphi^{\prime}}{d u} \frac{d u}{d t}\left(c_1+2 c_2 t+3 c_3 t^2+\cdots\right)+\varphi^{\prime}\left(2 c_2+6 c_3 t+\cdots\right)\right]=\frac{d}{d t} 6 t \\ &\begin{aligned}\frac{d}{d t}\left[\varphi^{\prime \prime}\right.\left(c_1+2 c_2 t+3 c_2 t^2\right.&\left.+\cdots\right)\left(c_1+2 c_2 t+3 c_3 t^2+\cdots\right)\\ & \left.+\varphi^{\prime}\left(2 c_2+6 c_3t+\cdots\right)\right]=\frac{d}{d t} 6 t \\ \end{aligned}\\ & \frac{d}{d t}\left[\varphi^{\prime \prime}\left[c_1^2+4 c_1 c_2 t+\cdots\right)+\varphi^{\prime}\left(2 c_2+6 c_1 t+\cdots\right)\right]=6 \\ &\begin{aligned} \frac{d \varphi^{\prime \prime}}{d u} \frac{d u}{d t}&\left(c_1^2+4 c_1 c_2 t+\cdots\right)+\varphi^{\prime \prime}\left(4 c_1 c_2+\cdots\right) \\ & +\frac{d \varphi^{\prime}}{d u} \frac{d u}{d t}\left(2 c_2+6 c_3 t+\cdots\right)+\varphi^{\prime}\left(6 c_3+\cdots\right)=6 \\ \end{aligned} \end{aligned}\quad(33) \] as \(\varphi=a \quad \varphi^{\prime \prime}(a)= \varphi^{\prime}(a)=\varphi(a)=0\) \[ \begin{aligned} & \rightarrow \varphi^{\prime \prime \prime}(a)\left(c_1^2+4 c_1 c_2 t+\cdots\right)\left(c_1+2 c_1 t+\cdots\right)=6 \\ & \rightarrow \varphi^{\prime \prime \prime}(a) c_1^3+\sum_1^{\infty} l_r t^n=6 \end{aligned}\quad(34) \] then \(c_1^3 \varphi^{\prime \prime \prime}(a)=6\) as \(t=0\).

Therefore: \[ c_1=\left(\frac{b}{\varphi^{\prime \prime \prime}(a)}\right)^{\frac{1}{3}} \quad(\neq 0)\quad(35) \] end

Bank to the integral \((19)\) \[ I=e^{i k \varphi(a)} \int_{t_2}^{t_1} e^{i k t^3} \psi(a+u) \frac{d u}{d t} d t\quad(36) \] where \[ \begin{aligned} &t_1=\left[w\left(u_1\right)\right]^{\frac{1}{3}}=\left[w\left(\varepsilon_1\right)\right]^{\frac{1}{3}}\\ &t_2=\left[w\left(u_2\right)\right]^{\frac{1}{3}}=\left[w\left(-\varepsilon_1\right)\right]^{\frac{1}{3}} \end{aligned}\quad(37) \]

It follows that \[ I=e^{i k \varphi(a)}\left[\int_0^{t_1} e^{i k t^3} \psi(a+u) \frac{d u}{d t} d t+\int_{t_2}^0 e^{i k t^3} \psi(a+u) \frac{d u}{d t} d t\right]\quad(38) \]

Given that \(\psi(x)\) is convergent power series and \(x \in(a-\delta, a+\delta)\), hence \(\psi(a+u)\) may be written as (taylor series) \[ \begin{aligned} \psi(a+u) & =\psi(a)+\psi^{\prime}(a) u^{\prime}+\frac{\psi^{\prime \prime}(a)}{2} u^2+\cdots \\ & =\psi(a)+\lambda_1 u^{\prime}+\lambda_2 u^2+\cdots \\ & =\psi(a)+\sum_{n=1}^{\infty} \lambda_n u^n \end{aligned}\quad(39) \]

