Laplace's method with nontrivial parameter dependency

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.

I think you can't apply the Laplace Method directly, at least no in the strong form described in the first equation here: Real approximation to the maximum using Laplace's method integral. If the goal is to compute the limit of the integral you can first bound the integral:

$$\int_0^{\infty} \frac{x^{\lambda} \lambda^{-x}}{(1+x^2)^\lambda} dx < \int_0^{\infty} \frac{x^{\lambda}}{(1+x^2)^\lambda} dx$$

Then do the proposed manipulation: $$\int_0^{\infty} \exp\left(\lambda (\log(x) - \log(1+x^2))\right) dx$$

and only then apply the Laplace method.

At this point is looks rather trivial because the argument of the exponential is always negative. That is $M = \max(\log(x) - \log(1+x^2)) < 0$, but let's continue.

Laplace Method tells that the limit of the integral is proportional to $\sqrt{2\pi/\lambda}\exp(\lambda M)$ times a constant that depends on the curvature at the maximum (which is finite). Since $M$ is negative the limit is 0. The integral is positive and bounded by a function that tends to zero. So the answer is zero.

The maximum is $M = -\log(2)$ at $x_0 = 1$ and the second derivative is $-1$. Therefore a naive or weak form of the Laplace Method (but probably correct) points to the following approximation of the original integral:

$$\int_0^{\infty} \frac{x^{\lambda} \lambda^{-x}}{(1+x^2)^\lambda} dx \simeq \sqrt{\frac{2\pi}{\lambda}} \lambda^{-1} 2^{-\lambda}$$

Althougt at first guess, I think that since the kernel function depends on $\lambda$ the order of the approximation is worst than the normal $O(\exp(-\lambda))$ (see below however).

For what it is worth, the original integral can be done exactly.

$$-\frac{\log (\lambda ) \Gamma \left(\frac{\lambda }{2}-1\right) \Gamma \left(\frac{\lambda }{2}+1\right) \, _1F_2\left(\frac{\lambda }{2}+1;\frac{3}{2},2-\frac{\lambda }{2};-\frac{1}{4} \log ^2(\lambda )\right)}{2 \Gamma (\lambda )}+\frac{\sqrt{\pi } 2^{-\lambda } \Gamma \left(\frac{\lambda -1}{2}\right) \, _1F_2\left(\frac{\lambda }{2}+\frac{1}{2};\frac{1}{2},\frac{3}{2}-\frac{\lambda }{2};-\frac{1}{4} \log ^2(\lambda )\right)}{\Gamma \left(\frac{\lambda }{2}\right)}+\frac{\pi \log ^{\lambda -1}(\lambda ) \csc (\pi \lambda ) \, _1F_2\left(\lambda ;\frac{\lambda }{2}+\frac{1}{2},\frac{\lambda }{2};-\frac{1}{4} \log ^2(\lambda )\right)}{\Gamma (\lambda )}$$

(source: symbolic Mathematica) and the limit is zero (source: numeric Mathematica). Additionally, numerically it seems that the order of the approximation analytically obtained above is actually worst than exponential but better than polynomial.

Error estimation: As per my conversation with @AntonioVargas and my numerical experiments, the error is confirmed to be indeed larger than the one expected from the Laplace Method:

$$\int = \sqrt{2\pi}\lambda^{-3/2}2^{-\lambda} + O(\lambda^{-5/2}2^{-\lambda}\log^2\lambda)$$

The other answer shows all the explicit terms up to order $O(\lambda^{-7/4})$.

• Hi alfC, you may be interested in the answer I just posted. – Antonio Vargas Mar 27 '15 at 21:17
• @AntonioVargas, Thanks, I think your answer is more formal. I also get to the same approximation and also find that some weaker variant of the Laplace method needs to be used. What I couldn't find is the "order" of the approximation. Did you? – alfC Mar 27 '15 at 21:38
• Yes, it's possible to deduce from the double series that the relative error is $O(\lambda^{-1} \log \lambda)$. I'd be surprised if it was better, though I could have overlooked something that might improve it. So, the absolute error looks like $$\int = \sqrt{2\pi} \lambda^{-3/2} 2^{-\lambda} + O(2^{-\lambda} \lambda^{-5/2} \log \lambda).$$ – Antonio Vargas Mar 27 '15 at 22:21
• @AntonioVargas, so if it is $O(2^{-\lambda}\dots)$ then it is much better than $O(\lambda^-1 \log\lambda)$, no?. Also, from my numerical experiment I get that $O(2^{-\lambda}\lambda^{-5/2}\log\lambda)$ still overestimates the error. It looks like the error is in practice $O(2^{-\lambda}\lambda^{-2}\log\lambda$. – alfC Mar 27 '15 at 22:56
• It has a relative error of $O(\lambda^{-1} \log \lambda)$, which means that $$\int = \sqrt{2\pi} \lambda^{-3/2} 2^{-\lambda} \left[ 1 + O(\lambda^{-1} \log\lambda) \right].$$ And $2^{-\lambda} \lambda^{-5/2} \log \lambda$ is smaller than $2^{-\lambda} \lambda^{-2} \log \lambda$, so what do you mean when you say it's an overestimate? – Antonio Vargas Mar 28 '15 at 1:24

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}