Finding a series expansion for this integral, for large $\lambda > 0$ Given the integral $$Z(\lambda ) = \frac{1}{\sqrt{2 \pi}} \int_{- \infty}^{\infty} dx e^{- \frac{x^2}{2!} - \frac{\lambda}{4!} x^4} $$ in quantum field theory, I need to find a series expansion for $Z(\lambda)$ when $\lambda >> 1$, of the form $$ Z_N ( \lambda) = \sum_{n=0}^N d_n \lambda^{-(2n+1)/4}. $$ 
I was thinking of writing $e^{- \frac{\lambda}{4!} x^4} \approx 1 - \frac{ \lambda}{4!} x^4$, because only the first two terms will dominate when $\lambda$ is large. But this doesn't seem to give me the right expansion when I plug it in the integral and do the Gaussian integration. 
Anyone has any ideas/advice? Thanks in advance. 

Using the substitution $y=\frac{\lambda}{24}x^4$, we find that
\begin{align*}
&\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \exp\left( -\frac{x^2}{2} - \frac{\lambda}{24}x^4 \right) \, dx \\
&\hspace{4em} = \frac{1}{2\sqrt{\pi}} \int_{0}^{\infty} \frac{1}{y^{3/4}} \left(\frac{6}{\lambda}\right)^{1/4} \exp\left( -\sqrt{\frac{6}{\lambda}y} - y \right) \, dx \\
&\hspace{4em} = \frac{1}{2\sqrt{\pi}} \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \left(\frac{6}{\lambda}\right)^{\frac{2n+1}{4}} \int_{0}^{\infty} y^{\frac{n}{2}-\frac{3}{4}} e^{-y} \, dx \\
&\hspace{4em} = \frac{1}{2\sqrt{\pi}} \sum_{n=0}^{\infty} (-1)^n \frac{\Gamma\left(\frac{n}{2}+\frac{1}{4}\right)}{n!} \left(\frac{6}{\lambda}\right)^{\frac{2n+1}{4}}.
\end{align*}
So you may let
$$ d_n = \frac{(-1)^n}{2\sqrt{\pi}} \cdot \frac{\Gamma\left(\frac{n}{2}+\frac{1}{4}\right)}{n!} \cdot 6^{\frac{2n+1}{4}}. $$
