Expansion of Integrals. 3


### Oscillatory integrals

We consider $$I(k)= \int_X e^{ik \phi(x)}f(x)\,dx \label{eq-2.3.1}$$ where $X=[a,b]$ and $f,\phi \in C^\infty(X)$, $\phi$ is a real-valued function. We are interested in the asymptotics of $I(k)$ as $k\to +\infty$. We will show that the contribution in delivered by stationary points of $\phi(x)$ and also end-points.

#### No stationary points

First we consider the simples case when $\phi$ has no stationary points at all: $\phi'(x)$ does not vanish on $X$.

Example 1. Let $\phi(x)=\alpha x$, $f(x)=\const$. Then \begin{equation*} I(k)=\int_a^b e^{i\alpha k x}f\, dx= i^{-1}\alpha^{-1}k^{-1} f e^{i\alpha bk} i^{-1}\alpha^{-1}k^{-1} f e^{i\alpha ak} \end{equation*} with both terms of the same magnitude $$\frac{1}{i\phi'(b)} e^{ik\phi (b)}f(b)k^{-1}-\frac{1}{i\phi'(a)} e^{ik\phi (a)}f(a)k^{-1} \label{eq-2.3.2}$$ and we expect that it will be the main part of the asymptotics in the general case.

Now consider the general case. Then integrating by parts we get \begin{multline} I(k)= \int_a^b e^{k \phi(x)}f(x)\,dx= k^{-1}\int_a^b (i\phi'(x))^{-1}\bigl(e^{ik \phi(x)}\bigr)'f(x)\,dx=\\ i^{-1}k^{-1}(\phi'(b))^{-1} e^{ik \phi(b)}f(b)- i^{-1}k^{-1}(\phi'(a))^{-1} e^{ik \phi(a)}f(a)-\\ k^{-1}\int^b e^{k \phi(x)}g(x)\,dx \label{eq-2.3.3} \end{multline} with \begin{equation*} g(x)=\bigl((i\phi'(x))^{-1} f(x)\bigr)'. \end{equation*} The first two terms in the right-hand expression is the guessed main part and the second term is again integral of the same type as $I(k)$ albeit with the different amplitude $g$ and an extra factor $k^{-1}$.

Continuing we can rewrite it in the same way and so on arriving to the following Theorem:

Theorem 1. Let $\phi$ have no stationary points on $X$. Then $$I(k) \sim e^{ik\phi (b)}\sum _{n=0}^\infty \kappa_{n}(b) k^{-1-n}+ e^{ik\phi (a)}\sum _{n=0}^\infty \kappa_{n}(a) k^{-1-n} \label{eq-2.3.4}$$ in the sense that $$|I(k)- I_N(k)|\le C_N k^{-N-1} \label{eq-2.3.5}$$ where $I_N(k)$ is the same sum but with $0\le n\le N-1$. Here the main coefficients are $$\kappa_{0}(x)=\pm \frac{1}{i\phi'(x)} f(x)\qquad\text{as }\ x=b,a \label{eq-2.3.6}$$

#### Single stationary point

Assume now that $\phi(x)$ has a single non-degenerate stationary at $c\in (a,b)$. Then $\phi'(c)=0$ and $\phi''(c)\ne 0$.

We need the following

Theorem 2. Let $\Re \beta \ge 0$. Then $$I:=\int_{-\infty}^\infty e^{-\beta x^2}\,dx= \pi^{\frac{1}{2}}\beta^{-\frac{1}{2}} \label{eq-2.3.7}$$ where $\beta^{\frac{1}{2}}$ is a square root defined on the complex plane with a cut $\bC\setminus (-\infty,0]$.

Proof. Obviously we can calculate integral over $\bR^+$ and then double result. Assume first that $|\beta|=1$. Consider the following contour in $\bC$:

Here $A=R$ and $B=Re^{i\sigma}$ $|\sigma|\le \pi/4$. Integral of $e^{-\beta z^2}\,dz$ over closed contour is $0$ (as we know from complex variables). Integral from $O$ to $A$ is $\int_0^R e^{-\beta x^2}\,dx$ and integral from $B$ to $O$ is $\int_0^R e^{-\beta e^{2i\sigma} x^2}\,e^{i\sigma} dx$.

