Question about steps/derivation regarding Laplace method.

Question

I am reading something on the Laplace method of integrals and I don't understand part of it's argument. It gives the integral $$\int_{-3}^4 e^{-\lambda x^2}\log(1+x^2)dx$$ and finding the leading term of asymptotics for $\lambda\to\infty$. It first argues

"Since only small $|x|$, such that $|x| \sim \frac{1}{\sqrt{\lambda}} \ll 1$ are important,

$$\log(1+x^2)\sim x^2$$

$$I(\lambda)\sim \int_{-3}^4 e^{-\lambda x^2}x^2 \, dx \sim \int_{-\infty}^{\infty} e^{-\lambda x^2}x^2 \, dx \sim 2\int_{0}^{\infty} e^{-\lambda x^2}x^2 \, dx." $$

I'm really not following that argument at all. Any help would be appreciated.

Answer 1

In your case you have something that looks like $$\int_a^b e^{\lambda f(x)} g(x)\,dx,$$ not just $$\int_a^b e^{\lambda f(x)}\,dx.$$ In this situation you need to replace both $f(x)$ and $g(x)$ with their approximations $f_0(x)$ and $g_0(x)$ near $x_0$.

In your example $f(x) = -x^2$ doesn't have a simpler approximation near $x_0 = 0$, so we can just take $$f_0(x) \stackrel{\text{def}}{=} f(x) = -x^2.$$ Now $g(x) = \log(1+x^2)$ and $g(x) \sim x^2 \stackrel{\text{def}}{=} g_0(x)$ near $x = x_0 = 0$, and the Laplace method says that $$\int_a^b e^{\lambda f(x)} g(x)\,dx \sim \int_{-\infty}^{\infty} e^{\lambda f_0(x)} g_0(x)\,dx.$$

Answer 2

Here, we give an approach that provides the full asymptotic series. We begin by writing

$$\int_{-3}^4 e^{-\lambda x^2}\log(1+x^2)dx=\frac{1}{\lambda^{1/2}}\int_{-3\sqrt{\lambda}}^{4\sqrt{\lambda}} e^{-x^2}\log(1+(x/\sqrt{\lambda})^2)dx$$

For a fixed $1>\delta >0$, one can split the integral as

$$\int_{-3}^4 e^{-\lambda x^2}\log(1+x^2)dx= \int_{-3}^{-1+\delta}e^{-\lambda x^2}\log(1+x^2)dx+\int_{-1+\delta}^{1-\delta}e^{-\lambda x^2}\log(1+x^2)dx+\int_{1-\delta}^4e^{-\lambda x^2}\log(1+x^2)dx$$

For the outer integrals, it is easy to see that they are less than $Ce^{-\lambda(1-\delta)^2}$ for some constant $C$. So, these terms behave like $O(e^{-\lambda(1-\delta)^2})$ as $\lambda \to \infty$

For the middle integral, we make use of the series expansion for $\log (1+x)$ to write

$$\log(1+(x/\sqrt{\lambda})^2)=\sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^{2n}}{n\,\lambda^n}$$

Thus, we have

$$\int_{-(1-\delta)}^{1-\delta} e^{-\lambda x^2}\log(1+x^2)dx=\frac{1}{\lambda^{1/2}} \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n\,\lambda^n} \int_{-(1-\delta)\sqrt{\lambda}}^{(1-\delta)\sqrt{\lambda}} x^{2n}\,e^{-x^2}dx$$

Now, it is important to understand that asymptotic series often are nowhere convergent! However, we don't despair inasmuch as truncated series can provide good approximations when taking the parameter (such as $\lambda$ here) large enough. To that end we note that if we take $\lambda$ large, the integral inside the series can be made arbitrarily close to

$$\int_{-\infty}^{\infty} x^{2n}\,e^{-x^2}dx=\sqrt{\pi} \frac{(2n)!}{4^n\,n!}$$

and thus we have the full asymptotic series

$$\int_{-3}^4 e^{-\lambda x^2}\log(1+x^2)dx \sim \frac{\sqrt{\pi}}{\lambda^{1/2}} \sum_{n=1}^{\infty} (-1)^{n+1} \left( \frac{(2n)!}{4^n\,n\,n!}\right) \frac{1}{\lambda^n}$$

which diverges everywhere. However, the leading term is

$$\int_{-3}^4 e^{-\lambda x^2}\log(1+x^2)dx \sim \frac{\sqrt{\pi}}{2\lambda^{3/2}}$$

which is a useful approximation for sufficiently large $\lambda$!