Then back to equation (36)

\[ \begin{aligned} & \psi(a+u) \frac{d u}{d t} \\ = & \left(\psi(a)+\sum_{n=1}^{\infty} \lambda_n u^n\right)\left(c_1+\sum_{n=1}^{\infty} \beta_n t^n\right) \\ = & \left\{\psi(a)+\sum_{n=1}^{\infty} \lambda_n\left(\sum_{n=1}^{\infty} c_n t^n\right)^n\right\}\left(c_1+\sum_{n=1}^{\infty} \beta_n t^n\right) \\ = & c_1 \psi(a)+c_1 \sum_{n=1}^{\infty} \lambda_n\left(\sum_{n=1}^{\infty} c_n t^n\right)^n+\psi(a) \sum_{n=1}^{\infty} \beta_n t^n \\ & \quad+\sum_{n=1}^{\infty} \lambda_n\left(\sum_{n=1}^{\infty} c_n t^n\right)^n \sum_{n=1}^{\infty} \beta_n t^n\\ =&c_1 \psi(a)+P_1(t) \end{aligned} \quad(40) \]

where \(P_1(t)\) is convergent power series and \(P_1(t)=\sum_{n=1}^{\infty} \eta_n t^n\)

the analysis of the \(\varphi(x)\) in neighborhood of \(a\) , \(\varphi(a)=\varphi^{\prime}(a)=\varphi^{\prime \prime}(a)=0\) and \(\varphi^{\prime \prime \prime}(a) \neq 0\). firstly, we assume \(\varphi^{\prime \prime \prime}(a)>0\), (the case \(\varphi^{\prime \prime \prime}(a)<0\) will discuss later) then \[ x \rightarrow a\left\{\begin{array}{l} \varphi^{\prime \prime \prime}\left(a^{-}\right)>0 \text { and } \varphi^{\prime \prime \prime}\left(a^{+}\right)>0 \\ \varphi^{\prime \prime}\left(a^{-}\right)<0 \text { and } \varphi^{\prime \prime}\left(a^{+}\right)>0 \\ \varphi^{\prime}\left(a^{-}\right)>0 \text { and } \varphi^{\prime}\left(a^{+}\right)>0 \\ \varphi\left(a^{-}\right)<0 \text { and } \varphi\left(a^{+}\right)>0 \end{array}\right.\quad(41) \] subsequently, determine the sign of \(t_2\) \[ t_2=\left[w\left(-\varepsilon_1\right)\right]^{\frac{1}{3}}=\left[w\left(u_2\right)\right]^{\frac{1}{3}}=\left(\varphi\left(a-\varepsilon_1\right)-\varphi(a)\right)^{\frac{1}{3}}<0\quad(42) \]

Let's rewrite equation \((36)\) \[ I=e^{i k \varphi(a)}\left[\int_0^{t_1} e^{i k t^3} \psi(a+u) \frac{d u}{d t} d t+\int_{t_2}^0 e^{i k t^3} \psi(a+u) \frac{d u}{d t} d t\right]\quad(43) \]

put \[ I=e^{i k \varphi(a)}\left(J_1+J_2\right) \quad(44) \] Substitute (40) int (44) and consider \(J_1\) firstly \[ \begin{aligned} & J_1=\int_0^{t_1} e^{i k t^3}\left[c_1 \psi(a)+P_1(t)\right] d t \\ & =c_1 \psi_1(a)\int_0 ^{t_1} e^{i k t^3} d t+\int_0^{t_1} e^{i k t^3} p_1(t) d t \\ & =J_1^1+J_1^2 \end{aligned}\quad(45) \] for \(J_1^1\) Let \(k t^3=\theta \quad d t=\frac{1}{3} \theta^{-\frac{2}{3}} k^{-\frac{1}{3}} d \theta \quad k t_1^3=\theta_1\) \[ \begin{aligned} J_1^{1} & =c_1 \psi(a) \int_0^{\theta_1} e^{i \theta} \frac{1}{3} \theta^{-\frac{2}{3}} k^{-\frac{1}{3}} d \theta \\ & =\frac{1}{3} k^{-\frac{1}{3}}c_1\psi(a) \int_0^{\theta_1} e^{i \theta} \theta^{-\frac{2}{3}} d \theta \end{aligned}\quad(46) \] ref From Asymptotic and special functions Lemma 12.1 [2] \[ \int_0^{\infty} e^{i x^\nu} \nu^{a-1} d \nu=\frac{e^{\frac{\pi}{2} a i} \Gamma(a)}{x^a}\quad(47) \]

