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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,

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3 Answers

up vote 1 down vote accepted

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 $\phi^{\prime\prime}(1)$...$)

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Thank you very much. I don't see how turn the integral in watson integrals. In order to obtain the $$int_0^1 exp(-xt)\phi(t)dt$$ on would substitute $x=(\frac{b}{a})^{a/(a-b)}u^a-(\frac{b}{a})^{b/(a-b)}u^b$ which does not work. What is the correct subtitution. Thank you –  warsaga Aug 13 '12 at 10:57
    
See the edit. In my mind this is "generalized Watson", but I guess it is not really... –  Igor Rivin Aug 13 '12 at 17:34
    
Thanks so much! –  warsaga Aug 13 '12 at 18:04
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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

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

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If $ca=b$ then the denominator turns into zero. –  Andrew Jan 19 at 12:07
    
@Andrew I agree. It should be solved in some different way. That's is enough for the OP to try himself. Thanks. –  Felix Marin Jan 19 at 21:24
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