Integral from $A$ to $B$ is $\int_0^\sigma e^{-\beta R^2 e^{2i\theta}} \,d Re^{i\theta}$. Observe that in this integral $\Re e^{2i\theta}\ge 0$ and integrating once by parts as we did before (using $e^{-\beta R^2 e^{2i\theta}} d Re^{i\theta}= -\frac{1}{2}\beta^{-1}R^{-1} e^{-i\theta} de^{-\beta R^2 e^{2i\theta}}$) and integrating by parts we obtain that this integral does not exceed $CR^{-1}$.

Therefore as $R\to +\infty$ we arrive to $\int_0^\infty e^{-\beta x^2}\,dx = \int_0^\infty e^{-\beta e^{2i\sigma} x^2}\,e^{i\sigma} dx$ which in turn equals $\beta^{-\frac{1}{2}}\int_0^\infty e^{- x^2} dx=\frac{1}{2}\pi^{\frac{1}{2}}\beta^{-\frac{1}{2}}$ as $e^{i\sigma}=\beta^{-\frac{1}{2}}$.

So, for $|\beta|=1$ (\ref{eq-2.3.7}) has been proven. The general case is reduced to this by substitution $x:= |\beta|^{-\frac{1}{2}}y$ (check it!).

Theorem 3. Let $\pm \alpha >0$. Then $$I:=\int_{-\infty}^\infty e^{i\alpha x^2}\,dx= \pi^{\frac{1}{2}}\alpha^{-\frac{1}{2}}e^{\pm i\frac{\pi}{4}} \label{eq-2.3.8}$$

Proof. Just using (\ref{eq-2.3.7}) with $\beta = |\alpha|e^{\mp i\frac{\pi}{2}}$.

#### Single stationary point inside

Theorem 4.

1. Let $\phi$ have a single non-degenerate stationary point $c\in (a,b)$ and $f$ be supported in $(a,b)$. As $\pm \phi''(c)>0$ $$I(k) \sim e^{ik\phi (c)}\sum _{n=0}^\infty \kappa_{2n} k^{-\frac{1}{2}-n} \label{eq-2.3.9}$$ in the sense that $$|I(k)- e^{ik\phi (c)}\sum _{n=0}^{N-1} \kappa_{2n}k^{-\frac{1}{2}-n}|\le C_N k^{-N-\frac{1}{2}} . \label{eq-2.3.10}$$ Here the main coefficient is $$\kappa_0=\frac{\sqrt{2\pi}}{\sqrt{|\phi''(c )|}} e^{\pm i\frac{\pi}{4}} f(c ). \label{eq-2.3.11}$$
2. Let $\phi$ have a single non-degenerate stationary point $c\in [a,b]$ and let $c=a$ or $c=b$ be an end-point. Then $$I(k) \sim e^{k\phi (c)}\sum _{n=0}^\infty \kappa_{n} k^{-\frac{1}{2}(n+1)} \label{eq-2.3.12}$$ Here the main coefficient is $$\kappa_0=\frac{1}{2} \frac{\sqrt{2\pi}}{\sqrt{|\phi''(c )|}} e^{\pm i\frac{\pi}{4}} f(c ). \label{eq-2.3.13}$$

Proof. Clearly, without any loss of the generality we can assume that $c=0$ and $\phi(c)=0$ and also $\phi''(x)>0$ (otherwise we can just complex-conjugate $I(k)$). Also without any loss of the generality we can assume that $f(x)$ is supported in $[-\epsilon,\epsilon]$ and $\phi(x)=x^2$. Indeed we can reach it introducing new variable $t=\pm \sqrt{|\phi(x)|}$ instead of $x\gtrless 0$.