Based on equation \((46)\) we have \[ J_1^{1}=\frac{1}{3} k^{-\frac{1}{3}} c_1 \psi(a)\left[\int_0^{\infty} e^{i \theta} \theta^{-\frac{2}{3}} d \theta-\int_{\theta_1}^{\infty} e^{i \theta} \theta^{-\frac{2}{3}} d \theta\right] \quad(48) \] consider the last term of this equation \((48)\), consider the last term of this equation \((48)\) \[ \begin{aligned} & \int_{\theta_1}^{\infty} e^{i \theta} \theta^{-\frac{2}{3}} d \theta, \quad \theta_1=k t_1^3 \\ = & \left.\frac{e^{i \theta}}{2} \theta^{-\frac{2}{3}}\right|_{\theta_1} ^{\infty}+\int_{\theta_1}^{\infty} \frac{e^{i \theta}}{i}\left(\frac{2}{3} \theta^{-\frac{5}{3}}\right) d \theta \\ = & i e^{i \theta} \theta_1^{-\frac{2}{3}}+\int_{\theta_1}^{\infty} \frac{e^{i \theta}}{i}\left(\frac{2}{3} \theta^{-\frac{5}{3}}\right) d \theta\\ \leq& \theta_1^{-\frac{2}{3}}+\int_{\theta_1}^{\infty}\left|\frac{2}{3} \theta^{-\frac{5}{3}}\right| d \theta \\ \leq & \theta_1^{-\frac{2}{3}}-\theta^{-\frac{2}{3}}\left|^{\infty}_{\theta_1}\right. \\ \leq& 2 \theta_1^{-\frac{2}{3}}\left\{\rightarrow O\left(\theta_1^{-\frac{2}{3}}\right)\right\} \end{aligned}\quad(49) \] Also consider the first term of this equation \((48)\) \[ \int_0^{\infty} e^{i \theta} \theta^{-\frac{2}{3}} d \theta=e^{\frac{\pi}{6} i} \Gamma\left(\frac{1}{3}\right)\quad(50) \] combine the two parts. It becomes \[ \begin{aligned} J_1^1 & =\frac{1}{3} k^{-\frac{1}{3}}c_1 \psi(a)\left[e^{\frac{\pi}{6} i} \Gamma\left(\frac{1}{3}\right)+O\left(\theta_1^{-\frac{2}{3}}\right)\right] \\ & =\frac{1}{3} k^{-\frac{1}{3}}\left(\frac{6}{\varphi^{\prime \prime \prime}(a)}\right)^{\frac{1}{3}} \psi(a)\left[e^{\frac{\pi}{6} i} \Gamma\left(\frac{1}{3}\right)+O\left(k^{-\frac{2}{3}}\right)\right] \\ & =\psi(a) \frac{\Gamma\left(\frac{1}{3}\right)}{3}\left(\frac{6}{k \varphi^{\prime \prime \prime}(a)}\right)^{\frac{1}{3}} e^{\frac{\pi}{6} i}+O\left(\frac{1}{k}\right) \end{aligned}\quad(51) \] Let's return to equation \((46)\) , consider the second terms \[ \begin{aligned} & J_1^2=\int_0^{t_1} e^{i k t^3} P_1(t) d t \\ & =\int_0^{t_1} e^{i k t^3} \sum_{n=1}^{\infty} \eta_n t^n d t \\ & =\int_0^{\zeta_1} e^{i k \zeta}\left[\sum_{n=1}^{\infty} \eta_n\left(\zeta^{\frac{1}{3}}\right)^n\right] \frac{1}{3} \zeta^{-\frac{2}{3}} d \zeta \\ & =\frac{1}{3} \int_0^{\zeta_1} e^{i k \zeta} \sum_{n=1}^{\infty} \eta_n \zeta^{\frac{n-2}{3}} d \zeta \end{aligned}\quad(52) \] The laurent expansion \(\sum_{n=1}^{\infty} \eta_n \zeta^{\frac{n-2}{3}}\) has a pole of order at \(\zeta=0\). To avoid the appearance of regularities in the integrand during integration by parts. The first four terms of series are split, Then equation \((52)\) becomes \[ \begin{aligned} =\frac{1}{3}&\left\{\int_0^{\zeta_1} e^{i k \zeta} \eta_1 \zeta^{-\frac{1}{3}} d \zeta+\int_0^{\zeta_1} e^{i k \zeta} \eta_2 d \zeta \right. \\ &+\int_0^{\zeta_1} e^{i k \zeta} \eta_3 \zeta^{\frac{1}{3}} d \zeta+\int_0^{\zeta_1} e^{i k \zeta} \eta_4 \zeta^{\frac{2}{3}} d \zeta \\ &\left.+\int_0^{\zeta_1} e^{i k \zeta} \sum_{n=5}^{\infty} \eta_n \zeta^{\frac{n-2}{3}} d \zeta\right\} \end{aligned}\quad(53) \]

