I'm struggling with the following integral:
$$\int_{-\infty}^{\infty} \frac{d x}{\sqrt{2 \pi}} \operatorname{sech}^4 (a x+b) \exp \left(-\frac{1}{2} x^2\right)=1-3b^2+4a^6$$
I believe I'm supposed to be solving it by power expanding the hyperbolic function (other methods are appreciated too). So I've first recast this integral as:
$$=\int_{-\infty}^{\infty} \frac{d m}{a \sqrt{2 \pi}} \operatorname{sech}^4 m \exp \left(-\frac{1}{2 a^2}(m-b)^2\right)$$
And then Taylor expanded the hyperbolic up to 2nd order:
$$=\int_{-\infty}^{\infty} \frac{d m}{a \sqrt{2 \pi}}\left(1-2 m^2\right) \exp \left(-\frac{1}{2 a^2}(m-b)^2\right)$$
For simplicity used $k = m-b$ and:
$$=\int_{-\infty}^{\infty} \frac{d k}{a \sqrt{2 \pi}}\left(1-2(k+b)^2\right) \exp \left(-\frac{1}{2 a^2} k^2\right)$$
Evaluating it I got:
$$=1-2 b^2-2 \int_{-\infty}^{\infty} \frac{d k}{a \sqrt{2 \pi}} k^2 \exp \left(-\frac{1}{2 a^2} k^2\right)$$
Which is:
$$=1-2 b^2-2 a^2$$
But I'm supposed to be getting:
$$=1-3b^2+4a^6$$
I'm so close but I just can't see where my mistake is. Any ideas?
After the first substitution, we have (as you wrote it)
$$I=\frac{1}{\sqrt{2 \pi } a}\int_{-\infty}^{+\infty} \text{sech}^4(m) e^{-\frac{(m-b)^2}{2 a^2}}\,dm$$
By Taylor $$ \text{sech}^4(m)=\sum_{n=0}^\infty a_n\,m^{2n}$$ where the first coefficients are
$$\left\{1,-2,\frac{7}{3},-\frac{94}{45},\frac{502}{315},-\frac{221 2}{2025},\frac{324419}{467775},-\frac{17698598}{42567525},\frac{21746206}{91216125}\right\}$$
Just for your curiosity
$$I_n=\int_{-\infty}^{+\infty} m^{2n}\,e^{-\frac{(m-b)^2}{2 a^2}}\,dm$$ $$I_n=2^{n+\frac{1}{2}} a^{2 n+1} \Gamma \left(n+\frac{1}{2}\right) \, _1F_1\left(-n;\frac{1}{2};-\frac{b^2}{2 a^2}\right)$$ where appears Kummer's hypergeometric function.
Anyhow, computing and summing the first terms, we have
$$I=\color{red}{(1-2a^2-2b^2)}+14 a^2 b^2+7a^4+\frac{7}{3} b^4+\cdots$$ which is your result for a first approximation when $(a,b)$ are small.
Using $a=b=\frac 1{10}$ your formula gives for the integral $\frac{24}{25}=0.96$ while numerical integration leads to $0.962186$. Using what I wrote would give $\frac{2887}{3000}=0.962333$ which is a marginal improvement.