Then $f(x)$ will be replaced by $g(t)= f(x)\frac{dx}{dt}$; observe that $\frac{dx}{dt}= 1/\sqrt{|\phi''(c)/2|}$. Then \begin{equation*} I(k)= \int e^{i kt^2}g(t)\,dt. \end{equation*} In such integral we can assume that $g$ and its derivatives have no more than a polynomial growth and take integral over $\bR$ in Statement (a) or over $\bR^\pm$ as Statement (b). Decomposing $g(t)$ into Taylor series we get after change of variables $y=k^{\frac{1}{2}}t$ that \begin{equation*} I(k)\sim\sum_{n=0}^\infty \frac{g^{(n)}(c)}{n!} k^{-\frac{1}{2}(n+1)} \int e^{it^2} t^n \,dt \end{equation*} and we arrive to decomposition (\ref{eq-2.3.49}) with \begin{equation*} \kappa_n= \frac{g^{(n)}(c)}{n!} \int e^{it^2} t^n \,dt. \end{equation*} Observe that if we integrate over $\bR$ then $\kappa_n=0$ for odd $n$. Also observe that $\kappa_0= g(0)\sqrt{\pi}e^{i\frac{\pi}{4}}$ in Statement (a) but only half of it in Statement (b).

Remark 1. We can calculate $\int_0^\infty e^{\pm it^2} t^n \,dt$ by integrations by part. Also to calculate $\int_0^\infty e^{\pm it^2} t^n \,dt$ we can change variables $z=\mp it^2$ and deforming contour as in the proof of Theorem 3 arriving to $$e^{\pm i\frac{\pi}{4}(n+1)}\frac{1}{2}\int_0^\infty e^{-z} z^{(n-1)/2} \,dz= \frac{1}{2}e^{\pm i\frac{\pi}{4}(n+1)}\Gamma ((n+1)/2) \label{eq-2.3.14}$$ where $\Gamma$ is Euler's $\Gamma$–function which we discuss later.

#### Single degenerate stationary point

Assume that there is a single degenerate maximum at $c$: $\phi'(c)=\ldots=\phi^{(m-1)}(c)=0$, $\phi^{(m)}(c)\ne 0$. There are two possibilities: $c$ is either inner or end-point.

Without any loss of the generality one can assume that $\phi(x)=-x^m$. Sign could be any as we always can use complex conjigation. Then we arrive to \begin{equation*} I(k) \sim \sum_{n=0}^\infty \frac{g^{(n)}(0)}{n!} \int e^{-i k x^m} x^n\,dx= \sum_{n=0}^\infty g^{(n)}(0) k^{-(n+1)/m} \int e^{- ix^m} x^n\,dx \end{equation*} and we arrive to

Theorem 5. Let $\phi$ reach its single maximum at $c\in [a,b]$ and $\phi'(c)=\ldots=\phi^{(m-1)}(c)=0$, $\pm \phi^{(m)}(c)> 0$.

1. Then $$I(k) \sim e^{ik\phi (c)}\sum _{n=0}^\infty \kappa_{n} k^{-\frac{1}{m}(n+1)} \label{eq-2.3.15}$$
2. If $c$ is an inner point and $m$ is even then $\kappa_n-0$ for odd $n$.
3. Here the main coefficient is $$\kappa_0=2\Gamma((m+1)/m)\bigl|\phi^{(m)}(c)/m!|^{-\frac{1}{m}} f(c )e^{\pm i\frac{\pi}{2m}}. \label{eq-2.3.16}$$ if $c$ is an inner point but only half of it if $c$ is an end-point.

Proof. Is similar to the proof of Theorem 4 but we need to consider \begin{equation*} \int_0^\infty e^{-\beta t^m}t^n\,dt \end{equation*} with $\beta =-i\alpha$. However we consider as in Theorem 2 with $\beta \in \bC$, $\Re \beta >0$ and deform contour so that we get instead \begin{equation*} \beta^{-(n+1)/m}\int_0^\infty e^{- t^m}t^n\,dt \end{equation*} which after change of variable $z=t^m$ becomes \begin{equation*} \frac{1}{m}\beta^{-(n+1)/m}\int_0^\infty e^{- z}z^{(n+1)/m-1}\,dt= \frac{1}{m}\beta^{-(n+1)/m}\Gamma ((n+1)/m). \end{equation*}

#### Multiple stationary points

Let now $\phi$ has several stationary points on $X$: $a\le x_1<\ldots < x_K\le b$ each of the type considered above; $\phi(x_1)=\ldots =\phi(x_K)$ (because we are looking only for absolute maxima). Then asymptotics of $I(k)$ is given by the sum of the contributions of all these points.