Asymptotic behaviour/upper bound for $\int_0^{\infty} \exp(-c x^a+K x^b)dx$ for $a>b>0$ as $K\rightarrow \infty$?

Question

What is the asymptotic behaviour or an upper bound for $\int_0^{\infty} \exp(-c x^a+K x^b) \, dx$, for $a>b>0,$ as $K\rightarrow \infty$? Or any good reference for tools to tackle this question?

I think the growth in $K$ should be polynomial because $-c x^a+K x^b=0$ yields $x=(K/C)^\frac{1}{a-b}$ on the range $[0,(K/C)^\frac{1}{a-b}]$ the maximum value of the integrand is again a power of K (take derivative and set 0) the product yields an upper bound on $\int_0^{(K/C)^\frac{1}{a-b}} \exp(-c x^a+K x^b) \, dx$.

On the other hand $\int_{(K/C)^\frac{1}{a-b}}^{\infty} \exp(-c x^a+K x^b) \, dx$ should be decreasing in $K$, for large $K$.

Thank you,

Answer 1

One can get a full asymptotic expansion as an application of Watson's Lemma. One need only observe that the integrand is maximized at $x_0 = \left(\frac{Kb}{ac}\right)^{1/(a-b)}.$

Substituting $x = x_0 u,$ one gets(where $I$ is the original integral) $I = x_0 \int_0^\infty \exp(-c^{b/(a-b)} K^{a/(a-b)}((b/a)^{a/(a-b)} u^a - (b/a)^{b/(a-b)} u^b)) d u.$ Letting $t = c^{b/(a-b)} K^{a/(a-b)},$ the integral breaks up into two Watson Lemma integrals, one from $0$ to $1,$ the second from $1$ to $\infty.$ I leave the final computation of the asymptotics to the interested reader.

EDIT Actually, this is not quite Watson's lemma. You approximate the function $\phi(u) = (b/a)^{a/(a-b)} u^a - (b/a)^{b/(a-b)} u^b)$ by its Taylor series (at the maximum point $1).$ Since it is the maximum, this will look like $\phi(1) - (u-1)^2 \phi^{\prime\prime}(1)/2.$ This means that the integral is asymptotically approximated by a Gaussian integral (I am too lazy to compute $\phi^{\prime\prime}(1)$...$)

Answer 2

This is a simple example for the Laplace Method of asymptotic evaluation of integrals. The essence of the method is that the main contribution to the integral comes from a small neighborhood of the critical point of the function under the exponent. Laplace method is explained in every serious calculus book. For more comprehensive treatment see the books of Fedoryuk, for example, Fedoryuk, M. V. (1987), Asymptotic: Integrals and Series, or here: http://www.encyclopediaofmath.org/index.php?title=Saddle_point_method

Answer 3

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Following $\verb=@Alexandre Eremenko=$ above answer:

$\ds{\int_{0}^{\infty}\exp\pars{-cx^{a} + Kx^{b}}\,\dd x =\int_{-\xi}^{\infty} \exp\pars{-c\bracks{\xi + x}^{a} + K\bracks{\xi + x}^{b}}\,\dd x}$ where $\xi$ satisfies $\ds{-c\xi^{a} + K\xi^{b} = 0}$. The $\large{\rm OP}$ already calculated $\xi = \pars{K \over c}^{1/\pars{a - b}}$. Then, \begin{align} &\int_{0}^{\infty}\exp\pars{-cx^{a} + Kx^{b}}\,\dd x \approx \int_{-\xi}^{\infty}\exp\pars{\bracks{-ca\xi^{a - 1} + Kb\xi^{b - 1}}x}\,\dd x \\[3mm]&= {-\exp\pars{ca\xi^{a} - Kb\xi^{b}} \over -ca\xi^{a - 1} + Kb\xi^{b - 1}} = {\exp\pars{ca\xi^{a} - Kb\xi^{b}} \over ca\xi^{a} - Kb\xi^{b}}\,\xi \end{align}

Also, $\ds{\xi^{a} \sim K^{a/\pars{a - b}}\,,\quad}$ $\ds{\xi^{b} \sim K^{b/\pars{a - b}}\,,\quad}$ $\ds{K\xi^{b} \sim K^{a/\pars{a - b}}}$. Then, \begin{align} &ca\xi^{a} - Kb\xi^{b} = ca\bracks{\pars{K \over c}^{1/\pars{a - b}}}^{a} - Kb\bracks{\pars{K \over c}^{1/\pars{a - b}}}^{b} \\[3mm]&=\bracks{c^{-a/\pars{a - b}}\,a - c^{-b/\pars{a - b}}\,b}K^{a/\pars{a - b}} \end{align}

\begin{align} \int_{0}^{\infty}\exp\pars{-cx^{a} + Kx^{b}}&\approx {\exp\pars{\bracks{c^{-a/\pars{a - b}}\,a - c^{-b/\pars{a - b}}\,b}K^{a/\pars{a - b}}} \over \bracks{c^{-a/\pars{a - b}}\,a - c^{-b/\pars{a - b}}\,b}K^{a/\pars{a - b}}} \,c^{-1/\pars{a - b}}K^{1/\pars{a - b}} \end{align} \begin{align} &\color{#00f}{\large\int_{0}^{\infty}\exp\pars{-cx^{a} + Kx^{b}}}\\[3mm]& \approx \color{#00f}{\large{1 \over c^{\pars{1 - a}/\pars{a - b}}\,a - c^{\pars{1 - b}/\pars{a - b}}\,b}\times} \\[3mm]&\color{#00f}{\large% \exp\pars{\bracks{c^{-a/\pars{a - b}}\,a - c^{-b/\pars{a - b}}\,b}K^{a/\pars{a - b}}} K^{\pars{1 - a}/\pars{a - b}}} \end{align}