Asymptotic expansion (Laplace Method) when Taylor series does not exist

Question

So I've been struggling on calculating the asymptotic behaviour of the integral $$I(\alpha) = \int_0^\infty e^{-n(t+t^\alpha)}dt$$ as $n\rightarrow\infty$ and where $\alpha$ is a real number and $\alpha\geq0$. Following the Laplace method and noting that the minima of $q(t)=t+t^\alpha$ is at $t=0$ over the integration interval, I attempted to Taylor expand only to realize that the series does not exist in general.

As far as I know of, one typically tries to form a power series and use this to evaluate the integral, but how does one do it when the Taylor series does not exist at the minimum? Or are there any another approaches which are reasonable at handling this kind of issues? Any help in either the specified integral or how to tackle these kind of questions is greatly appreciated!

Answer 1

Clearly, $$ I(0) = \int_0^{ + \infty } {{\rm e}^{ - n(t + 1)} {\rm d}t} = {\rm e}^{ - n} \int_0^{ + \infty } {{\rm e}^{ - nt} {\rm d}t} = \frac{{{\rm e}^{ - n} }}{n} $$ and $$ I(1) = \int_0^{ + \infty } {{\rm e}^{ - n(t + t)} {\rm d}t} = \int_0^{ + \infty } {{\rm e}^{ - 2nt} {\rm d}t} = \frac{1}{{2n}}. $$ Thus, I will assume that $\alpha\neq 0,1$. Consider first the case that $\alpha>1$. We make the change of variables $$ t = s^\beta , \quad \beta = \frac{1}{{\alpha - 1}} , $$ and apply Laplace's method to deduce $$ I(\alpha ) = \beta \int_0^{ + \infty } {\exp\left( { - n(s^\beta + s^{1 + \beta } )} \right)s^{\beta - 1} {\rm d}s} \sim \frac{\beta }{n}\sum\limits_{k = 0}^\infty {\Gamma\! \left( {\frac{k}{\beta } + 1} \right)\frac{{b_k }}{{n^{k/\beta } }}} $$ as $n\to +\infty$, with $$ b_k = \frac{{( - 1)^k }}{\beta }\binom{k/\beta + k}{k}. $$ After simplification, we can write this expansion as $$ I(\alpha ) \sim \frac{1}{n}\sum\limits_{k = 0}^\infty {( - 1)^k \frac{{\Gamma (\alpha k + 1)}}{{\Gamma (k + 1)}}\frac{1}{{n^{(\alpha - 1)k} }}} . $$ If $0<\alpha<1$, then we make the change of variables $$ t = s^{\gamma + 1} ,\quad \gamma = \frac{\alpha }{{1 - \alpha }}, $$ and apply Laplace's method to obtain $$ I(\alpha ) = (\gamma + 1)\int_0^{ + \infty } {\exp\left( { - n(s^\gamma + s^{1 + \gamma } )} \right)s^\gamma {\rm d}s} \sim \frac{{\gamma + 1}}{n}\sum\limits_{k = 0}^\infty {\Gamma\! \left( {\frac{{k + 1}}{\gamma } + 1} \right)\frac{{c_k }}{{n^{(k + 1)/\gamma } }}} $$ as $n\to +\infty$, with $$ c_k = \frac{{( - 1)^k }}{\gamma }\binom{(k + 1)/\gamma + k}{k}. $$ After simplification, we can write this expansion as $$ I(\alpha ) \sim \frac{1}{\alpha }\frac{1}{{n^{1/\alpha } }}\sum\limits_{k = 0}^\infty {( - 1)^k \frac{{\Gamma \big( {\frac{{k + 1}}{\alpha }} \big)}}{{\Gamma (k + 1)}}\frac{1}{{n^{(1 - \alpha )k/\alpha } }}} . $$ To leading order, $$ I(\alpha ) \sim \frac{1}{n} $$ for $\alpha>1$ and $$ I(\alpha ) \sim \Gamma\! \left( {\frac{1}{\alpha } + 1} \right)\frac{1}{{n^{1/\alpha } }} $$ for $0<\alpha<1$.

Answer 2

$$\begin{align} I(\alpha)&=\int_0^\infty e^{-n(t+t^\alpha)}dt\\ &=\int_0^\infty t'e^{-n(t+t^\alpha)}dt\\ &=\int_0^\infty t.n.(1+\alpha t^{\alpha-1})e^{-n(t+t^\alpha)}dt\\ \end{align}$$

Make a change of variable $x = n(t+t^\alpha)$.

We denote $t(x,n)$ the solution of the equation $t+t^\alpha = \frac{x}{n}$. Then $$\begin{align} I(\alpha)&=\int_0^\infty t(x,n)e^{-x}dx\\ \end{align}$$

In general, $t(x,n)$ doesn't have analytical solution. So, we estimate an upper bound for $t(x,n)$. As the function $f(t) = t+t^{\alpha}$ is increasing, we have $$t+t^\alpha = \frac{x}{n} \Longrightarrow t+t^\alpha<\frac{x}{n} + \left( \frac{x}{n}\right)^\alpha \Longrightarrow t<\frac{x}{n} \tag{1}$$ and $$t+t^\alpha =\left( \left( \frac{x}{n}\right)^{\frac{1}{\alpha}}\right)^\alpha \Longrightarrow t+t^\alpha<\left( \frac{x}{n}\right)^{\frac{1}{\alpha}} + \left( \left( \frac{x}{n}\right)^{\frac{1}{\alpha}}\right)^\alpha \Longrightarrow t<\left( \frac{x}{n}\right)^{\frac{1}{\alpha}}\tag{2}$$

So,

$$(1) \Longrightarrow I(\alpha)< \frac{1}{n}\int_0^\infty xe^{-x}dx=\frac{1}{n}$$ $$(2) \Longrightarrow I(\alpha)< \frac{1}{n^{1/ \alpha}}\int_0^\infty x^{1/ \alpha}e^{-x}dx=\frac{1}{n^{1/ \alpha}}\Gamma\left(1+ \frac{1}{\alpha}\right)$$

Finally, we can conclude $$I(\alpha) = \mathcal{O}\left( \frac{1}{n^{\max{\left\{1,\frac{1}{\alpha}\right\}}}} \right)$$