寻求将含双曲余弦对数的高斯积分展开为渐近级数的方法

Alright, let's break down how to derive this integral expansion step by step. The core idea is to use a Taylor series approximation for $\ln(\cosh(z))$ when $z$ is small (since we're focusing on small $b$, $a\sqrt{b}x$ will be small for most $x$ where the Gaussian weight is significant), then plug that into the integral and compute the resulting Gaussian moments.

First, recall the Taylor series for $\cosh(z)$ around $z=0$:
$$\cosh(z) = 1 + \frac{z^2}{2!} + \frac{z^4}{4!} + \frac{z^6}{6!} + \cdots$$
Then use the expansion for $\ln(1+t)$ (valid when $|t|<1$):
$$\ln(1+t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \cdots$$

Let $t = \cosh(z) - 1 = \frac{z^2}{2} + \frac{z^4}{24} + \frac{z^6}{720} + \cdots$. Substitute this into the $\ln$ expansion, and truncate at terms up to $z^4$ (since higher terms will contribute to $O(b^3)$ later):
$$
\begin{align*}
\ln(\cosh(z)) &= \ln(1 + t) \
&= t - \frac{t^2}{2} + O(t^3) \
&= \left(\frac{z^2}{2} + \frac{z^4}{24}\right) - \frac{1}{2}\left(\frac{z^2}{2}\right)2 + O(z^6) \
&= \frac{z^2}{2} + \frac{z^4}{24} - \frac{z^4}{8} + O(z^6) \
&= \frac{z^2}{2} - \frac{z^4}{12} + O(z^6)
\end{align*}
$$

Now substitute $z = a\sqrt{b}x$ into this result:
$$
\ln(\cosh(a\sqrt{b}x)) = \frac{a^2 b x^2}{2} - \frac{a^4 b^2 x^4}{12} + O(a^6 b^3 x^6)
$$

The original integral becomes:
$$
I = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-x2/2} \left( \frac{a^2 b x^2}{2} - \frac{a^4 b^2 x^4}{12} + O(a^6 b^3 x^6) \right) dx
$$

We can split this into three separate integrals (linearity of integration):
$$
I = \frac{a^2 b}{2} \cdot J_2 - \frac{a^4 b^2}{12} \cdot J_4 + O\left(b^3 \cdot J_6\right)
$$
where $J_n = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty x^n e^{-x2/2} dx$ are the moments of the standard normal distribution.

For the standard normal distribution ($\mu=0$, $\sigma=1$), we know:

• Second moment: $J_2 = 1$ (this is the variance)
• Fourth moment: $J_4 = 3$ (a standard result for normal distributions: $E[X^4] = 3\sigma^4$)
• Sixth moment: $J_6 = 15$ (we only need to know this is a finite constant, so the last term scales with $b^3$)

Substitute these moment values back into the expression for $I$:
$$
I = \frac{a^2 b}{2} \cdot 1 - \frac{a^4 b^2}{12} \cdot 3 + O(b^3)
$$
Simplify the second term:
$$
I = \frac{a^2 b}{2} - \frac{a^4 b^2}{4} + O(b^3)
$$

A quick note: You mentioned thinking the $b^2$ coefficient is $-a^2/4$, which seems to be a small typo. The correct coefficient should include $a^4$ since the fourth power of $z = a\sqrt{b}x$ introduces $a^4$. If the original integral had a different form (e.g., $\ln(\cosh(a x))$ multiplied by $b$ terms), the coefficient might change, but based on the integral you wrote, the $a^4$ factor is correct.

The $O(b^3)$ term comes from the truncated $z^6$ term in the $\ln(\cosh(z))$ expansion. Since $\int_{-\infty}^\infty x^6 e^{-x2/2} dx$ is a finite constant (as we noted, it's $15\sqrt{2\pi}$), multiplying by $b^3$ gives a term that scales like $b^3$, which matches the required $O(b^3)$ remainder.

内容的提问来源于stack exchange，提问作者Sarah