TITLE: Laplace's method with nontrivial parameter dependency
QUESTION [6 upvotes]: I need to approximate the following integral using Laplace's method: 
$$
\int_0^{\infty} \frac{x^{\lambda} \lambda^{-x}}{(1+x^2)^\lambda} dx \\ = 
\int_0^{\infty} \exp\left(\lambda \log(x) - x\log(\lambda)-\lambda \log(1+x^2)\right) dx
$$
as $\lambda \to \infty$. However, I cannot directly apply Laplace's method as there's an untrivial dependency on $\lambda$. So I guess I need to do some change of variables or something like that but I didn't come up with anything useful.

REPLY [4 votes]: The "method" version of the Laplace method can be applied directly with some care.
The basic intuition is that the factor $\lambda^{-x}$ does not vary much in the region dominated by the saddle point coming from the other factor $x^\lambda/(1+x^2)^\lambda$.  Away from the saddle point, $\lambda^{-x}$ is dominated by the the exponential smallness coming from $x^\lambda/(1+x^2)^\lambda$.
We begin by writing
$$
\frac{x^\lambda}{(1+x^2)^{\lambda}} = \exp\!\left[\lambda \log\!\left(\frac{x}{1+x^2}\right)\right] = e^{\lambda \varphi(x)}.
$$
The function $\varphi(x)$ has a maximum at $x = 1$, and near there we have
$$
\varphi(x) = -\log 2 - \frac{1}{2}(x-1)^2 + O((x-1)^3),
$$
so the width of the interval in which this critical point contributes to the integral is on the order of $1/\sqrt{\lambda}$.  To give ourselves some breathing room we'll consider the slightly larger interval $(1-\lambda^{-1/4},1+\lambda^{-1/4})$.  We thus split the integral as
$$
\int_0^{\infty} e^{\lambda \varphi(x)}\lambda^{-x} \,dx = \int_{1-\lambda^{-1/4}}^{1+\lambda^{-1/4}} e^{\lambda \varphi(x)}\lambda^{-x}\,dx + \left( \int_0^{1-\lambda^{-1/4}} + \int_{1+\lambda^{-1/4}}^\infty \right) e^{\lambda \varphi(x)}\lambda^{-x}\,dx.
$$
We'll first bound these last two integrals.  To bound the first we proceed like
$$
\begin{align}
\int_0^{1-\lambda^{-1/4}} e^{\lambda \varphi(x)}\lambda^{-x}\,dx &< \left. \frac{x^{\lambda}}{(1+x^2)^\lambda} \right|_{x = 1-\lambda^{-1/4}} \int_0^\infty \lambda^{-x}\,dx \\
&= \frac{2^{-\lambda}}{\log \lambda} \left(1 - \lambda^{-1/4}\right)^\lambda \left(1 - \lambda^{-1/4} + \frac{1}{2}\lambda^{-1/2}\right)^{-\lambda} \\
&= \frac{2^{-\lambda}}{\log \lambda} \exp\!\left[-\frac{1}{2}\lambda^{1/2} + O(\lambda^{1/4})\right]
\tag{1}
\end{align}
$$
as $\lambda \to \infty$.  For the second we have
$$
\begin{align}
\int_{1+\lambda^{-1/4}}^\infty e^{\lambda \varphi(x)}\lambda^{-x}\,dx &< \left. \frac{x^{\lambda}}{(1+x^2)^\lambda} \right|_{x = 1+\lambda^{-1/4}} \int_0^\infty \lambda^{-x}\,dx \\
&= \frac{2^{-\lambda}}{\log \lambda} \exp\!\left[-\frac{1}{2}\lambda^{1/2} + O(\lambda^{1/4})\right]
\tag{2}
\end{align}
$$
also as $\lambda \to \infty$.
For $x$ in the interval $(1-\lambda^{-1/4},1+\lambda^{1/4})$ we have
$$
\lambda^{-x} = \lambda^{-1} \sum_{k=0}^{\infty} \frac{(\log \lambda)^k}{k!} (1-x)^k
$$
uniformly in $x$.  It follows that
$$
\int_{1-\lambda^{-1/4}}^{1+\lambda^{-1/4}} e^{\lambda \varphi(x)}\lambda^{-x}\,dx = \lambda^{-1} \sum_{k=0}^{\infty} \frac{(\log \lambda)^k}{k!} \int_{1-\lambda^{-1/4}}^{1+\lambda^{-1/4}} e^{\lambda \varphi(x)} (1-x)^k\,dx. \tag{3}
$$
Using the usual techniques associated with the Laplace method we can then calculate asymptotic series for each of these these integrals then recombine them to conclude that we have an asymptotic series of the form
$$
\int_{1-\lambda^{-1/4}}^{1+\lambda^{-1/4}} e^{\lambda \varphi(x)}\lambda^{-x}\,dx \approx \lambda^{-3/2} 2^{-\lambda} \sum_{j=0}^{\infty} \lambda^{-j} \sum_{k=0}^{2j} c_{j,k} (\log \lambda)^k,
$$
valid as $\lambda \to \infty$.  The estimates in $(1)$ and $(2)$ are exponentially small compared to this, so we obtain for the original integral
$$
\int_0^{\infty} e^{\lambda \varphi(x)}\lambda^{-x} \,dx \approx \lambda^{-3/2} 2^{-\lambda} \sum_{j=0}^{\infty} \lambda^{-j} \sum_{k=0}^{2j} c_{j,k} (\log \lambda)^k.
$$

In particular we have the leading order behavior
  $$
\int_0^{\infty} \frac{x^\lambda \lambda^{-x}}{(1+x^2)^\lambda} \,dx \sim \int_{-\infty}^{\infty} e^{-\lambda \log 2 - \lambda/2 (x-1)^2} \lambda^{-1}\,dx = \sqrt{2\pi} \lambda^{-3/2} 2^{-\lambda}
$$
  as $\lambda \to \infty$.

The next few terms of the asymptotic series can be computed directly from $(3)$ - we get
$$
\begin{align}
&\int_0^{\infty} \frac{x^\lambda \lambda^{-x}}{(1+x^2)^\lambda} \\
&\quad = \sqrt{2\pi} \lambda^{-3/2} 2^{-\lambda} \left( 1 + \frac{2(\log\lambda)^2 - 6\log\lambda + 3}{4\lambda} \right. \\
&\qquad\qquad \left. + \frac{4(\log\lambda)^4 - 40(\log\lambda)^3 + 120(\log\lambda)^2 - 120\log\lambda + 25}{32\lambda^2} + \cdots \right)
\end{align}
$$