Expansion of Integrals. 1


### Laplace integrals

We consider $$I(k)= \int_X e^{k \phi(x)}f(x)\,dx \label{eq-2.1.1}$$ 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 $k\to +\infty$. Naturally we expect that the main contribution to $I(k)$ is delivered by the vicinities of the points $x\in X$ in which $\phi(x)$ reaches its maximum. Among these points could be boundary points $a$ and $b$ and inner points $x_1,\ldots,x_m$ with $a<x_1<\ldots < x_m<b$.

#### Single maximum at the end-point

First we consider the simples case when $\phi$ reaches its maximum at $b$ and $\phi'(b)>0$ (obviously $\phi'(b)\ge 0$ so we assume only that $\phi'(b)\ne 0$).

Example 1. Let $\phi(x)=\alpha x$, $f(x)=\const$. Then \begin{equation*} I(k)=\int_a^b e^{\alpha k x}f\, dx= \underbracket{\alpha^{-1}k^{-1} f e^{\alpha bk}}- \alpha^{-1}k^{-1} f e^{\alpha ak} \end{equation*} with main term $\alpha^{-1}k^{-1} f e^{\alpha bk}$ which equals $$\frac{1}{\phi'(b)} e^{k\phi (b)}f(b)k^{-1} \label{eq-2.1.2}$$ and we expect that it will be the main term of the asymptotics in the general case.

Now consider the general case. Without any loss of the generality we can assume that $f=0$ as $x < b-\epsilon$ and $\phi'(x)>0$ on $[a-\epsilon, b]$. Indeed, contribution of $[a,b-\epsilon]$ is exponentially small in comparison with $e{k\phi(b)}$. Then the left end of the interval is of no importance and integrating by parts we get \begin{multline} I(k)= \int^b e^{k \phi(x)}f(x)\,dx= k^{-1}\int^b (\phi'(x))^{-1}\bigl(e^{k \phi(x)}\bigr)'f(x)\,dx=\\ k^{-1}(\phi'(b))^{-1} e^{k \phi(b)}f(b)- k^{-1}\int^b e^{k \phi(x)}g(x)\,dx \label{eq-2.1.3} \end{multline} with \begin{equation*} g(x)=\bigl((\phi'(x))^{-1} f(x)\bigr)'. \end{equation*} The first term in the right-hand expression is the guessed main term 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 Statement (a) of the following Theorem; Statement (b) is proven in the same way:

Theorem 1.

1. Let $\phi$ reach its single maximum at the right end $b$ and $\phi'(b)>0$. Then $$I(k) \sim e^{k\phi (b)}\sum _{n=0}^\infty \kappa_n k^{-1-n} \label{eq-2.1.4}$$ in the sense that $$|I(k)- e^{k\phi (b)}\sum _{n=0}^{N-1} \kappa_n k^{-1-n}|\le C_N k^{-N-1} e^{k\phi(b)}. \label{eq-2.1.5}$$ Here the main coefficient is $$\kappa_0=\frac{1}{\phi'(b)} f(b). \label{eq-2.1.6}$$
2. Let $\phi$ reach its single maximum at the left end $a$ and $\phi'(a)<0$. Then $$I(k) \sim e^{k\phi (a)}\sum _{n=0}^\infty \kappa_n k^{-1-n} \label{eq-2.1.7}$$ with the main coefficient $$\kappa_0=-\frac{1}{\phi'(a)} f(a). \label{eq-2.1.8}$$

#### Single maximum inside

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

Example 2. Let $\phi (x)=-\alpha x^2$ with $\alpha>0$ (then $c=0$) and $f=\const$. Let $X=\bR$ (otherwise the error will be exponentially small in comparison with the main term). Then \begin{multline} I(k) =\int_{-\infty}^\infty e^{-\alpha k x^2}f\,dx = \alpha^{-1/2}k^{-1/2}\int _{-\infty}^\infty e^{- t^2}f\,dt=\pi^{1/2}\alpha^{-1/2}k^{-1/2} f=\\ \frac{\sqrt{2\pi}}{\sqrt{-\phi''(c )}} f(c )e^{k\phi ()c )} k^{-\frac{1}{2}}. \label{eq-2.1.9} \end{multline}

Again we expect that this will be the main term of asymptotics in the general case.

Theorem 2. Let $\phi$ reach its single maximum at $c\in [a,b]$ and $\phi'(c)=0$, $\phi''(c)<0$.

