Steepest descent with slow variation

Question

Suppose that one has an integral of the form $$ \int_{\mathbb{R}} g(x) e^{\lambda f(x;\lambda)} $$ where $f(x;\lambda) = f_0(x) + f_1(x;\log(\lambda))/\lambda $ and $f_1$ depends slowly on $\lambda$ in the sense that it can grow at most polynomially in its argument. Is it true that, as $\lambda\to \infty$, $$ \int_{\mathbb{R}} g(x) e^{\lambda f(x;\lambda)}dx \sim \sqrt{\frac{2\pi}{\lambda |f_0''(x_0)|}}g(x_0)e^{\lambda f_0(x_0)}, $$ where $x_0$ is a unique stationary point of $f_0(x)$, i.e. the asymptotics agree with those replacing $f(x;\lambda)$ with $f_0(x)$?

Answer 1

You may appeal to the following theorem by Olver (Asymptotics and Special Functions, Chapter 9, Section 2, Theorem 2.1). You have to shift the stationary point $x_0$ to the origin, split the integral at $x=0$ into two separate integrals, and finally cut off the integrals at finite points and show (in the usual way) that the contributions from the tails are negligible.

Theorem. Let $k$ and $\Lambda$ be fixed positive numbers, and $$ I(\lambda ) = \int_0^k {g(x,\lambda) e^{\lambda f_0 (x) + f_1 (x,\lambda )} dx} . $$ Assume that

(i) $f'_0 (x)$ is continuous and negative in $(0,k]$, and as $x\to 0+$ $$ f_0 (x) = f_0(0) + F_0 x^\alpha + \mathcal{O}(x^{\alpha _1 } ),\quad f'_0 (x) = \alpha F_0x^{\alpha - 1} + \mathcal{O}(x^{\alpha _1 - 1} ), $$ where $F_0<0$ and $\alpha _1 > \alpha > 0$.

(ii) For each $\lambda \in [\Lambda,\infty)$, the real or complex functions $f_1(x,\lambda)$ and $g(x,\lambda)$ are continuous in $0 < x \leq k$. Moreover $$ \left| {f_1 (x,\lambda )} \right| \le F_1 x^\beta \lambda ^\varepsilon ,\quad \left| {g(x,\lambda ) - G_0 x^{\gamma - 1} } \right| \le G_1 x^{\gamma _1 - 1} \lambda ^\delta , $$ where $F_1,\beta,\varepsilon,G_0,\gamma,G_1,\gamma_1$ and $\delta$ are independent of $x$ and $\lambda$, and $$ \beta \geq 0,\quad \gamma>0, \quad \gamma_1>0,\quad \varepsilon <\min(1,\beta/\alpha),\quad \delta < (\gamma_1-\gamma)/\alpha. $$

Then $$ I(\lambda ) = \frac{{G_0 }}{\alpha }\Gamma\! \left( {\frac{\gamma }{\alpha }} \right)\frac{{e^{\lambda f_0 (0)} }}{{(|F_0| \lambda )^{\gamma /\alpha } }}\left( {1 + \mathcal{O}\!\left( {\frac{1}{{\lambda ^{\varpi /\alpha } }}} \right)} \right) $$ as $\lambda \to +\infty$, where $$ \varpi = \min(\alpha_1-\alpha,\beta-\alpha \varepsilon,\gamma_1-\gamma-\alpha\delta).$$