Recall Lemma 12.1 and Lemma 12.2 in Asymptotic and special function [3], it follow that the first term of (53) becomes

\[ \begin{aligned} \frac{1}{3} \int_0^{\zeta_1} e^{i k \zeta} \eta_1 \zeta^{-\frac{1}{3}} d \zeta& =\frac{\eta_1}{3} \frac{e^{\frac{1}{3} \pi i} \Gamma\left(\frac{2}{3}\right)}{k^{\frac{2}{3}}}-O\left(\frac{1}{k}\right) \\ & =O\left(k^{-\frac{2}{3}}\right) \\ \end{aligned}\quad(54) \]

and the second terms becomes

\[ \begin{aligned} \frac{1}{3} \int_0^{\xi_1} e^{i k\zeta} \eta_2 d \zeta & =\frac{1}{3}\left[\frac{e^{i k \zeta}}{i k} \eta_2\right]_0^{\xi_1} \\ & =\frac{1}{3}\left(\frac{e^{i k \xi_1}}{i k} \eta_2-\frac{1}{i k} \eta_2\right) \\ & =O\left(k^{-1}\right) \end{aligned}\quad(55) \]

The third terms becomes

\[ \begin{aligned} \frac{1}{3} \int_0^{\zeta_1} e^{i k \zeta} \eta_3 \zeta^{\frac{1}{3}} d \zeta&=\frac{1}{3}\left[\frac{e^{i k \zeta}}{i k} \eta_3 \zeta^{\frac{1}{3}}\right]_0^{\zeta_1} -\frac{1}{3} \int \frac{e^{i k \zeta}}{i k} \frac{1}{3} \eta_3 \zeta^{-\frac{2}{3}} d \zeta \\ & =\frac{1}{3} \frac{e^{i k \zeta_1}}{i k} \eta_3 \zeta_1-\frac{\eta_3}{9 i k} \frac{e^{\frac{\pi_i}{6}} \Gamma\left(\frac{1}{3}\right)}{k^{\frac{1}{3}}} \\ & =O\left(k^{-1}\right) \end{aligned}\quad(56) \]