1. Let $c\in (a,b)$ be an inner point. Then $$I(k) \sim e^{k\phi (c)}\sum _{n=0}^\infty \kappa_{2n} k^{-\frac{1}{2}-n} \label{eq-2.1.10}$$ in the sense that $$|I(k)- e^{k\phi (c)}\sum _{n=0}^{N-1} \kappa_{2n}k^{-\frac{1}{2}-n}|\le C_N k^{-N-\frac{1}{2}} e^{k\phi(c)}. \label{eq-2.1.11}$$ Here the main coefficient is $$\kappa_0=\frac{\sqrt{2\pi}}{\sqrt{-\phi''(c )}} f(c ). \label{eq-2.1.12}$$
2. Let $c=a$ or $c=b$ be an end-point. Then $$I(k) \sim e^{k\phi (c)}\sum _{n=0}^\infty \kappa_{n} k^{-\frac{1}{2}(n+1)} \label{eq-2.1.13}$$ Here the main coefficient is $$\kappa_0=\frac{\sqrt{2\pi}}{2\sqrt{-\phi''(c )}} f(c ). \label{eq-2.1.14}$$

Proof. Clearly, without any loss of the generality we can assume that $c=0$ and $\phi(c)=0$. 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^{-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^{-t^2} t^n \,dt \end{equation*} and we arrive to decomposition (\ref{eq-2.1.4}) with \begin{equation*} \kappa_n= \frac{g^{(n)}(c)}{n!} \int e^{-t^2} t^n \,dt. \end{equation*} Observe that if we integrate over $\bR$ then $\kappa_n=0$ for odd $n$. Also observe that $\kappa_0= g(0)\sqrt{\pi}$ in Statement (a) but only half of it in Statement (b).

Remark 1. We can calculate $\int_0^\infty e^{-t^2} t^n \,dt$ by integrations by part. Also to calculate $\int_0^\infty e^{-t^2} t^n \,dt$ we can change variables $z=t^2$ arriving to $$\frac{1}{2}\int_0^\infty e^{-z} z^{(n-1)/2} \,dz= \frac{1}{2}\Gamma ((n+1)/2) \label{eq-2.1.15}$$ where $\Gamma$ is Euler's $\Gamma$–function which we discuss later.

#### Single degenerate maximum

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: $m$ is even then $\phi^{(m)}(c)<0$ and maximum could be either at an end-point or in an inner point. Without any loss of the generality one can assume that $\phi(x)=-x^m$. Then we arrive to \begin{equation*} I(k) \sim \sum_{n=0}^\infty \frac{g^{(n)}(0)}{n!} \int e^{-k x^m} x^n\,dx= \sum_{n=0}^\infty g^{(n)}(0) k^{-(n+1)/m} \int e^{- x^m} x^n\,dx \end{equation*} and we arrive to

Theorem 3. Let $\phi$ reach its single maximum at $c\in [a,b]$ and $\phi'(c)=\ldots=\phi^{(m-1)}(c)=0$, $\phi^{(m)}(c)< 0$ with even $m$.

1. Let $c\in (a,b)$ be an inner point. Then $$I(k) \sim e^{k\phi (c)}\sum _{n=0}^\infty \kappa_{2n} k^{-\frac{1}{m}-2n/m} \label{eq-2.1.16}$$ Here the main coefficient is $$\kappa_0=2\Gamma((m+1)/m)\bigl|\phi^{(m)}(c)/m!|^{-\frac{1}{m}} f(c ). \label{eq-2.1.17}$$
2. Let $c=a$ or $c=b$ be an end-point. Then $$I(k) \sim e^{k\phi (b)}\sum _{n=0}^\infty \kappa_{n} k^{-\frac{1}{m}(n+1)} \label{eq-2.1.18}$$ Here the main coefficient is $$\kappa_0=\Gamma((m+1)/m)\bigl|\phi^{(m)}(c)/m!\bigr|^{-\frac{1}{m}} f(c ). \label{eq-2.1.19}$$

Alternatively, $m$ is odd. Then either $c=a$, $\phi^{(m)}(a)<0$ or $c=b$, $\phi^{(m)}(b)>0$.

Theorem 4. Let $\phi$ reach its single maximum at $c\in [a,b]$ and $\phi'(c)=\ldots=\phi^{(m-1)}(c)=0$, $\phi^{(m)}(c)\ne 0$ with odd $m$. Then (\ref{eq-2.1.18}) holds and $$\kappa_0=\Gamma((m+1)/m)\bigl(\phi^{(m)}(c)/m!\bigr)^{-\frac{1}{m}} f(c ). \label{eq-2.1.20}$$

#### Multiple maxima

Let now $\phi$ has several maxima 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.