Asymptotics of an integral by two methods

Question

I want to compute the asymptotic behavior of the integral $$ f(K,a)=\int_0^1 (1-x)^Ke^{iKa\frac{x}{1-x}}x^2dx$$ when $K$ is large and $0<a<1$. I tried two different approaches.

1) My first idea was that the exponential, a fast-oscillating function around $x=1$, is killed by the $(1-x)^K$, and the integral should be dominated by the vicinity of $x=0$. Therefore, I put $x=y/K$ and approximate $(1-y/K)^K\approx e^{-y}$ and $\frac{x}{1-x}\approx \frac{y}{K}$ to get

$$f(K,a)\approx \frac{1}{K^3}\int_0^\infty e^{-y+iay}y^2dy=\frac{2}{K^3(1-ia)^3}.$$

2) On the other hand, the stationary phase approximation should be valid. If I write $$f(K,a)=\int_0^1 e^{KS(x)}x^2dx,$$ with $S(x)=\log(1-x)+iax/(1-x)$, the equation $S'(x_0)=0$ gives $x_0=1-ia$. Second derivative is $S''(x_0)=-1/a^2$. Hence, this idea leads to $$f(K,a)\approx e^{KS(x_0)}x_0^2\sqrt{\frac{\pi a^2}{K}}=(ia)^Ke^{K(1-ia)}(1-ia)^2a\sqrt{\frac{\pi}{K}}.$$

These two results are completely different! I need help understanding this.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\mrm{f}\pars{K,a} \equiv \int_{0}^{1}\pars{1 - x}^{K} \expo{\ic Ka x/\pars{1 - x}}x^{2}\,\dd x\,,\qquad 0 < a < 1\,,\quad K\ \mbox{large}}$.

With $\ds{t \equiv x/\pars{1 - x} \iff x = t/\pars{1 + t}}$ :

\begin{align} \mrm{f}\pars{K,a} &\equiv \int_{0}^{1}\pars{1 - x}^{K}\expo{\ic Ka x/\pars{1 - x}}x^{2}\,\dd x = \int_{0}^{\infty}{t^{2} \over \pars{t + 1}^{K + 4}}\expo{\ic Kat}\,\dd t \\[5mm] & = \overbrace{-\lim_{R \to \infty}\int_{0}^{\pi/2}\!\!\!\!\!{R^{2}\expo{2\ic\theta} \over \pars{R\expo{\ic\theta} + 1}^{K + 4}}\exp\pars{\ic KaR\expo{\ic\theta}} R\expo{\ic\theta}\ic\,\dd\theta}^{\ds{\to\ 0\ \mrm{as}\ R\ \to\ \infty}} \\[2mm] & -\int_{\infty}^{0}{-y^{2} \over \pars{\ic y + 1}^{K + 4}}\,\expo{-Kay}\,\ic \,\dd y = -\ic\int_{0}^{\infty}{y^{2} \over \pars{1 + \ic y}^{K + 4}}\,\expo{-Kay}\dd y \\[5mm] & = -\ic\int_{0}^{\infty}y^{2}\exp\pars{-Kay - \bracks{K + 4}\ln\pars{1 + \ic y}} \,\dd y \end{align}

With Laplace's Method:

\begin{align} \mrm{f}\pars{K,a} & \,\,\,\stackrel{\mrm{as}\ K\ \to\ \infty}{\sim}\,\,\, -\ic\int_{0}^{\infty}y^{2}\exp\pars{-\braces{Ka + \bracks{K + 4}\ic}y}\,\dd y = -\,{2\ic \over \bracks{Ka + \pars{K + 4}\ic}^{\, 3}} \\[5mm] & = {2 \over \bracks{K + 4 - Ka\ic}^{\, 3}} \implies \bbx{\mrm{f}\pars{K,a} \sim {2 \over K^{3}\bracks{1 + 4/K -a\ic}^{3}}\quad\mbox{as}\ K \to \infty} \end{align}

which agrees with your 'first attempt' as $\ds{K \to \infty}$.

I guess you can not make the replacement $\ds{y^2 \mapsto \pars{1 - \ic a}^{2}}$ in your second attempt. If you follow my answer you'll see it's 'somehow' equivalent to the replacement $\ds{\pars{1 + \ic y}^{K + 4} \mapsto 1}$.