The fourth terms becomes

\[ \small\begin{aligned} & \frac{1}{3} \int_0^{\zeta_1} e^{i k} \zeta \eta_4 \zeta^{\frac{2}{3}} d \zeta=\frac{1}{3}\left[\frac{e^{i k \zeta}}{i k} \eta_4 \zeta^{\frac{2}{3}}\right]_0^{\zeta_1} -\frac{1}{3} \int_0^{\zeta_1} \frac{e^{i k \zeta}}{i k} \frac{2}{3} \eta_4 \zeta^{-\frac{1}{3}} d \zeta\\ & =\frac{1}{3} \frac{e^{i k \zeta_1}}{i k} \eta_4 \zeta_1^{\frac{2}{3}}-\frac{2 \eta_4}{9i k} \frac{e^{\frac{1}{3} \pi i}}{k^{\frac{2}{3}}} \Gamma\left(\frac{2}{3}\right) \\ & =O\left(k^{-1}\right) \end{aligned}\quad(57) \]

The last terms become

\[ \begin{aligned} \frac{1}{3} \int_0^{\zeta_1} e^{i k} \sum_{n=5}^{\infty} \eta_n \zeta^{\frac{n-2}{3}} d \zeta & =\frac{1}{3}\left[\frac{e^{i k \zeta}}{i k} \sum_{n=5}^{\infty} \eta_n \zeta^{\frac{n-2}{3}}\right]_0^{\xi_1} \\ & -\frac{1}{3} \int_0^{\zeta_1} \frac{e^{i k \zeta}}{i k} \sum_{n=5}^{\infty} \eta_n \frac{n-2}{3} \zeta^{\frac{n-5}{3}} d \zeta \end{aligned}\quad(58) \] It's easy to see that \(\sum_{n=5}^{\infty} \eta_n \zeta^{\frac{n-2}{3}}\) is still convergent and its convergent radius is \(R^3\), as \(R\) is the convergent radius of \(\sum_{n=1}^{\infty} \eta_n t^n\). similarly the convergent radius of \(\sum_{n=5}^\infty \eta_n \frac{n-2}{3} \zeta^{\frac{n-5}{3}}\) is \(\left(\frac{n-2}{n-1} R\right)^3\) and readily to see

\[ \lim _{n \rightarrow \infty}\left(\frac{n-2}{n-1}\right)^3 R^3=R^3 \]

therefore it can be obtained that two series in Equation \((58)\) are both convergent, then it becomes

\[ \begin{aligned} & \frac{1}{3} \frac{e^{i k \zeta_1}}{i k} \sum_{n=5}^{\infty} \eta_n \zeta_1^{\frac{n-2}{3}}+\frac{1}{9 k} \int_0^{\zeta_1} i\frac{e^{i k \zeta}}{k} \sum_{n=5}^{\infty}(n-2) \zeta^{\frac{n-5}{3}} d \zeta \\ & \leq O\left(k^{-1}\right)+\frac{1}{9 k} \int_0^{\zeta_1} \sum_{n=5}^{\infty}(n-2) \zeta^{\frac{n-5}{3}} d \zeta \\ & \leq O\left(k^{-1}\right) \end{aligned} \]

finally. \(J_1^2\) can be written as

\[ J_1^2=O\left(k^{-\frac{2}{3}}\right) \]

therefore

\[ \begin{aligned} J_1 & =J_1^1+J_1^2 \\ & =\psi(a) \frac{\Gamma\left(\frac{1}{3}\right)}{3}\left(\frac{6}{k \psi^{\prime \prime \prime}(a)}\right)^{\frac{1}{3}} e^{\frac{\pi i}{6} i}+O\left(k^{-\frac{2}{3}}\right) \end{aligned} \]

and

\[ \begin{aligned} I & =e^{i k \psi(a)} J_1 \quad \text { (we will discuss } J_2 \text { further below) } \\ & =\psi(a) \frac{\left[\left(\frac{1}{3}\right)\right.}{3}\left[\frac{6}{k \psi^{\prime \prime \prime}(a)}\right]^{\frac{1}{3}} e^{i k \varphi(a)+\frac{2}{6} i}+O\left(k^{-\frac{2}{3}}\right) \end{aligned} \]

Now let's consider \(J_2\), we can rewrite it as \[ \int_{-t_2}^0 e^{i k} t^3 \psi(a+u) \frac{d u}{d t} d t \quad\text{ where } t_2>0 \] and we already had \[ \begin{aligned} & \psi(a+u) \frac{d u}{d t}=\int_{-t_2}^0 e^{i kt^3}\left(c_1 \psi(a)+P_1(t)\right) d t \\ & \int_{-t_2}^0 e^{i k t^3} c_1 \psi(a) d t+\int_{-t_2}^0 e^{i k t^3} P_1(t) d t \end{aligned} \] Let first terms as \(J_2^1\) the second as \(J_2^2\), to solve \(J_2^1\) let $k t³⁼3 $ \[ \begin{aligned} & \qquad\left\{\begin{array}{l} t=k^{-\frac{1}{3}} \theta, d t=k^{-\frac{1}{3}} d \theta \\ t_2=k^{-\frac{1}{3}} \theta_1, \theta_1 \rightarrow-\infty \\ J_2^{1}=c_1 \psi(a) \int_{-\infty}^0 e^{i \theta^3} k^{-\frac{1}{3}} d \theta \end{array}\right. \end{aligned} \] It follows that \[ J_2^{1}=\frac{c_1 \psi(a)}{k^{\frac{1}{3}}} \int_{-\infty}^0 e^{i \theta^3} d \theta \] To evaluate the integral of \(J_2^{1}\), we consider the new one \(\int_c e^{-z^3} d z\) where \(c\) is the closed contour, The integral can be divided into three regions. as show in the fig

Therefore \[ \begin{aligned} \int_c e^{-z^3} d z= & \int_0^R e^{-r^3} d r+\int_0^{-\frac{\pi}{2}} e^{-\left[R e^{-\theta i}\right]^3} d\left[R e^{-\theta_i}\right] \\ & +\int_{-R}^0 e^{-\left(r i\right)^3} d\left(r i\right) \end{aligned} \]

Show: \[ \Gamma\left(\frac{n+1}{n}\right)=\int_0^{\infty} e^{-x^n} d x \quad(n>0) \] A: form the equation \[ \Gamma(x)=\int_0^{\infty} t^{x-1} e^{-t} d t \] then we have \[ \Gamma\left(\frac{n+1}{n}\right)=\int_0^{\infty} t^{\frac{1}{n}} e^{-t} d t \]

Let \(t=y^n \rightarrow t^{\frac{1}{n}}=y\) then \(d t=n y^{n-1} d y\) \[ \begin{aligned} & {\Gamma\left(\frac{n+1}{n}\right)=\int_0^{\infty} y e^{-y^n} n y^{n-1} d y} \\ & =\int_0^{\infty}(-y)\left(-e^{-y^n} n y^{n-1}\right) d y \\ & =\left[e^{-y^n}(-y)\right]_0^{\infty}-\int_0^{\infty} e^{-y^n}(-1) d y \\ & =\int_0^{\infty} e^{-y^n} d y \end{aligned} \] Q.E.D

we have \[ \int_0^{\infty} e^{-r^3} d r=\Gamma\left(\frac{4}{3}\right) \] thus \[ \begin{aligned} i \int_{-\infty}^0 e^{i r^3} d r & =\Gamma\left(\frac{4}{3}\right) \\ \Rightarrow \int_{-\infty}^0 e^{i r^3} d r & =-i \Gamma\left(\frac{4}{3}\right) \end{aligned} \] As for \(J_2^2\), Its value is same as the previous one \[ J_2^2=O\left(k^{-\frac{2}{3}}\right) \text {. } \]